TANGENT (CALCULUS DERIVATIVE)
Slope touching 1 point = instantaneous velocity derivative.
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 4
Linear Functions
SLIDE 06
If position is modeled as a linear function:
\(x(t) = vt + x_0\)
The calculus derivative of a linear function with respect to \(t\) is simply its slope coefficient \(v\), which is completely constant over time: \(\frac{dx}{dt} = v \quad (\text{constant})\)
Case 1 vs Case 2 Derivatives
CASE 1: RISE \(\frac{dx}{dt} = +v\)
Steady Positive Velocity
CASE 2: FALL \(\frac{dx}{dt} = -v\)
Steady Negative Velocity
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 5
Concept Buster 2
SLIDE 07
Curved Road Illusion Packet Page 6
Does a curved line on a position-time graph mean the vehicle is physically driving along a curved, winding road in space?
No! The student is on a completely straight sidewalk. The curvature represents a variable first derivative \(\frac{dx}{dt}\), which implies a non-zero second derivative, acceleration: \(a = \frac{d^2x}{dt^2}\).
Understanding Concavity
CONCAVE UP (🌳) \(\frac{d^2x}{dt^2} > 0\)
Positive Acceleration
CONCAVE DOWN (🍎) \(\frac{d^2x}{dt^2} < 0\)
Negative Acceleration
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 6
Speeding Up
SLIDE 08
In calculus physics, an object is speeding up if and only if its velocity derivative \(v = \frac{dx}{dt}\) and acceleration derivative \(a = \frac{dv}{dt}\) share the same sign.
Case 3: Speeding Up Positive
\(v = \frac{dx}{dt} > 0\) and \(a = \frac{d^2x}{dt^2} > 0 \implies \text{Speeding Up}\)
Case 5: Speeding Up Negative
\(v = \frac{dx}{dt} < 0\) and \(a = \frac{d^2x}{dt^2} < 0 \implies \text{Speeding Up}\)
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 7 & 8
Slowing Down
SLIDE 09
In calculus physics, an object is slowing down if and only if its velocity derivative \(v = \frac{dx}{dt}\) and acceleration derivative \(a = \frac{dv}{dt}\) have opposite signs.
Case 4: Slowing Down Positive
\(v = \frac{dx}{dt} > 0\) and \(a = \frac{d^2x}{dt^2} < 0 \implies \text{Slowing Down}\)
Case 6: Slowing Down Negative
\(v = \frac{dx}{dt} < 0\) and \(a = \frac{d^2x}{dt^2} > 0 \implies \text{Slowing Down}\)
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 7 & 8
Concept Buster 3
SLIDE 10
Direction vs. Speed Trap Packet Page 9
If a ball rolls backward, accelerating from \(v = -1\text{ m/s}\) to \(v = -8\text{ m/s}\), is it slowing down because the numerical value is decreasing?
No! The negative sign purely represents direction. Speed is the absolute magnitude of the position derivative, \(|v(t)| = |\frac{dx}{dt}|\), which increased from \(1\text{ m/s}\) to \(8\text{ m/s}\). It is gaining speed backward!
Absolute Value Bridge
VELOCITY VALUE \(v = -8\text{ m/s}\)
PHYSICAL SPEED \(|v| = 8\text{ m/s}\)
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 9
Synthesis Reference
SLIDE 11
1. POS CONSTANT \(v = \frac{dx}{dt} > 0\) (const), \(a = 0\)
3. POS SPEEDING UP \(v = \frac{dx}{dt} > 0, \ a > 0\)
4. POS SLOWING DOWN \(v = \frac{dx}{dt} > 0, \ a < 0\)
2. NEG CONSTANT \(v = \frac{dx}{dt} < 0\) (const), \(a = 0\)
5. NEG SPEEDING UP \(v = \frac{dx}{dt} < 0, \ a < 0\)
6. NEG SLOWING DOWN \(v = \frac{dx}{dt} < 0, \ a > 0\)
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 10
Concept Buster 4
SLIDE 12
The Passing Paradox Packet Page 11
If Drone A and Drone B's position graphs cross perfectly at \(t = 4\text{ s}\), does that imply they are moving at the exact same speed?
No! The intersection means they share the same location: \(x_A(4) = x_B(4)\). Their instantaneous position derivatives \(\frac{dx_A}{dt} \neq \frac{dx_B}{dt}\) (slopes) remain entirely different. They are just passing each other!
Crossing Slopes t x t = 4s Drone A Drone B
Equal position, entirely different velocity slopes.
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 11
Synthesis Practice
SLIDE 13
Stage A Stage B Stage C Stage D Time (t) Pos (x)
Stage C: \(v = \frac{dx}{dt} < 0\) and \(a = \frac{d^2x}{dt^2} < 0\) (Speeding Up backward)
Stage D: \(v = \frac{dx}{dt} > 0\) and \(a = \frac{d^2x}{dt^2} > 0\) (Speeding Up forward)
UNIT 1 // 1D KINEMATICS BRIDGE USE WITH PACKET PAGE 12
Think: My Defense & Explanation
Share: Partner Consensus Explanation
Golden Rule of Position Graphs
In physics, coordinate direction (\(+\) or \(-\)) and moving rate (speeding up vs. slowing down) are completely distinct mechanics. On a position-time graph, the **direction** is determined by whether the line slopes upward or downward, while **speeding up** is represented by the slope getting **steeper** (more vertical), regardless of whether it trends up or down.
MOTION MECHANICS KINEMATICS COOPERATIVE WORKBOOK PAGE 2 OF 2
Concept Check: Explain why you can't simply find the instantaneous velocity at exactly \(t = 2\text{ s}\) by plugging \(\Delta t = 0\) directly into the formula \(v = \frac{\Delta x}{\Delta t}\). (What happens to the math?)
If we plug in \(\Delta t = 0\), the equation becomes...
AVERAGE VS. INSTANTANEOUS PAGE 2 OF 8
Kinematics Transition // Algebra to Calculus
PAGE 3 OF 8
Before we ever touch calculus symbols, we can find the exact instantaneous velocity of a curving object geometrically by drawing a tangent line at a single point and finding its linear slope (\(\frac{\text{rise}}{\text{run}}\)).
Graphical Lab: The Roller Coaster Climb
A B Time (t) Pos (x)
Look at the graph representing an object speeding up. We have drawn tangent lines at two separate points, A and B.
Point A: The curve is still quite flat. The slope of the blue tangent line is low.
Point B: The curve is much steeper. The slope of the red tangent line is high.
POINT A ESTIMATE
Estimate the slope at A. If the run is \(5\text{ s}\) and the rise is \(2\text{ m}\), calculate the velocity:
\(v_A = \frac{\text{rise}}{\text{run}} = \)
POINT B ESTIMATE
Estimate the slope at B. If the run is \(5\text{ s}\) and the rise is \(8\text{ m}\), calculate the velocity:
\(v_B = \frac{\text{rise}}{\text{run}} = \)
Geometric Rule of Curves: Summarize how the visual steepness of tangent lines tells you whether a curving object is speeding up or slowing down.
READING TANGENTS GEOMETRICALLY PAGE 3 OF 8
Kinematics Transition // Active learning
PAGE 4 OF 8
The Turning Point: When Slope Flattens to Zero
Think-Pair-Share
The Situation: Throwing a Ball Upwards
A student throws a tennis ball vertically straight up. It rises to its peak height, stops for an instantaneous moment, and falls back down. Its position-time graph is a smooth curve that arches like a dome. Student A says: "At the absolute peak of its curve, the ball's position is high, so its speed must also be at its maximum value."
Student B says: "No, at the absolute peak, the tangent line to the curve is completely horizontal, which means its speed is exactly zero."
Think: My Defense & Explanation
Identify which student is correct. Defend your choice by drawing what a horizontal tangent line implies about rate of change:
Share: Partner Consensus Discussion
Compare arguments. Draft a shared physical description of how an object turns around in one dimension:
💡 Key Kinematic Connection: A peak or valley on a position graph is a turning point! A flat tangent line always denotes an instantaneous velocity of zero (\(v = 0\)).
CONCEPT BUSTER: TURNING POINTS PAGE 4 OF 8
Kinematics Transition // Algebra to Calculus
PAGE 5 OF 8
Drawing manual tangent lines and counting grid squares is slow and imprecise. What if we had a magical "slope-finder" equation that could instantly calculate the exact slope of a curve at any second?
In mathematics, that slope-finder function is called the derivative. The symbol for the derivative of position with respect to time is written as: \(\frac{dx}{dt}\) or simply \(v(t)\).
The Input-Output Bridge
POSITION FUNCTION \(x(t)\) Input: Time \(t\) \(\implies\) Output: Location
Tells you exactly where the object is on the line.
VELOCITY DERIVATIVE \(v(t) = \frac{dx}{dt}\) Input: Time \(t\) \(\implies\) Output: Slope (Speed)
Tells you how fast and in what direction it is moving.
The most famous shortcut for finding a derivative of any simple polynomial term is the Power Rule:
The Simple Power Rule
\(\frac{d}{dt}\left( c t^n \right) = n \cdot c t^{n-1}\)
Multiply the constant coefficient (\(c\)) by the current exponent (\(n\)), then subtract \(1\) from the power of \(t\).
Guided Demonstration
If an accelerating vehicle has a position formula of: \(x(t) = 5 t^2\). Let's use the Power Rule:
1. Bring down the power of 2: multiply it by 5: \((2 \times 5) = 10\)
2. Subtract 1 from the exponent: \(2 - 1 = 1\) (which is just \(t^1\) or \(t\))
3. This gives the velocity derivative: \(\mathbf{v(t) = 10t}\).
Now try it: If a projectile has a position formula of \(x(t) = 4 t^3\), write down its velocity equation \(v(t)\):
\(v(t) = \frac{dx}{dt} = \)
THE INTUITIVE DERIVATIVE PAGE 5 OF 8
Kinematics Transition // Algebra to Calculus
PAGE 6 OF 8
Let's practice our transition skills. Note that the derivative of any constant (like \(+5\)) is always **zero**, because a constant location has zero slope (no motion!).
PROBLEM 1: The Rolling Lab Cart
A small laboratory cart roll along a level metal track. Its curving position is described by the equation: \[x(t) = 3t^2 + 2t + 4\]
A. Derive the velocity equation \(v(t)\):
B. Calculate the velocity at exactly \(t = 3\text{ s}\):
PROBLEM 2: The Braking Vehicle
A vehicle traveling down a road slams on its brakes. Its position-time equation is described by: \[x(t) = 24t - 3t^2\]
A. Derive the velocity equation \(v(t)\):
B. Find the time \(t\) when the vehicle comes to a complete rest (\(v = 0\)):
POWER RULE WORKBOOK LAB PAGE 6 OF 8
Kinematics Transition // Algebra to Calculus
PAGE 7 OF 8
We have seen how taking the slope (derivative) of a position graph gives us velocity. But what if we start with a velocity graph and want to go backward to find out how far we traveled?
In physics, the area under a velocity-time graph equals the net displacement (\(\Delta x\)). Geometrically reversing the slope process is called integration.
Constant Velocity Area
When velocity is constant, the area is a simple rectangle: \(\text{Area} = \text{base} \times \text{height} = t \times v\).
v × t
Variable Velocity Area
When velocity increases steadily, the area under the curve is a triangle: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\).
1/2 v × t
Concept Check: Area Accumulation
An athlete starts running from the starting line. Their velocity increases steadily over time, modeled by \(v(t) = 3t\). At \(t = 4\text{ seconds}\), their velocity is \(12\text{ m/s}\).
Calculate their displacement during these first \(4\text{ seconds}\) using the triangular area formula:
\(\Delta x = \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \)
ACCUMULATING MOTION AREAS PAGE 7 OF 8
Kinematics Transition // Algebra to Calculus
PAGE 8 OF 8
Just like the derivative has the Power Rule shortcut for slopes, the integral has an antiderivative shortcut for calculating exact curving areas.
The Simple Integral Power Rule (Anti-Derivative):
\(\int c t^n dt = \frac{c t^{n+1}}{n+1}\)
How to use it: Add exactly \(1\) to the exponent, then divide the term by this new exponent value. This rule mathematically calculates the area accumulation under your curve!
Simple Derivation: Constant Acceleration Formula
An object moves with a velocity of \(v(t) = a \cdot t\) (starts from rest, accelerates steadily). Let's find its position equation \(x(t)\) by integrating velocity:
\[x(t) = \int a t^1 dt = \frac{a t^{1+1}}{1+1} = \mathbf{\frac{1}{2} a t^2}\] This is exactly where the famous kinematic formula comes from! Calculus derived it from simple area accumulation!
Final Bridge Challenge: Reverse the Rule
A vehicle starts from rest at the origin. Its velocity equation is \(v(t) = 6t^2\). Use the integral power rule to find its position equation \(x(t)\):
1. Add 1 to the exponent of \(t^2 \implies t^3\)
2. Divide the term by this new exponent of 3:
\(x(t) = \int 6 t^2 dt = \)
INTRODUCING THE INTEGRAL PAGE 8 OF 8
Page 2: Average vs. Instantaneous
Concept Check Solution:
If we plug in a time interval of \(\Delta t = 0\), the displacement \(\Delta x\) is also \(0\). The formula becomes \(v = \frac{0}{0}\), which is mathematically undefined. We cannot divide by zero, meaning algebra alone cannot calculate an exact split-second rate of change.
Page 3: Reading Tangents Geometrically
POINT A ESTIMATE: \(v_A = \frac{\text{rise}}{\text{run}} = \frac{2\text{ m}}{5\text{ s}} = \mathbf{0.4\text{ m/s}}\)
POINT B ESTIMATE: \(v_B = \frac{\text{rise}}{\text{run}} = \frac{8\text{ m}}{5\text{ s}} = \mathbf{1.6\text{ m/s}}\)
Stop & Jot Summary: As the tangent lines tilt more vertically (steeper slope), speed is increasing. As they tilt more horizontally (flatter slope), speed is decreasing.
Page 4: Concept Buster (Turning Point) Key
Correct Student: Student B is correct.
Explanation: Speed is the slope of the position graph. At the absolute peak of the trajectory, the ball's position graph changes from rising (positive slope) to falling (negative slope). To transition from positive to negative, the slope must pass through exactly zero. Thus, the tangent line is flat, meaning instantaneous speed is exactly zero (\(v = 0\)) at that split second.
KINEMATICS TEACHER RESOURCES PAGE 2 OF 5
Teacher Resource // Kinematics Answer Key
PAGE 3 OF 5
Page 5: Introducing the Derivative
Try It Practice Answer:
\(v(t) = \frac{dx}{dt} = \frac{d}{dt}\left( 4 t^3 \right) = 3 \times 4 t^{3-1} = \mathbf{12t^2}\)
Page 6: Power Rule Workbook Lab
PROBLEM 1 (Lab Cart)
A. \(v(t) = \frac{d}{dt}(3t^2 + 2t + 4) = \mathbf{6t + 2}\)
B. At \(t=3\text{ s}\): \(v(3) = 6(3) + 2 = \mathbf{20\text{ m/s}}\)
PROBLEM 2 (Braking Vehicle)
A. \(v(t) = \frac{d}{dt}(24t - 3t^2) = \mathbf{24 - 6t}\)
B. Set \(v(t) = 0 \implies 24 - 6t = 0 \implies \mathbf{t = 4\text{ seconds}}\)
Page 7: Reversing the Process (Areas)
Area Accumulation Practice Answer:
\(\Delta x = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\text{ s} \times 12\text{ m/s} = \mathbf{24\text{ meters}}\)
Page 8: Introducing the Integral
Final Bridge Challenge Answer:
\(x(t) = \int 6 t^2 dt = \frac{6 t^{2+1}}{2+1} = \frac{6 t^3}{3} = \mathbf{2t^3}\) Note: Since the cart starts from rest at the origin, there is no constant of integration offset (initial position is zero).
KINEMATICS TEACHER RESOURCES PAGE 3 OF 5
Teacher Resource // Kinematics Answer Key
PAGE 4 OF 5
Concept Buster 1 (Reading Packet Page 3): The Turnaround Trick
Student B is correct. Net displacement is \(\Delta x = x_f - x_i = +4\text{ m} - 0\text{ m} = +4\text{ m}\). Total distance is \(10\text{ m} + 6\text{ m} = 16\text{ m}\). At turnaround peak (\(10\text{ m}\)), the continuous position derivative \(\frac{dx}{dt} = 0\) as velocity changes sign.
Concept Buster 2 (Reading Packet Page 6): Curved Road Illusion
Student B is correct. A 1D position graph curve indicates a changing derivative \(\frac{dx}{dt} \neq \text{const}\), representing accelerating motion along a completely straight sidewalk, not a 2D spatial winding road.
Concept Buster 3 (Reading Packet Page 9): Direction vs. Speed
Student B is correct. Negative sign denotes backward direction. Speed is the absolute derivative magnitude \(|v(t)|\), which increases from \(1\text{ m/s}\) to \(8\text{ m/s}\). This indicates speeding up backward.
Concept Buster 4 (Reading Packet Page 11): The Passing Paradox
Student B is correct. An intersection of \(x(t)\) represents identical coordinates (being at the same spot). However, speed is represented by the local derivatives \(\frac{dx_A}{dt} \neq \frac{dx_B}{dt}\) (different tangent slopes), so they pass each other with different velocities.
KINEMATICS TEACHER RESOURCES PAGE 4 OF 5
Teacher Resource // Kinematics Answer Key
PAGE 5 OF 5
Stage C Derivative Analysis
Answers: Velocity derivative \(v = \frac{dx}{dt}\) is negative (\(-\)), and acceleration \(a = \frac{d^2x}{dt^2}\) is negative (\(-\)).
Explanation: The graph slopes downward, so \(v < 0\). The curve is concave down (bending steeper downward), which means acceleration is negative (\(a < 0\)). Since both signs are negative, the drone is speeding up moving backward.
Stage D Derivative Analysis
Answers: Velocity derivative \(v = \frac{dx}{dt}\) is positive (\(+\)), and acceleration \(a = \frac{d^2x}{dt^2}\) is positive (\(+\)).
Explanation: The graph slopes upward, so \(v > 0\). The curve is concave up (bending steeper upward), which means acceleration is positive (\(a > 0\)). The drone is speeding up moving forward.
The Ultimate Slope Comparison Answer
An algebra student calculating average velocity (\(\frac{\Delta x}{\Delta t}\)) only uses the endpoints (\(x_f\) and \(x_i\)). Because the drone ended at a higher position than it started, the overall displacement is positive, resulting in a positive average velocity. This algebraic calculation completely fails to capture the fact that during Stage C, the position was decreasing and the drone was flying backward because algebra averages smooth out the entire path, hiding any intermediate negative derivative intervals.
KINEMATICS TEACHER RESOURCES PAGE 5 OF 5
If we find the average speed between a wide interval (like \(t = 1\text{ s}\) to \(t = 5\text{ s}\)), we get a crude estimate. But if we shrink that window to a fraction of a second (like \(t = 1\text{ s}\) to \(t = 1.01\text{ s}\)), the secant line is practically indistinguishable from the exact tangent line at that single moment.
Concept Check Solution (In Red):
Explain why you can't simply find the instantaneous velocity at exactly \(t = 2\text{ s}\) by plugging \(\Delta t = 0\) directly into the formula \(v = \frac{\Delta x}{\Delta t}\). (What happens to the math?)
If we plug in \(\Delta t = 0\), the displacement \(\Delta x\) is also \(0\). The formula becomes \(v = \frac{0}{0}\), which is mathematically undefined. We cannot divide by zero, meaning algebra alone cannot calculate an exact split-second rate of change.
AVERAGE VS. INSTANTANEOUS — KEY PAGE 2 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 3 OF 8
Before we ever touch calculus symbols, we can find the exact instantaneous velocity of a curving object geometrically by drawing a tangent line at a single point and finding its linear slope (\(\frac{\text{rise}}{\text{run}}\)).
Graphical Lab: The Roller Coaster Climb
A B Time (t) Pos (x)
Look at the graph representing an object speeding up. We have drawn tangent lines at two separate points, A and B.
Point A: The curve is still quite flat. The slope of the blue tangent line is low.
Point B: The curve is much steeper. The slope of the red tangent line is high.
POINT A ESTIMATE (SOLUTION)
Estimate the slope at A. If the run is \(5\text{ s}\) and the rise is \(2\text{ m}\), calculate the velocity:
\(v_A = \frac{\text{rise}}{\text{run}} = \frac{2\text{ m}}{5\text{ s}} = \mathbf{0.4\text{ m/s}}\)
POINT B ESTIMATE (SOLUTION)
Estimate the slope at B. If the run is \(5\text{ s}\) and the rise is \(8\text{ m}\), calculate the velocity:
\(v_B = \frac{\text{rise}}{\text{run}} = \frac{8\text{ m}}{5\text{ s}} = \mathbf{1.6\text{ m/s}}\)
Geometric Rule of Curves Solution: Summarize how the visual steepness of tangent lines tells you whether a curving object is speeding up or slowing down.
As the tangent line's slope becomes steeper (more vertical), the object's speed is increasing. As it becomes flatter (more horizontal), the speed is decreasing.
READING TANGENTS GEOMETRICALLY — KEY PAGE 3 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 4 OF 8
The Turning Point: When Slope Flattens to Zero
Think-Pair-Share Key
The Situation: Throwing a Ball Upwards
A student throws a tennis ball vertically straight up. It rises to its peak height, stops for an instantaneous moment, and falls back down. Its position-time graph is a smooth curve that arches like a dome. Student A says: "At the absolute peak of its curve, the ball's position is high, so its speed must also be at its maximum value."
Student B says: "No, at the absolute peak, the tangent line to the curve is completely horizontal, which means its speed is exactly zero."
Think: My Defense Solution
Identify which student is correct. Defend your choice by drawing what a horizontal tangent line implies about rate of change:
Student B is correct.
At the absolute peak of the ball's trajectory, the position-time graph transitions from rising (positive slope) to falling (negative slope). To change direction smoothly, the slope must pass through exactly zero. A completely flat tangent line always denotes an instantaneous velocity of zero (v = 0). Height represents location, but speed is slope.
Share: Partner Consensus Key
Compare arguments. Draft a shared physical description of how an object turns around in one dimension:
When an object reverses direction in 1D kinematics, its velocity must change sign (e.g., from positive to negative). For velocity to change sign continuously, the object must briefly come to rest. This turning point is represented graphically as a vertex (maximum or minimum) where the instantaneous rate of change is zero.
💡 Key Kinematic Connection: A peak or valley on a position graph is a turning point! A flat tangent line always denotes an instantaneous velocity of zero (\(v = 0\)).
CONCEPT BUSTER: TURNING POINTS — KEY PAGE 4 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 5 OF 8
Drawing manual tangent lines and counting grid squares is slow and imprecise. What if we had a magical "slope-finder" equation that could instantly calculate the exact slope of a curve at any second?
In mathematics, that slope-finder function is called the derivative. The symbol for the derivative of position with respect to time is written as: \(\frac{dx}{dt}\) or simply \(v(t)\).
The Input-Output Bridge
POSITION FUNCTION \(x(t)\) Input: Time \(t\) \(\implies\) Output: Location
Tells you exactly where the object is on the line.
VELOCITY DERIVATIVE \(v(t) = \frac{dx}{dt}\) Input: Time \(t\) \(\implies\) Output: Slope (Speed)
Tells you how fast and in what direction it is moving.
The most famous shortcut for finding a derivative of any simple polynomial term is the Power Rule:
The Simple Power Rule
\(\frac{d}{dt}\left( c t^n \right) = n \cdot c t^{n-1}\)
Multiply the constant coefficient (\(c\)) by the current exponent (\(n\)), then subtract \(1\) from the power of \(t\).
Guided Demonstration Solution
If an accelerating vehicle has a position formula of: \(x(t) = 5 t^2\). Let's use the Power Rule:
1. Bring down the power of 2: multiply it by 5: \((2 \times 5) = 10\)
2. Subtract 1 from the exponent: \(2 - 1 = 1\) (which is just \(t^1\) or \(t\))
3. This gives the velocity derivative: \(\mathbf{v(t) = 10t}\).
Now try it: If a projectile has a position formula of \(x(t) = 4 t^3\), write down its velocity equation \(v(t)\):
\(v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^3) = 3 \times 4t^{3-1} = \mathbf{12t^2}\)
THE INTUITIVE DERIVATIVE — KEY PAGE 5 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 6 OF 8
Let's practice our transition skills. Note that the derivative of any constant (like \(+5\)) is always **zero**, because a constant location has zero slope (no motion!).
PROBLEM 1: The Rolling Lab Cart (SOLUTION) Key
A small laboratory cart rolls along a level metal track. Its curving position is described by the equation: \[x(t) = 3t^2 + 2t + 4\]
A. Derive the velocity equation \(v(t)\):
\(v(t) = \frac{d}{dt}(3t^2 + 2t + 4) = \mathbf{6t + 2}\)
B. Calculate the velocity at exactly \(t = 3\text{ s}\):
\(v(3) = 6(3) + 2 = \mathbf{20\text{ m/s}}\)
PROBLEM 2: The Braking Vehicle (SOLUTION) Key
A vehicle traveling down a road slams on its brakes. Its position-time equation is described by: \[x(t) = 24t - 3t^2\]
A. Derive the velocity equation \(v(t)\):
\(v(t) = \frac{d}{dt}(24t - 3t^2) = \mathbf{24 - 6t}\)
B. Find the time \(t\) when the vehicle comes to a complete rest (\(v = 0\)):
\(24 - 6t = 0 \implies 6t = 24 \implies \mathbf{t = 4\text{ s}}\)
POWER RULE WORKBOOK LAB — KEY PAGE 6 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 7 OF 8
We have seen how taking the slope (derivative) of a position graph gives us velocity. But what if we start with a velocity graph and want to go backward to find out how far we traveled?
In physics, the area under a velocity-time graph equals the net displacement (\(\Delta x\)). Geometrically reversing the slope process is called integration.
Constant Velocity Area
When velocity is constant, the area is a simple rectangle: \(\text{Area} = \text{base} \times \text{height} = t \times v\).
v × t
Variable Velocity Area
When velocity increases steadily, the area under the curve is a triangle: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\).
1/2 v × t
Concept Check Solution: Area Accumulation
An athlete starts running from the starting line. Their velocity increases steadily over time, modeled by \(v(t) = 3t\). At \(t = 4\text{ seconds}\), their velocity is \(12\text{ m/s}\).
Calculate their displacement during these first \(4\text{ seconds}\) using the triangular area formula:
\(\Delta x = \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\text{ s} \times 12\text{ m/s} = \mathbf{24\text{ meters}}\)
ACCUMULATING MOTION AREAS — KEY PAGE 7 OF 8
Kinematics Transition Key // Teacher Copy
PAGE 8 OF 8
Just like the derivative has the Power Rule shortcut for slopes, the integral has an antiderivative shortcut for calculating exact curving areas.
The Simple Integral Power Rule (Anti-Derivative):
\(\int c t^n dt = \frac{c t^{n+1}}{n+1}\)
How to use it: Add exactly \(1\) to the exponent, then divide the term by this new exponent value. This rule mathematically calculates the area accumulation under your curve!
Simple Derivation: Constant Acceleration Formula
An object moves with a velocity of \(v(t) = a \cdot t\) (starts from rest, accelerates steadily). Let's find its position equation \(x(t)\) by integrating velocity:
\[x(t) = \int a t^1 dt = \frac{a t^{1+1}}{1+1} = \mathbf{\frac{1}{2} a t^2}\] This is exactly where the famous kinematic formula comes from! Calculus derived it from simple area accumulation!
Final Bridge Challenge Solution
A vehicle starts from rest at the origin. Its velocity equation is \(v(t) = 6t^2\). Use the integral power rule to find its position equation \(x(t)\):
1. Add 1 to the exponent of \(t^2 \implies t^3\)
2. Divide the term by this new exponent of 3:
\(x(t) = \int 6 t^2 dt = \frac{6 t^{2+1}}{2+1} = \frac{6 t^3}{3} = \mathbf{2t^3}\)
INTRODUCING THE INTEGRAL — KEY PAGE 8 OF 8
The Core Rate Divide: Algebra vs. Calculus
While distance and displacement measure where you go, speed and velocity measure how fast. How we calculate this rate marks the boundary between Algebra and Calculus:
Calculates Average Velocity (\(v_{\text{avg}} = \frac{\Delta x}{\Delta t}\)) over a finite interval of time. This treats the journey as a single constant rate, ignoring stop-and-go details.
Calculates Instantaneous Velocity (\(v(t) = \frac{dx}{dt}\)), which is the derivative of position with respect to time. It gives the velocity at an exact split second.
Concept Check Solution (In Red)
An athlete starts at position \(x_i = +2\text{ m}\), runs to the right to \(x = +10\text{ m}\), and then turns around and runs to the left to finish at \(x_f = -4\text{ m}\).
A. Total Distance Travelled (d):
d = 8m (right) + 14m (left) = 22 meters
B. Net Displacement (\(\Delta x\)):
\(\Delta x = x_f - x_i = -4\text{ m} - (+2\text{ m}) = -6\text{ meters}\)
AVERAGE SPEED VS. DERIVATIVE VELOCITY — KEY MOTION MECHANICS READING PACKET
Kinematics Key // Teacher Copy
PAGE 3 OF 12
The Turnaround Trick: Distance vs. Displacement & Turnaround Derivatives
Active Debate Key
The Situation: Tracking Coordinate Reversals & Infinitesimal Slopes
A runner starts at coordinate \(x = 0\text{ m}\), sprints out to \(+10\text{ m}\), then immediately spins around and runs back to \(+4\text{ m}\). Student A says: "Their net displacement is \(+14\text{ m}\) because they traveled a total of \(14\text{ m}\) of distance."
Student B says: "No, their net displacement is only \(+4\text{ m}\). Also, at the exact split-second peak at \(+10\text{ m}\) where they changed direction, their continuous velocity derivative \(\frac{dx}{dt}\) must have passed through zero."
Think: My Defense Solution
Is Student B correct about both displacement and the velocity derivative? Prove your reasoning mathematically and conceptually:
Student B is correct about both.
1. Displacement is solely the change in position: \(\Delta x = x_f - x_i = +4\text{ m} - 0\text{ m} = \mathbf{+4\text{ m}}\). Total distance is indeed \(10 + 6 = 16\text{ m}\), but they are distinct.
2. At the turning point, coordinate direction reverses. To transition from positive velocity (moving right) to negative velocity (moving left), a continuous velocity function must pass through exactly zero. Thus, \(\frac{dx}{dt} = 0\) at that instant.
Share: Partner Consensus Key
Compare arguments. Draft a shared explanation detailing how turning around mathematically forces a derivative \(\frac{dx}{dt} = 0\):
We agree that Student B's explanation is correct. Displacement is path-independent and measures net change in location. Geometrically, on a position-time graph, the turnaround point represents a maximum/vertex. At any local maximum on a differentiable curve, the tangent line is horizontal. A horizontal tangent line has a mathematical slope of zero, so the instantaneous velocity derivative must be \(\frac{dx}{dt} = 0\).
💡 Key Kinematic Metric: Displacement represents only the net change in coordinate position, matching the direct path from the start line to the end line! In calculus terms, changing direction means velocity (\(\frac{dx}{dt}\)) flips sign, passing through zero.
COOPERATIVE DISPLACEMENT BREAK — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 4 OF 12
Over an extended interval of time (\(\Delta t = t_f - t_i\)), an object's instantaneous velocity is constantly changing. To look at the big picture of a journey, we calculate averages. However, geometrically and mathematically, these two rates are represented completely differently on a position-time graph.
Algebra: Secant Slope
The slope of a secant line that cuts across a curved graph, connecting two separate, distinct points.
t₁ t₂
\(v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}\)
Calculus: Tangent Slope
The slope of a tangent line that balances perfectly against the curved graph, touching at exactly one point.
t
\(v(t) = \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\)
Problem 1 Solution (In Red): Solution
A continuous position function is given as \(x(t) = t^2\). Find the average velocity between \(t_1 = 1\text{ s}\) and \(t_2 = 3\text{ s}\).
\(x(1) = 1^2 = 1\text{ m}, \quad x(3) = 3^2 = 9\text{ m}\).
\(v_{\text{avg}} = \frac{x(3) - x(1)}{3 - 1} = \frac{9 - 1}{3 - 1} = \frac{8}{2}\)
Ans: 4 m/s
STOP & JOT Solution
The speedometer calculates instantaneous speed by dividing a tiny distance by a near-zero time interval (\(\Delta t \to 0\)), which is equivalent to the slope of a tangent line touching the curve at exactly one split-second.
AVERAGE VS. INSTANTANEOUS RATES — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 5 OF 12
In 1D kinematics, when velocity is constant, the position graph is a completely straight line. In non-calculus physics, we call the slope constant. In calculus physics, we see this as the derivative of a linear equation!
Case 1: Positive Constant Velocity (\(v = \text{const } > 0\)) Derivative: \(\frac{dx}{dt} = v > 0\)
t x
A straight line rising to the right. In calculus, if \(x(t) = vt + x_0\), then taking the time derivative yields a positive constant: \(\frac{dx}{dt} = v\). The slope is constant and positive!
Case 2: Negative Constant Velocity (\(v = \text{const } < 0\)) Derivative: \(\frac{dx}{dt} = v < 0\)
t x
A straight line falling to the right. In calculus, if \(x(t) = -vt + x_0\), then its time derivative is negative: \(\frac{dx}{dt} = -v\). The slope is constant and negative, meaning steady backward motion!
STOP & JOT Solution
The derivative value \(\frac{dx}{dt} = 0\). This makes physical sense because a stationary brick is completely at rest, meaning its position coordinate is constant and not changing over time. Its flat graphical slope of zero directly mirrors its physical velocity.
LINEAR POSITION DERIVATIVES — KEY MOTION MECHANICS READING PACKET
Kinematics Key // Teacher Copy
PAGE 6 OF 12
The Curved Road Illusion: Graph Curves vs. Non-Zero Derivatives
Active Debate Key
Decoding Plot Shape vs. Physical Coordinate Space
A skateboarder’s position-time graph is a beautiful, smooth curve bending upwards. Student A says: "The curved shape of this graph proves the skateboarder is physically riding along a curved, winding street."
Student B says: "No, they are riding along a straight sidewalk. The curve means their position derivative \(\frac{dx}{dt}\) is changing over time, which mathematically means they are accelerating!"
Think: My Defense Solution
Explain who is correct. Detail how a curved graph represents a variable derivative rather than a 2D physical path:
Student B is correct.
A position-time graph is not a spatial coordinate map. This is 1D kinematics, meaning the object moves strictly in a straight line along the x-axis. A curved line on an \(x-t\) graph indicates that the tangent slope, which is the instantaneous velocity derivative \(\frac{dx}{dt}\), is changing at each moment. Since the slope is not constant, the skateboarder is accelerating along their straight sidewalk.
Share: Partner Consensus Key
Compare. Why does acceleration represent the second derivative of position, \(a = \frac{d^2x}{dt^2}\)? Define this conceptually:
We agree that Student B is correct. In physics, acceleration is the rate of change of velocity: \(a(t) = \frac{dv}{dt}\). Since velocity itself is already the first derivative of position, \(v(t) = \frac{dx}{dt}\), acceleration is mathematically the second derivative of position: \(a = \frac{d^2x}{dt^2}\). Geometrically, this second derivative represents the concavity of the position curve. An upward curve (concave up) represents a positive second derivative, meaning positive acceleration.
💡 Key Kinematic Metric: Graphs are records of single-axis coordinate positions over time! A curved line represents velocity changing at each moment, meaning the position derivative \(\frac{dx}{dt}\) is changing.
COOPERATIVE CURVE BREAK — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 7 OF 12
When an object moves in the positive direction, position increases. Curvature displays whether the rate of increase is gaining magnitude (speeding up) or flinging flat (slowing down).
Case 3: Positive Direction, Speeding Up Convex / Concave Up
t x
The graph starts flat and curves increasingly steep. In calculus, this means the continuous derivative \(\frac{dx}{dt}\) (tangent slope) is positive and getting larger over time (\(v > 0, a = \frac{dv}{dt} > 0\)).
Case 4: Positive Direction, Slowing Down Concave Down
t x
The graph starts steep and flattens out. In calculus, this means the position derivative \(\frac{dx}{dt}\) is positive but getting smaller, approaching zero (\(v > 0, a = \frac{dv}{dt} < 0\)).
STOP & JOT Solution
In Case 3, the tangent slope is getting steeper, so velocity is increasing, making acceleration positive (\(a > 0\)). In Case 4, the slope is flattening horizontally towards zero, so velocity is decreasing, making acceleration negative (\(a < 0\)).
ACCELERATION: POSITIVE DERIVATIVES — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 8 OF 12
When an object moves in the negative direction, position coordinates decrease. The derivative \(\frac{dx}{dt}\) is always negative, but the curvature dictates acceleration!
Case 5: Negative Direction, Speeding Up Concave Down
t x
Starts slow (flat slope) and gains negative speed (steeper downwards curve). Mathematically, \(\frac{dx}{dt} < 0\) and is getting increasingly negative, so acceleration is negative (\(v < 0, a = \frac{dv}{dt} < 0\)).
Case 6: Negative Direction, Slowing Down Concave Up
t x
Starts steep downward and flattens horizontally, approaching zero speed. Since the negative slope is getting less steep, the derivative is approaching zero, meaning acceleration is positive (\(v < 0, a = \frac{dv}{dt} > 0\)).
STOP & JOT Solution
In Case 5, both signs are negative (\(v < 0, a < 0\)), which matches the rule because the object speeds up moving backward. In Case 6, velocity is negative and acceleration is positive (\(v < 0, a > 0\)), so they oppose, causing the object to slow down backward.
ACCELERATION: NEGATIVE DERIVATIVES — KEY MOTION MECHANICS READING PACKET
Kinematics Key // Teacher Copy
PAGE 9 OF 12
The Direction vs. Speed Trap: Signs vs. Absolute Magnitudes
Active Debate Key
Parsing Velocity Signs & Acceleration Directions
A bowling ball rolls backward in the negative direction, starting with a position derivative \(v = \frac{dx}{dt} = -1\text{ m/s}\) and speeding up to a derivative of \(v = -8\text{ m/s}\). Student A says: "Since its numeric velocity derivative values are getting more negative (\(-1\) to \(-8\)), the value is decreasing, so the bowling ball is slowing down."
Student B says: "No, they confuse numeric value with physical speed. The negative sign only indicates backward direction. The absolute magnitude of the derivative, \(|v| = | \frac{dx}{dt} |\), represents speed, which went from \(1\text{ m/s}\) to \(8\text{ m/s}\)!"
Think: My Defense Solution
Write out which student is correct. Define how derivative signs differ from speed magnitudes:
Student B is correct.
Student A is making a classic mathematical error by confusing physical speed with numerical scalar values. In physics, velocity is a vector quantity. The positive/negative sign only denotes spatial direction (moving forward or backward relative to the origin). Speed is the absolute magnitude of the velocity vector: \(|v| = |\frac{dx}{dt}|\). Gaining speed backward results in velocity changing from \(-1\text{ m/s}\) to \(-8\text{ m/s}\). This is a physical speeding up.
Share: Partner Consensus Key
Compare arguments. Draft a unified statement explaining why a negative velocity derivative with a negative acceleration derivative means speeding up:
We agree that Student B is correct. In kinematics, an object speeds up whenever its velocity and acceleration share the exact same sign. When both are negative (\(v < 0, a < 0\)), they work together. Acceleration is pushing the ball in the same negative direction it is already moving. This causes the magnitude of the position derivative to increase from \(1\text{ m/s}\) to \(8\text{ m/s}\) in the backward direction, which represents a clear increase in speed.
💡 Key Kinematic Metric: Physical speed is the absolute magnitude of the position derivative, \(|v(t)|\). A derivative changing from \(-1\text{ m/s}\) to \(-8\text{ m/s}\) indicates gaining speed backward!
COOPERATIVE SPEED BREAK — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 10 OF 12
Use this reference table to immediately decode any motion profile on a position-time graph. All 6 curves represent the fundamental relationship between position functions, velocity derivatives, and acceleration second derivatives.
Case 1: Positive Dir.
Constant Velocity v = dx/dt > 0 (const)
a = dv/dt = 0
Case 3: Positive Dir.
Speeding Up v = dx/dt > 0
a = dv/dt > 0
Case 4: Positive Dir.
Slowing Down v = dx/dt > 0
a = dv/dt < 0
Case 2: Negative Dir.
Constant Velocity v = dx/dt < 0 (const)
a = dv/dt = 0
Case 5: Negative Dir.
Speeding Up v = dx/dt < 0
a = dv/dt < 0
Case 6: Negative Dir.
Slowing Down v = dx/dt < 0
a = dv/dt > 0
Decoding Challenge Solutions (In Red)
1. A skateboarder accelerating naturally down a positive hill (\(v > 0, a > 0\)).
Case 3
2. A driver slamming on brakes in the positive direction (\(v > 0, a < 0\)).
Case 4
3. A bicyclist traveling backward but braking to slow down (\(v < 0, a > 0\)).
Case 6
STOP & JOT Solution
Acceleration is the second position derivative, representing concavity. A concave down shape bends like a dome, indicating that the slope (velocity) is continuously decreasing mathematically, which corresponds directly to a negative acceleration rate.
KINEMATICS MATRIX ANALYSIS — KEY MOTION MECHANICS READING PACKET
Kinematics Key // Teacher Copy
PAGE 11 OF 12
The Passing Paradox: Line Intersections vs. Speed Derivatives
Active Debate Key
Evaluating Graph Intersections & Dynamic Slopes
On a position-time graph, the line for Drone A (constant positive slope) and the line for Drone B (steep constant positive slope) cross perfectly at time \(t = 4\text{ s}\). Student A says: "Since their graph lines intersect at exactly \(4\text{ s}\), both drones must have the exact same speed at that instant."
Student B says: "No, they are at the exact same location, but their instantaneous position derivatives \(\frac{dx}{dt}\) are completely different!"
Think: My Defense Solution
Identify who is correct. Contrast what an "intersection" means (equal function value) versus a "derivative" (equal slope):
Student B is correct.
An intersection of two curves on a position graph indicates they share the exact same spatial coordinate: \(x_A(t) = x_B(t)\). In physics, this means they pass each other at that exact moment. However, speed is represented by the slope of the curve (the derivative \(\frac{dx}{dt}\)). Since Drone B's line has a much steeper slope than Drone A's line, Drone B has a higher instantaneous velocity and is flying past Drone A.
Share: Partner Consensus Key
Debate. Explain how two objects can be at the exact same coordinate without traveling at the same velocity derivative:
We agree that Student B is correct. A practical example is overtaking a slower car on the highway. At the exact split-second you pass alongside the other vehicle, you both share the same physical position coordinate on the road. However, you do not share the same speed. If you did, you would not be passing them; you would remain side-by-side. The act of passing mathematically necessitates that the two derivatives (\(\frac{dx}{dt}\)) are different.
💡 Key Kinematic Metric: Crossing lines mean identical coordinates (passing each other). Speed is represented purely by the derivative \(\frac{dx}{dt}\) (the slope of the lines), which remain entirely different!
COOPERATIVE INTERSECTION BREAK — KEY MOTION MECHANICS READING PACKET
Physics Mechanics Key // Teacher Copy
PAGE 12 OF 12
Analyze the multi-stage position function \(x(t)\) graphed below representing the continuous flight of a scientific drone.
Stage A Stage B Stage C Stage D Time (t) Position (x)
STAGE C DERIVATIVE (SOLUTION)
Describe the signs of \(v = \frac{dx}{dt}\) and \(a = \frac{d^2x}{dt^2}\) during Stage C.
Signs: v is negative (-), a is negative (-) because the curve slopes downward (moving backward) and curves concave down (speeding up backward).
STAGE D DERIVATIVE (SOLUTION)
Describe the signs of \(v = \frac{dx}{dt}\) and \(a = \frac{d^2x}{dt^2}\) during Stage D.
Signs: v is positive (+), a is positive (+) because the curve slopes upward (moving forward) and curves concave up (speeding up forward).
The Ultimate Slope Comparison Challenge Solution
Compare: An algebra student calculates average velocity over the entire flight (t=0 to end) and gets a positive value. Why does this average fail to represent that the drone flew backward (negative derivative) for an entire segment?
An algebraic average only uses the endpoints (\(\frac{x_f - x_i}{\Delta t}\)). Since the drone finished at a higher coordinate than it started, the average is positive. This completely hides Stage C because algebraic averaging smooths out and hides all intermediate negative position derivative intervals.
COMPREHENSIVE DERIVATIVE EXAM — KEY MOTION MECHANICS READING PACKET
Motion Mechanics // Student Practice
PAGE 3 OF 5
PROBLEM 07 Instantaneous Velocity
The continuous position of a particle traveling along the x-axis is modeled by the function: \(x(t) = 2t^3 - t^2\) (with \(x\) in meters and \(t\) in seconds). Derive the velocity function \(v(t)\), and calculate the particle's exact instantaneous velocity at \(t = 4\text{ seconds}\).
PROBLEM 08 Finding Turnaround Point (v = 0)
A braking freight train’s position is modeled by the equation: \(x(t) = 30t - 5t^2\). Derive its velocity equation \(v(t)\), and calculate exactly how many seconds it takes for the train to decelerate to a complete stop (\(v = 0\)).
PROBLEM 09 Second Derivative (Acceleration)
An accelerating laboratory cart moves with a position equation of: \(x(t) = 4t^3 - 3t\). Calculate the exact second derivative of position, \(a(t) = \frac{d^2x}{dt^2}\), and evaluate the cart's acceleration at \(t = 2\text{ seconds}\).
KINEMATICS PRACTICE HANDOUT // 15 PROBLEMS PAGE 3 OF 5
Motion Mechanics // Student Practice
PAGE 4 OF 5
PROBLEM 10 Conceptual Derivative Signs
A physical body moves with instantaneous velocity \(v(t) < 0\) (negative velocity) and acceleration \(a(t) > 0\) (positive acceleration). Conceptualize this motion: is the body speeding up or slowing down? Explain why, referencing what opposite derivative signs imply physically.
PROBLEM 11 Geometric Tangent Estimates
A curved position-time graph rises slowly at first, passes through an inflection point, peaks at \(t = 5\text{ s}\), and then falls steeply. Explain how to estimate the instantaneous velocity derivative geometrically at: (A) the rising section, (B) the absolute peak, and (C) the falling section. What are the signs of these tangent slopes?
PROBLEM 12 Velocity & Acceleration Profiles
A heavy shipping crate is hoisted vertically with a position-time equation of: \(x(t) = t^3 - 6t^2 + 9t\). Derive: (A) the velocity equation \(v(t) = \frac{dx}{dt}\), and (B) the acceleration equation \(a(t) = \frac{dv}{dt}\). Evaluate both variables at \(t = 2\text{ seconds}\).
KINEMATICS PRACTICE HANDOUT // 15 PROBLEMS PAGE 4 OF 5
Motion Mechanics // Student Practice
PAGE 5 OF 5
PROBLEM 13 Critical Points (v = 0)
An automated laboratory track robot moves along the x-axis with a continuous position function modeled by: \(x(t) = t^3 - 12t\). Apply the derivative to find the velocity equation \(v(t) = \frac{dx}{dt}\). Calculate the exact time \(t > 0\) when the robot momentarily halts to change direction (\(v(t) = 0\)).
PROBLEM 14 Double Differentiation & Jerk
A high-speed bullet train accelerates from the station. Its position-time formula is modeled by: \(x(t) = \frac{1}{6}t^4 - 2t^2\). Calculate the exact velocity derivative function \(v(t) = \frac{dx}{dt}\) and the acceleration derivative function \(a(t) = \frac{dv}{dt}\). Evaluate both at \(t = 3\text{ seconds}\).
PROBLEM 15 Conceptual Secant vs. Tangent
Synthesize: Explain why the algebraic average velocity formula \(v_{\text{avg}} = \frac{\Delta x}{\Delta t}\) fails to calculate exact speed at a specific split-second, while the derivative limit \(\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\) succeeds. Reference tangent lines, secant lines, and shrinking time intervals.
KINEMATICS PRACTICE HANDOUT // 15 PROBLEMS — ONLY DERIVATIVES PAGE 5 OF 5
A physical therapy patient walks \(40\text{ meters}\) East, turns around and walks \(60\text{ meters}\) West, and finally turns East again and walks \(10\text{ meters}\). Determine: (A) the total physical distance walked by the patient, and (B) the net coordinate displacement (\(\Delta x\)) from their initial starting position.
Let East be positive (+) and West be negative (-). Paths are: \(s_1 = +40\text{ m}\), \(s_2 = -60\text{ m}\), \(s_3 = +10\text{ m}\). (A) Total Physical Distance (scalar, path-dependent magnitude): \(d = |s_1| + |s_2| + |s_3| = 40\text{ m} + 60\text{ m} + 10\text{ m} = \mathbf{110\text{ meters}}\). (B) Net Coordinate Displacement (vector, path-independent coordinate change): \(\Delta x = s_1 + s_2 + s_3 = +40\text{ m} - 60\text{ m} + 10\text{ m} = \mathbf{-10\text{ meters}}\) (or \(10\text{ m West}\)).
PROBLEM 05 (SOLUTION) Key
An object moves with constant velocity according to the linear algebraic formula: \(x(t) = -4t + 15\) (where position \(x\) is in meters and time \(t\) is in seconds). Determine the coordinate position of the object at \(t = 0\text{ s}\), its position at \(t = 5\text{ s}\), and write down its constant velocity.
1. Initial position at \(t = 0\text{ s}\): \(x(0) = -4(0) + 15 = \mathbf{+15\text{ meters}}\). 2. Final position at \(t = 5\text{ s}\): \(x(5) = -4(5) + 15 = -20 + 15 = \mathbf{-5\text{ meters}}\). 3. Velocity is the slope of the linear function: \(v = \frac{dx}{dt} = \mathbf{-4\text{ m/s}}\) (constant). Note: The negative sign signifies the object is moving at 4 m/s in the negative direction (leftward).
PROBLEM 06 (SOLUTION) Key
A miniature test rocket's position-time function is modeled continuously by: \(x(t) = 5t^2 + 3t + 2\). Apply the Power Rule derivative shortcut to find the exact velocity equation \(v(t) = \frac{dx}{dt}\).
Velocity is the first derivative of position with respect to time, \(v(t) = \frac{dx}{dt}\): \(v(t) = \frac{d}{dt}\left( 5t^2 + 3t + 2 \right) = \frac{d}{dt}(5t^2) + \frac{d}{dt}(3t) + \frac{d}{dt}(2)\) \(v(t) = (2 \cdot 5)t^{2-1} + (1 \cdot 3)t^{1-1} + 0 = \mathbf{10t + 3}\).
KINEMATICS PRACTICE HANDOUT — KEY PAGE 2 OF 5
Motion Mechanics Key // Teacher Copy
PAGE 3 OF 5
PROBLEM 07 (SOLUTION) Key
The continuous position of a particle traveling along the x-axis is modeled by the function: \(x(t) = 2t^3 - t^2\) (with \(x\) in meters and \(t\) in seconds). Derive the velocity function \(v(t)\), and calculate the particle's exact instantaneous velocity at \(t = 4\text{ seconds}\).
1. Derivative of position function to obtain velocity equation: \(v(t) = \frac{dx}{dt} = \frac{d}{dt}\left( 2t^3 - t^2 \right) = (3 \cdot 2)t^{3-1} - (2 \cdot 1)t^{2-1} = \mathbf{6t^2 - 2t}\) 2. Evaluate the velocity equation at exactly \(t = 4\text{ seconds}\): \(v(4) = 6(4)^2 - 2(4) = 6(16) - 8 = 96 - 8 = \mathbf{88\text{ m/s}}\).
PROBLEM 08 (SOLUTION) Key
A braking freight train’s position is modeled by the equation: \(x(t) = 30t - 5t^2\). Derive its velocity equation \(v(t)\), and calculate exactly how many seconds it takes for the train to decelerate to a complete stop (\(v = 0\)).
1. Derivative of position function to obtain velocity equation: \(v(t) = \frac{dx}{dt} = \frac{d}{dt}\left( 30t - 5t^2 \right) = 30 - 10t\) 2. Set the velocity equation to zero to solve for the complete stopping time: \(v(t) = 0 \implies 30 - 10t = 0 \implies 10t = 30 \implies \mathbf{t = 3\text{ seconds}}\).
PROBLEM 09 (SOLUTION) Key
An accelerating laboratory cart moves with a position equation of: \(x(t) = 4t^3 - 3t\). Calculate the exact second derivative of position, \(a(t) = \frac{d^2x}{dt^2}\), and evaluate the cart's acceleration at \(t = 2\text{ seconds}\).
1. First derivative of position gives velocity: \(v(t) = \frac{dx}{dt} = 12t^2 - 3\). 2. Second derivative of position (or first derivative of velocity) gives acceleration: \(a(t) = \frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{d}{dt}\left( 12t^2 - 3 \right) = \mathbf{24t}\). 3. Evaluate acceleration equation at \(t = 2\text{ seconds}\): \(a(2) = 24(2) = \mathbf{48\text{ m/s}^2}\).
KINEMATICS PRACTICE HANDOUT — KEY PAGE 3 OF 5
Motion Mechanics Key // Teacher Copy
PAGE 4 OF 5
PROBLEM 10 (SOLUTION) Key
A physical body moves with instantaneous velocity \(v(t) < 0\) (negative velocity) and acceleration \(a(t) > 0\) (positive acceleration). Conceptualize this motion: is the body speeding up or slowing down? Explain why, referencing what opposite derivative signs imply physically.
The body is slowing down (decelerating). Opposite signs between velocity (\(v < 0\)) and acceleration (\(a > 0\)) mean they oppose each other. Velocity indicates the body is moving backward (to the left), while positive acceleration acts like a continuous forward push (to the right). This opposing vector reduces the speed magnitude toward zero.
PROBLEM 11 (SOLUTION) Key
A curved position-time graph rises slowly at first, passes through an inflection point, peaks at \(t = 5\text{ s}\), and then falls steeply. Explain how to estimate the instantaneous velocity derivative geometrically at: (A) the rising section, (B) the absolute peak, and (C) the falling section. What are the signs of these tangent slopes?
(A) Rising section: Tangent slope trends upward, so velocity derivative is **positive (\(v > 0\))**. (B) Absolute peak: Tangent line is completely flat and horizontal, so velocity is **exactly zero (\(v = 0\))**. (C) Falling section: Tangent slope trends downward, so velocity derivative is **negative (\(v < 0\))**. Geometrical Estimate: Draw a tangent line at the point and calculate: \(\text{slope} = \frac{\Delta x}{\Delta t} = \frac{\text{rise}}{\text{run}}\).
PROBLEM 12 (SOLUTION) Key
A heavy shipping crate is hoisted vertically with a position-time equation of: \(x(t) = t^3 - 6t^2 + 9t\). Derive: (A) the velocity equation \(v(t) = \frac{dx}{dt}\), and (B) the acceleration equation \(a(t) = \frac{dv}{dt}\). Evaluate both variables at \(t = 2\text{ seconds}\).
1. Equations: \(v(t) = \frac{dx}{dt} = \mathbf{3t^2 - 12t + 9}\) and \(a(t) = \frac{dv}{dt} = \mathbf{6t - 12}\). 2. Evaluate at \(t = 2\text{ seconds}\): Velocity: \(v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = \mathbf{-3\text{ m/s}}\) (moving downward). Acceleration: \(a(2) = 6(2) - 12 = 12 - 12 = \mathbf{0\text{ m/s}^2}\) (velocity is momentarily constant).
KINEMATICS PRACTICE HANDOUT — KEY PAGE 4 OF 5
Motion Mechanics Key // Teacher Copy
PAGE 5 OF 5
PROBLEM 13 (SOLUTION) Key
An automated laboratory track robot moves along the x-axis with a continuous position function modeled by: \(x(t) = t^3 - 12t\). Apply the derivative to find the velocity equation \(v(t) = \frac{dx}{dt}\). Calculate the exact time \(t > 0\) when the robot momentarily halts to change direction (\(v(t) = 0\)).
1. Differentiate position function to find velocity: \(v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 12t) = \mathbf{3t^2 - 12}\). 2. Find the turnaround time where instantaneous velocity reaches exactly zero: \(v(t) = 0 \implies 3t^2 - 12 = 0 \implies 3t^2 = 12 \implies t^2 = 4\). Since time must be positive, \(t > 0 \implies \mathbf{t = 2\text{ seconds}}\).
PROBLEM 14 (SOLUTION) Key
A high-speed bullet train accelerates from the station. Its position-time formula is modeled by: \(x(t) = \frac{1}{6}t^4 - 2t^2\). Calculate the exact velocity derivative function \(v(t) = \frac{dx}{dt}\) and the acceleration derivative function \(a(t) = \frac{dv}{dt}\). Evaluate both at \(t = 3\text{ seconds}\).
1. Velocity derivative: \(v(t) = \frac{dx}{dt} = \frac{d}{dt}\left( \frac{1}{6}t^4 - 2t^2 \right) = \frac{4}{6}t^3 - 4t = \mathbf{\frac{2}{3}t^3 - 4t}\). 2. Acceleration derivative: \(a(t) = \frac{dv}{dt} = \frac{d}{dt}\left( \frac{2}{3}t^3 - 4t \right) = 3 \cdot \frac{2}{3}t^2 - 4 = \mathbf{2t^2 - 4}\). 3. Evaluate both functions at \(t = 3\text{ seconds}\): Velocity: \(v(3) = \frac{2}{3}(3^3) - 4(3) = \frac{2}{3}(27) - 12 = 18 - 12 = \mathbf{6\text{ m/s}}\). Acceleration: \(a(3) = 2(3)^2 - 4 = 2(9) - 4 = 18 - 4 = \mathbf{14\text{ m/s}^2}\).
PROBLEM 15 (SOLUTION) Key
Synthesize: Explain why the algebraic average velocity formula \(v_{\text{avg}} = \frac{\Delta x}{\Delta t}\) fails to calculate exact speed at a specific split-second, while the derivative limit \(\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\) succeeds. Reference tangent lines, secant lines, and shrinking time intervals.
The algebraic formula \(v_{\text{avg}} = \frac{\Delta x}{\Delta t}\) calculates the slope of a secant line connecting two distinct points. This measures only the overall average rate, smoothing out any variable changes. Conversely, the derivative limit \(\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}\) shrinks the time interval (\(\Delta t\)) towards zero, bringing the two secant points closer and closer together until they merge into a single point. This transforms the secant line into a tangent line, whose slope mathematically defines the exact, split-second instantaneous velocity.
KINEMATICS PRACTICE HANDOUT — KEY PAGE 5 OF 5
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 2 OF 5
Honors Physics // 1D Kinematics
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PROBLEM 05 Incline Roll: Velocity & Acceleration Vectors
A small steel marble is launched rolling up a smooth ramp. It steadily decelerates, momentarily comes to rest at its highest point (turning point), and then rolls back down the ramp, steadily speeding up.
Drawing Challenge: Construct a complete 2-tier horizontal motion diagram (or incline motion diagram) showing: (1) position dots at equal \(\Delta t\), (2) velocity vectors \(\vec{v}\), and (3) acceleration vectors \(\vec{a}\) for both the upward and downward trips.
Questions: Compare the direction of \(\vec{v}\) relative to \(\vec{a}\) during: (A) the upward motion, (B) the instant the ball is stopped at the peak, and (C) the downward motion.
PROBLEM 06 Comparative Motion Diagrams (Two Objects)
Two runners, Runner A and Runner B, start at the exact same starting line \(x = 0\text{ m}\) at time \(t = 0\text{ s}\). Runner A runs at a **constant velocity** of \(+4.0\text{ m/s}\). Runner B starts from rest (\(v_0 = 0\)) and **accelerates at a constant rate**, covering increasing distances each second until passing Runner A at \(t = 4.0\text{ s}\).
Tasks: (A) Draw side-by-side (parallel) motion diagrams for Runner A and Runner B from \(t = 0\text{ s}\) to \(t = 4\text{ s}\) at \(\Delta t = 1.0\text{ s}\) intervals. (B) Explain how the dot spacing and vector lengths illustrate the exact moment Runner B catches up to Runner A.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 3 OF 5
Honors Physics // 1D Kinematics
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PROBLEM 07 Piecewise Linear Slopes & Direction Reversals
A remote-controlled rover moves along a straight laboratory track. Its position \(x\) over a \(10.0\text{-second}\) interval is shown in the position-time graph below.
20 15 10 5 0 0 2 4 6 8 10 x (m) t (s)
TASKS:
PROBLEM 08 Curvature, Concavity & Tangent Slopes
An object moves along the x-axis. Its position-time graph displays a smooth curve starting at coordinate \((0\text{ s}, 0\text{ m})\) and curving upward with increasing steepness (concave up). (A) Explain why an upward curvature represents acceleration (speeding up in the positive direction) by analyzing the changing slope of tangent lines. (B) Describe step-by-step how an honors physics student can determine the instantaneous velocity at exactly \(t = 3.0\text{ s}\) using a ruler and tangent line method without calculus.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 4 OF 5
Honors Physics // 1D Kinematics
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+8 +6 +4 +2 0 -2 -4 -6 -8 0 2 4 6 8 10 12 Position x (m) Time t (seconds)
MULTI-INTERVAL CART TRACK: The graph illustrates the position \(x\) over \(12.0\text{ s}\) for a motorized dynamics cart along a track with three distinct piecewise intervals:
PROBLEM 09 Comprehensive Multi-Interval Honors Synthesis
Using the multi-interval position-time graph above, answer each part completely, showing formulas and units:
(A) Instantaneous Segment Velocities:
Calculate the exact slope (velocity) for: Interval A (\(0-4\text{ s}\)), Interval B (\(4-8\text{ s}\)), and Interval C (\(8-12\text{ s}\)).
(B) Total Distance & Net Displacement:
Calculate the total scalar distance \(d\) and net displacement vector \(\Delta x\) for the full \(12.0\text{-second}\) trip.
(C) Trip Averages: Calculate the cart's overall average speed \(s_{\text{avg}}\) and average velocity vector \(\vec{v}_{\text{avg}}\) across the entire \(12.0\text{-second}\) motion.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 5 OF 5
(B) Average Speed vs. Average Velocity:
• Average Speed: \(s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{100.0\text{ m}}{55.0\text{ s}} \approx \mathbf{1.82\text{ m/s}}\).
• Average Velocity Vector: \(\vec{v}_{\text{avg}} = \frac{\Delta x_{\text{total}}}{t_{\text{total}}} = \frac{0.0\text{ m}}{55.0\text{ s}} = \mathbf{0.00\text{ m/s}}\).
PROBLEM 04 (SOLUTION) Key
Skateboarder motion diagram analysis: dot spacing at \(\Delta t = 1.0\text{ s}\).
Start
End
\(t=0\text{s}\)\(t=1\text{s}\)\(t=2\text{s}\)\(t=3\text{s}\)\(t=4\text{s}\)\(t=5\text{s}\)\(t=6\text{s}\)
Step-by-step Solution & Dot Spacing Justifications:
• Slowing Down (Decelerating): Interval \(\mathbf{t = 0\text{ s} \to 3\text{ s}}\). The distance between successive dots decreases progressively (\(14\text{ px} \to 9\text{ px} \to 5\text{ px}\)), proving \(\Delta x/\Delta t\) is dropping.
• Constant Speed: Interval \(\mathbf{t = 3\text{ s} \to 4\text{ s}}\). The dot spacing remains identical (\(5\text{ px}\)), proving zero acceleration.
• Speeding Up (Accelerating): Interval \(\mathbf{t = 4\text{ s} \to 6\text{ s}}\). The distance between consecutive dots increases progressively (\(9\text{ px} \to 14\text{ px}\)), proving the speed is increasing.
SPEED AND VELOCITY PROBLEMS — KEY PAGE 2 OF 5
Honors Motion Key // Teacher Copy
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PROBLEM 05 (SOLUTION) Key
Incline Ramp Roll: Motion diagram sketch and vector comparisons.
Diagram & Vector Analysis:
• Upward Motion Diagram: Dots spaced wide at bottom, getting progressively closer together toward the peak. Velocity vectors \(\vec{v}\) point UP the ramp with shrinking lengths. Acceleration vectors \(\vec{a}\) point DOWN the ramp with constant length.
• Peak: Single dot. Instantaneous velocity is zero (\(\vec{v} = 0\)). Acceleration vector \(\vec{a}\) is non-zero pointing DOWN the ramp.
(A) Upward Trip: \(\vec{v}\) and \(\vec{a}\) point in **opposite directions** (\(\vec{v}\) up-ramp, \(\vec{a}\) down-ramp), causing the marble to slow down.
(B) Instant at the Peak: \(\vec{v} = 0\), while \(\vec{a}\) remains **constant pointing down-ramp** (due to the parallel component of gravity \(g \sin\theta\)).
(C) Downward Trip: \(\vec{v}\) and \(\vec{a}\) point in the **same direction** (both pointing down-ramp), causing the marble to speed up.
PROBLEM 06 (SOLUTION) Key
Runner A (constant \(v = +4\text{ m/s}\)) and Runner B (constant acceleration from rest passing A at \(t = 4\text{ s}\)).
(A) Comparative Motion Diagram Structure:
• Runner A (Top Track): Dot positions at \(t = 0, 1, 2, 3, 4\text{ s}\) are at \(x = 0, 4, 8, 12, 16\text{ m}\). Dot spacing is uniform (\(4\text{ m}\) per tick), and velocity vectors are identical in length (\(4\text{ m/s}\)).
• Runner B (Bottom Track): Dot positions at \(t = 0, 1, 2, 3, 4\text{ s}\) are at \(x = 0, 1, 4, 9, 16\text{ m}\) (with \(a = 2.0\text{ m/s}^2\), \(x = \frac{1}{2}at^2\)). Dot spacing grows (\(1\text{ m} \to 3\text{ m} \to 5\text{ m} \to 7\text{ m}\)), and velocity vectors grow linearly (\(0 \to 2 \to 4 \to 6 \to 8\text{ m/s}\)).
(B) Meeting Point Explanation:
At \(t = 4.0\text{ s}\), both dots align at the exact same physical coordinate position \(x = 16.0\text{ m}\). At this instant of overtaking, Runner B's velocity vector (\(8.0\text{ m/s}\)) is twice as long as Runner A's velocity vector (\(4.0\text{ m/s}\)).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 3 OF 5
Honors Motion Key // Teacher Copy
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PROBLEM 07 (SOLUTION) Key
Rover position-time graph slopes and direction reversals.
20 15 10 5 0 0 2 4 6 8 10 x (m) t (s)
SLOPE CALCULATIONS:
Step-by-step Solution:
(A) Slopes: Interval 1: \(v_1 = \frac{15.0\text{ m} - 0.0\text{ m}}{4.0\text{ s} - 0.0\text{ s}} = \mathbf{+3.75\text{ m/s}}\). Interval 2: \(v_2 = \frac{0.0\text{ m} - 15.0\text{ m}}{10.0\text{ s} - 4.0\text{ s}} = \mathbf{-2.50\text{ m/s}}\).
(B) Physical Event at \(t = 4.0\text{ s}\): The rover reaches its maximum forward position (\(x = +15.0\text{ m}\)), instantly reverses direction (turning point), and heads backward toward the origin.
(C) Distance & Displacement: Total Distance: \(d = 15.0\text{ m} + 15.0\text{ m} = \mathbf{30.0\text{ m}}\). Net Displacement: \(\Delta x = 0.0\text{ m} - 0.0\text{ m} = \mathbf{0.0\text{ m}}\).
PROBLEM 08 (SOLUTION) Key
Curvature, concavity, and tangent slope visual estimation without calculus.
(A) Curvature and Acceleration Analysis:
On a position-time graph, the slope of the curve at any instant equals the instantaneous velocity (\(v = \text{slope}\)). When the curve is concave upward (curving up), the tangent lines become steeper and steeper as time passes (\(\text{slope}_2 > \text{slope}_1\)). This means the instantaneous velocity is continuously increasing, which by definition represents positive acceleration (\(a > 0\)).
(B) Tangent Line Step-by-Step Procedure:
1. Place a transparent straightedge tangent to the curved curve precisely at the point \((t = 3.0\text{ s}, x(3.0))\), matching the exact angle of the curve without crossing through it.
2. Draw an extended straight line along the straightedge.
3. Identify two easily readable coordinate points on this drawn tangent line: \((t_1, x_1)\) and \((t_2, x_2)\).
4. Compute the slope: \(v(3.0) = \frac{x_2 - x_1}{t_2 - t_1}\). This slope gives the instantaneous velocity in \(\text{m/s}\).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 4 OF 5
Honors Motion Key // Teacher Copy
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+8 +6 +4 +2 0 -2 -4 -6 -8 0 2 4 6 8 10 12 Position x (m) Time t (seconds)
SOLVED INTERVAL SUMMARY: Continuous piecewise functions on \(x\text{-}t\) grid:
PROBLEM 09 (SOLUTION) Key
Comprehensive step-by-step mathematical calculations:
(A) Instantaneous Segment Velocities (Slopes):
• Interval A (\(0-4\text{ s}\)): \(v_A = \frac{\Delta x_A}{\Delta t_A} = \frac{+6.0\text{ m} - (-2.0\text{ m})}{4.0\text{ s} - 0.0\text{ s}} = \frac{+8.0\text{ m}}{4.0\text{ s}} = \mathbf{+2.00\text{ m/s}}\) (moving right/forward).
• Interval B (\(4-8\text{ s}\)): \(v_B = \frac{\Delta x_B}{\Delta t_B} = \frac{+6.0\text{ m} - (+6.0\text{ m})}{8.0\text{ s} - 4.0\text{ s}} = \frac{0.0\text{ m}}{4.0\text{ s}} = \mathbf{0.00\text{ m/s}}\) (cart is at rest / stationary).
• Interval C (\(8-12\text{ s}\)): \(v_C = \frac{\Delta x_C}{\Delta t_C} = \frac{-6.0\text{ m} - (+6.0\text{ m})}{12.0\text{ s} - 8.0\text{ s}} = \frac{-12.0\text{ m}}{4.0\text{ s}} = \mathbf{-3.00\text{ m/s}}\) (moving left/backward).
(B) Total Scalar Distance & Net Displacement:
• Total Distance \(d = |+8.0\text{ m}| + |0.0\text{ m}| + |-12.0\text{ m}| = 8.0\text{ m} + 12.0\text{ m} = \mathbf{20.0\text{ meters}}\).
• Net Displacement \(\Delta x = x(12) - x(0) = -6.0\text{ m} - (-2.0\text{ m}) = \mathbf{-4.0\text{ meters}}\) (or \(4.0\text{ m in negative direction}\)).
(C) Trip Averages:
• Average Speed \(s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{20.0\text{ m}}{12.0\text{ s}} = \mathbf{1.67\text{ m/s}}\) (or \(\frac{5}{3}\text{ m/s}\)).
• Average Velocity Vector \(\vec{v}_{\text{avg}} = \frac{\Delta x_{\text{total}}}{t_{\text{total}}} = \frac{-4.0\text{ m}}{12.0\text{ s}} = \mathbf{-0.33\text{ m/s}}\) (or \(0.33\text{ m/s in negative direction}\)).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 5 OF 5
2. Case of Equal Distance Intervals (\(x_1 = x_2\))
If an object travels equal distances \(x_1\) and \(x_2\) (such as completing one lap down and back) at different speeds, rely on the universal formula:
\(v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_1 + x_2}{t_1 + t_2}\)
How to Solve:
1. Calculate the elapsed time for each stage first: \(t_1 = \frac{x_1}{v_1}\) and \(t_2 = \frac{x_2}{v_2}\).
2. Substitute these times and distances directly into the universal formula above.
SPEED VELOCITY REFERENCE SHEET // STUDY AID PAGE 1 OF 2
Honors Physics // Quick Reference Sheet
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Section 2: Labeled Graphs, Spacing Dictionary & Vector Relationships
Unit 1.1 Guide
Visual Translation
Constant Speed (+)
Straight line rising. Moving East steadily.
Constant Speed (-)
Straight line falling. Moving West steadily.
Speeding Up (+)
Curve gets steeper. Accelerating East.
Speeding Up (-)
Curve gets steeper down. Accelerating West.
Slowing Down (+)
Curve flattens out. Decelerating East.
Slowing Down (-)
Curve flattens out. Decelerating West.
Constant Speed (+)
Equal dot spacing. Solid arrows point East (Right).
Constant Speed (-)
Equal dot spacing. Solid arrows point West (Left).
Speeding Up (+)
Dots get wider. Velocity arrows grow pointing East.
Speeding Up (-)
Dots get wider. Velocity arrows grow pointing West.
Slowing Down (+)
Dots get closer. Velocity arrows shrink pointing East.
Slowing Down (-)
Dots get closer. Velocity arrows shrink pointing West.
Seamless Translation Matrix
Honors physicists translate between three modes. Below is a comparative roadmap of collinear translation behaviors:
Physical Motion
Moving constant speed East
Completely stationary
Moving constant speed West
Graph Slope Shape
Rising straight line
Flat horizontal line
Falling straight line
Motion Diagram Dots
Equal spacing, arrows East
Stacked dots, no arrows
Equal spacing, arrows West
SPEED VELOCITY REFERENCE SHEET // STUDY AID PAGE 2 OF 2
KINEMATICS ANTIDERIVATIVE HANDOUT // 15 PROBLEMS PAGE 2 OF 5
Motion Mechanics // Antiderivative Practice
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PROBLEM 07 Variable Acceleration (Power Rule)
The acceleration of an automated trolley is modeled by the polynomial function: \(a(t) = 12t^2\). At time \(t = 0\text{ s}\), the trolley is moving with an initial velocity of \(v(0) = -3\text{ m/s}\). Find the trolley's exact velocity function \(v(t)\) by integrating \(a(t)\).
PROBLEM 08 Polynomial Velocity to Position
A small drone flies horizontally with a velocity modeled by: \(v(t) = 3t^2 - 4t + 5\). Given that the drone passes coordinate \(x = +12\text{ m}\) at time \(t = 0\text{ s}\), integrate the velocity equation to find the drone's position function \(x(t)\).
PROBLEM 09 Double Integration with Multi-term Equations
A physical body is subjected to a time-varying acceleration of \(a(t) = 6t - 4\). At time \(t = 0\text{ s}\), the body starts with an initial velocity of \(v(0) = +8\text{ m/s}\) and an initial position of \(x(0) = +2\text{ m}\). Integrate acceleration twice to derive the position function \(x(t)\).
KINEMATICS ANTIDERIVATIVE HANDOUT // 15 PROBLEMS PAGE 3 OF 5
Motion Mechanics // Antiderivative Practice
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PROBLEM 10 The Fundamental Theorem in Kinematics
The velocity of a cargo glider is given by \(v(t) = 3t^2\). Show that calculating its displacement over the interval from \(t = 1\text{ s}\) to \(t = 3\text{ s}\) using the definite integral \(\Delta x = \int_{1}^{3} v(t) dt\) yields the exact same answer as evaluating the change in its position function: \(\Delta x = x(3) - x(1)\).
PROBLEM 11 Definite Integral with Constant Accel
A high-speed maglev sled accelerates down a linear track with velocity modeled by \(v(t) = 10t + 5\). Compute the exact physical displacement (\(\Delta x\)) of the maglev sled from time \(t = 2\text{ seconds}\) to \(t = 5\text{ seconds}\) using definite integration.
PROBLEM 12 Variable Speed Definite Integration
A wind turbine blade is decelerating. Its linear velocity profile at the tip is given by the function: \(v(t) = 12 - 3t^2\) (for \(0 \le t \le 2\)). Compute the tip's net displacement (\(\Delta x\)) over this entire interval by setting up and evaluating the definite integral: \(\Delta x = \int_{0}^{2} (12 - 3t^2) dt\).
KINEMATICS ANTIDERIVATIVE HANDOUT // 15 PROBLEMS PAGE 4 OF 5
Motion Mechanics // Antiderivative Practice
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PROBLEM 13 Solving for C with Boundary Conditions
The velocity function of a small probe is \(v(t) = 3t^2 - 6t\). At time \(t = 2\text{ seconds}\), the probe is physically located at coordinate \(x(2) = +10\text{ m}\). Integrate the velocity function and solve for the integration constant \(C\) using this non-zero time boundary condition to find the unique position function \(x(t)\).
PROBLEM 14 Integrating Fractional Power Profiles
An underwater automated vehicle experiences fluid resistance, resulting in a decaying acceleration function modeled by: \(a(t) = t^{-1/2}\) (for \(t \ge 1\)). Knowing that the vehicle's initial velocity is \(v(1) = +5\text{ m/s}\), integrate \(a(t)\) to derive its velocity equation \(v(t)\).
PROBLEM 15 Synthesis: Physical Constant of Integration
Synthesize: In algebra physics, we memorize the equation \(x(t) = \frac{1}{2}at^2 + v_0 t + x_0\). Explain how this kinematic equation can be derived from first principles using calculus double integration of a constant acceleration \(a(t) = a\), showing specifically how the initial velocity (\(v_0\)) and initial position (\(x_0\)) physically arise as the constants of integration.
KINEMATICS ANTIDERIVATIVE HANDOUT // 15 PROBLEMS — ONLY INTEGRATION PAGE 5 OF 5
\(v(t) = \int a(t) dt = \int 6 dt = 6t + C\)
Plug in initial condition \(t = 0, v(0) = 4\) to solve for \(C\):
\(v(0) = 6(0) + C = 4 \implies C = 4\)
Therefore, the velocity function is: \(\mathbf{v(t) = 6t + 4}\)
PROBLEM 05 Constant Velocity to Position
A bicyclist rides along a track with a velocity modeled by the linear function: \(v(t) = 4t + 3\) (where \(v\) is in m/s and \(t\) is in seconds). At \(t = 0\text{ s}\), the bicyclist passes the reference marker at position \(x(0) = +10\text{ m}\). Integrate the velocity function to derive the exact position-time function, \(x(t)\).
\(x(t) = \int v(t) dt = \int (4t + 3) dt = \frac{4t^2}{2} + 3t + C = 2t^2 + 3t + C\)
Plug in initial condition \(t = 0, x(0) = 10\) to solve for \(C\):
\(x(0) = 2(0)^2 + 3(0) + C = 10 \implies C = 10\)
Therefore, the position function is: \(\mathbf{x(t) = 2t^2 + 3t + 10}\)
PROBLEM 06 Double Integration (Acceleration to Position)
A laboratory cart starts from rest at the coordinate origin at time \(t = 0\text{ s}\) (meaning \(v(0) = 0\text{ m/s}\) and \(x(0) = 0\text{ m}\)). The cart is subjected to a constant acceleration of \(a(t) = +8\text{ m/s}^2\). Integrate the acceleration twice to derive: (A) the velocity function \(v(t)\), and (B) the position function \(x(t)\).
(A) \(v(t) = \int a(t) dt = \int 8 dt = 8t + C_1\). Since \(v(0) = 0 \implies C_1 = 0 \implies \mathbf{v(t) = 8t}\)
(B) \(x(t) = \int v(t) dt = \int 8t dt = \frac{8t^2}{2} + C_2 = 4t^2 + C_2\). Since \(x(0) = 0 \implies C_2 = 0 \implies \mathbf{x(t) = 4t^2}\)
KINEMATICS ANTIDERIVATIVE ANSWER KEY PAGE 2 OF 5
Motion Mechanics // Teacher Answer Key
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PROBLEM 07 Variable Acceleration (Power Rule)
The acceleration of an automated trolley is modeled by the polynomial function: \(a(t) = 12t^2\). At time \(t = 0\text{ s}\), the trolley is moving with an initial velocity of \(v(0) = -3\text{ m/s}\). Find the trolley's exact velocity function \(v(t)\) by integrating \(a(t)\).
\(v(t) = \int a(t) dt = \int 12t^2 dt = \frac{12t^3}{3} + C = 4t^3 + C\)
Plug in initial condition \(t = 0, v(0) = -3\) to solve for \(C\):
\(v(0) = 4(0)^3 + C = -3 \implies C = -3\)
Therefore, the velocity function is: \(\mathbf{v(t) = 4t^3 - 3}\)
PROBLEM 08 Polynomial Velocity to Position
A small drone flies horizontally with a velocity modeled by: \(v(t) = 3t^2 - 4t + 5\). Given that the drone passes coordinate \(x = +12\text{ m}\) at time \(t = 0\text{ s}\), integrate the velocity equation to find the drone's position function \(x(t)\).
\(x(t) = \int v(t) dt = \int (3t^2 - 4t + 5) dt = \frac{3t^3}{3} - \frac{4t^2}{2} + 5t + C = t^3 - 2t^2 + 5t + C\)
Plug in initial condition \(t = 0, x(0) = 12\) to solve for \(C\):
\(x(0) = 0^3 - 2(0)^2 + 5(0) + C = 12 \implies C = 12\)
Therefore, the position function is: \(\mathbf{x(t) = t^3 - 2t^2 + 5t + 12}\)
PROBLEM 09 Double Integration with Multi-term Equations
A physical body is subjected to a time-varying acceleration of \(a(t) = 6t - 4\). At time \(t = 0\text{ s}\), the body starts with an initial velocity of \(v(0) = +8\text{ m/s}\) and an initial position of \(x(0) = +2\text{ m}\). Integrate acceleration twice to derive the position function \(x(t)\).
1. \(v(t) = \int (6t-4)dt = 3t^2-4t+C_1\). Plug \(v(0)=8 \implies C_1=8 \implies v(t)=3t^2-4t+8\)
2. \(x(t) = \int (3t^2-4t+8)dt = t^3-2t^2+8t+C_2\). Plug \(x(0)=2 \implies C_2=2\)
Therefore, the position function is: \(\mathbf{x(t) = t^3 - 2t^2 + 8t + 2}\)
KINEMATICS ANTIDERIVATIVE ANSWER KEY PAGE 3 OF 5
Motion Mechanics // Teacher Answer Key
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PROBLEM 10 The Fundamental Theorem in Kinematics
The velocity of a cargo glider is given by \(v(t) = 3t^2\). Show that calculating its displacement over the interval from \(t = 1\text{ s}\) to \(t = 3\text{ s}\) using the definite integral \(\Delta x = \int_{1}^{3} v(t) dt\) yields the exact same answer as evaluating the change in its position function: \(\Delta x = x(3) - x(1)\).
Definite integral: \(\Delta x = \int_1^3 3t^2 dt = [t^3]_1^3 = 3^3 - 1^3 = 27 - 1 = \mathbf{26\text{ m}}\)
Evaluating position: \(x(t) = \int 3t^2 dt = t^3 + C \implies x(3) - x(1) = (3^3+C) - (1^3+C) = 27 - 1 = \mathbf{26\text{ m}}\).
Both methods yield exactly 26 m, verifying the Fundamental Theorem!
PROBLEM 11 Definite Integral with Constant Accel
A high-speed maglev sled accelerates down a linear track with velocity modeled by \(v(t) = 10t + 5\). Compute the exact physical displacement (\(\Delta x\)) of the maglev sled from time \(t = 2\text{ seconds}\) to \(t = 5\text{ seconds}\) using definite integration.
\(\Delta x = \int_2^5 (10t + 5) dt = [5t^2 + 5t]_2^5\)
Upper bound: \(5(5)^2 + 5(5) = 125 + 25 = 150\)
Lower bound: \(5(2)^2 + 5(2) = 20 + 10 = 30\)
Displacement: \(\Delta x = 150 - 30 = \mathbf{+120\text{ meters}}\)
PROBLEM 12 Variable Speed Definite Integration
A wind turbine blade is decelerating. Its linear velocity profile at the tip is given by the function: \(v(t) = 12 - 3t^2\) (for \(0 \le t \le 2\)). Compute the tip's net displacement (\(\Delta x\)) over this entire interval by setting up and evaluating the definite integral: \(\Delta x = \int_{0}^{2} (12 - 3t^2) dt\).
\(\Delta x = \int_0^2 (12 - 3t^2) dt = [12t - t^3]_0^2\)
Evaluate bounds: \((12(2) - 2^3) - (12(0) - 0^3) = (24 - 8) - (0) = \mathbf{+16\text{ meters}}\)
The blade tip traveled a net distance of 16 meters forward as it slowed down.
KINEMATICS ANTIDERIVATIVE ANSWER KEY PAGE 4 OF 5
Motion Mechanics // Teacher Answer Key
PAGE 5 OF 5
PROBLEM 13 Solving for C with Boundary Conditions
The velocity function of a small probe is \(v(t) = 3t^2 - 6t\). At time \(t = 2\text{ seconds}\), the probe is physically located at coordinate \(x(2) = +10\text{ m}\). Integrate the velocity function and solve for the integration constant \(C\) using this non-zero time boundary condition to find the unique position function \(x(t)\).
\(x(t) = \int v(t) dt = \int (3t^2 - 6t) dt = t^3 - 3t^2 + C\)
Plug in \(t = 2, x(2) = 10 \implies 2^3 - 3(2)^2 + C = 10\)
\(8 - 12 + C = 10 \implies -4 + C = 10 \implies C = 14\)
Therefore, the unique position function is: \(\mathbf{x(t) = t^3 - 3t^2 + 14}\)
PROBLEM 14 Integrating Fractional Power Profiles
An underwater automated vehicle experiences fluid resistance, resulting in a decaying acceleration function modeled by: \(a(t) = t^{-1/2}\) (for \(t \ge 1\)). Knowing that the vehicle's initial velocity is \(v(1) = +5\text{ m/s}\), integrate \(a(t)\) to derive its velocity equation \(v(t)\).
\(v(t) = \int t^{-1/2} dt = \frac{t^{1/2}}{1/2} + C = 2t^{1/2} + C = 2\sqrt{t} + C\)
Plug in \(t = 1, v(1) = 5 \implies 2(1)^{1/2} + C = 5 \implies 2 + C = 5 \implies C = 3\)
Therefore, the velocity function is: \(\mathbf{v(t) = 2\sqrt{t} + 3}\) or \(\mathbf{2t^{1/2} + 3}\)
PROBLEM 15 Synthesis: Physical Constant of Integration
Synthesize: In algebra physics, we memorize the equation \(x(t) = \frac{1}{2}at^2 + v_0 t + x_0\). Explain how this kinematic equation can be derived from first principles using calculus double integration of a constant acceleration \(a(t) = a\), showing specifically how the initial velocity (\(v_0\)) and initial position (\(x_0\)) physically arise as the constants of integration.
1. First integral: \(v(t) = \int a dt = at + C_1\). Let initial velocity \(v(0) = v_0 \implies v(0) = a(0) + C_1 = v_0 \implies C_1 = v_0\). Thus \(v(t) = at + v_0\).
2. Second integral: \(x(t) = \int (at + v_0) dt = \frac{1}{2}at^2 + v_0 t + C_2\). Let initial position \(x(0) = x_0 \implies x(0) = \frac{1}{2}a(0)^2 + v_0(0) + C_2 = x_0 \implies C_2 = x_0\).
This completes the derivation of: \(x(t) = \frac{1}{2}at^2 + v_0 t + x_0\).
KINEMATICS ANTIDERIVATIVE ANSWER KEY PAGE 5 OF 5
PART A (3 Points):
Calculate the velocity function \(v(t)\) and evaluate it at exactly \(t = 2\text{ seconds}\).
PART B (3 Points):
Calculate the acceleration function \(a(t)\) and evaluate it at exactly \(t = 2\text{ seconds}\).
Part A Work Area (Velocity)
Part B Work Area (Acceleration)
MOTION MECHANICS KINEMATICS QUIZ PAGE 2 OF 2
Part B Solutions & Grading
Step 1: Acceleration derivative function
\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 12t + 1) = \mathbf{12t - 12}\text{ m/s}^2\)
Step 2: Evaluate at t = 3s
\(a(3) = 12(3) - 12\)
\(a(3) = 36 - 12 = \mathbf{24.00\text{ m/s}^2}\)
Grading: +1.5 pts correct derivative function, +1.5 pts evaluation with units.
MOTION MECHANICS KINEMATICS QUIZ KEY PAGE 1 OF 2
Teacher Answer Key // Kinematics
SECTION 2 QUIZ KEY
3
A mechanical arm moves on a linear path with position coordinate: \(x(t) = -t^2 + 8t + 12\) (where position \(x\) is in meters, time \(t \ge 0\) is in seconds).
PART A (3 Points):
Calculate the velocity function \(v(t)\) and determine the exact time \(t\) when the arm momentarily stops moving.
PART B (3 Points):
Calculate the acceleration function \(a(t)\) and state whether it changes over time or remains constant.
Part A Solutions & Grading
Step 1: Differentiate to find v(t)
\(v(t) = \frac{dx}{dt} = \mathbf{-2t + 8}\text{ m/s}\)
Step 2: Solve where velocity is zero
\(-2t + 8 = 0 \implies 2t = 8\)
\(t = \mathbf{4.00\text{ seconds}}\)
Grading: +1.5 pts correct derivative function, +1.5 pts correct linear solution with units.
Part B Solutions & Grading
Step 1: Differentiate v(t) to find a(t)
\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(-2t + 8) = \mathbf{-2.00\text{ m/s}^2}\)
Step 2: Analyze acceleration behavior
Since the acceleration function is a constant numerical value and does not contain the variable \(t\), the acceleration is constant and does not change over time.
Grading: +1.5 pts derivative value, +1.5 pts correct qualitative constant analysis.
4
A research capsule accelerates along a test track with a position coordinates given by the equation: \(x(t) = \frac{1}{12}t^4 - t^2 + 8\) (where position \(x\) is in meters, time \(t \ge 0\) is in seconds).
PART A (3 Points):
Calculate the velocity function \(v(t)\) and evaluate it at exactly \(t = 2\text{ seconds}\).
PART B (3 Points):
Calculate the acceleration function \(a(t)\) and evaluate it at exactly \(t = 2\text{ seconds}\).
Part A Solutions & Grading
Step 1: Take derivative to find v(t)
\(v(t) = \frac{4}{12}t^3 - 2t = \mathbf{\frac{1}{3}t^3 - 2t}\text{ m/s}\)
Step 2: Evaluate at t = 2s
\(v(2) = \frac{1}{3}(2)^3 - 2(2) = \frac{8}{3} - 4\)
\(v(2) = 2.67 - 4 = \mathbf{-1.33\text{ m/s}}\)
Grading: +1.5 pts correct velocity derivative function, +1.5 pts evaluation with units.
Part B Solutions & Grading
Step 1: Take derivative to find a(t)
\(a(t) = \frac{d}{dt}\left( \frac{1}{3}t^3 - 2t \right) = \frac{1}{3}(3t^2) - 2\)
\(a(t) = \mathbf{t^2 - 2}\text{ m/s}^2\)
Step 2: Evaluate at t = 2s
\(a(2) = (2)^2 - 2 = 4 - 2 = \mathbf{2.00\text{ m/s}^2}\)
Grading: +1.5 pts correct acceleration derivative, +1.5 pts evaluation with units.
MOTION MECHANICS KINEMATICS QUIZ KEY PAGE 2 OF 2
| -10.0 |
| 2.0 | -5.0 |
| 4.0 | 0.0 (Origin) |
| 6.0 | +5.0 |
| 8.0 | +10.0 |
| 10.0 | +15.0 |
3. Describe how the ball's position is changing. Does it change by equal amounts during each 2.0-second interval?
4. Calculate the constant rate of change of position (the slope) for the entire 10-second trip using any two rows from the table.
5. What is the total change in position (\(\Delta x = x_f - x_i\)) from \(t = 0\text{ s}\) to \(t = 10\text{ s}\)? Show your calculation.
6. Based on the pattern in the table, predict the ball's exact coordinate position at a clock reading of \(t = 12.0\text{ s}\).
Dot diagrams map an object's position at equal intervals of time (e.g., a camera flash recording position every 1.0 second).
Diagram 1: Constant Spacing
t=0s
t=4s
Describe position changes...
Diagram 2: Growing Spacing
t=0s
t=4s
Describe position changes...
Diagram 3: Shrinking Spacing
t=0s
t=4s
Describe position changes...
7. Explain how the spacing between the dots in Diagram 2 reveals how the object's position changes over time compared to Diagram 1.
Unit: Tracking Position and Time Page 2 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name:
Date: Period:
An object undergoes a complex journey along a straight line over a period of 20 seconds. The position-time graph below represents this entire motion divided into five distinct segments labeled A, B, C, D, and E.
t (s) x (m) -15 -10 -5 0 5 10 15 0 4 8 12 16 20 Segment A Segment B Segment C Segment D Segment E
8. Read the coordinate values directly from the graph above and describe the qualitative features of each segment (state starting/ending positions, direction, and whether it is at rest, constant speed, speeding up, or slowing down):
Segment A:
Segment B:
Segment C:
Segment D:
Segment E:
Unit: Tracking Position and Time Page 3 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name:
Date: Period:
Refer back to the mixed linear-curved graph on Page 3. Complete a precise mathematical average slope calculation (\(\text{slope}_{\text{avg}} = \frac{x_f - x_i}{t_f - t_i}\)) for each of the five segments. For curved segments, this represents the average rate of change of position (secant line slope). Show your formula substitution and units.
Segment A Average Slope:
Segment B Average Slope:
Segment C Average Slope:
Segment D Average Slope:
Segment E Average Slope:
9. Calculate the total net change in position (\(\Delta x_{\text{net}} = x_{\text{final}} - x_{\text{initial}}\)) of the object for the entire 20-second trip.
10. Calculate the total cumulative path distance traveled by the object during these 20 seconds. (Add up the absolute sizes of each change in position).
Unit: Tracking Position and Time Page 4 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name:
Date: Period:
Directions: Read the story below about a model rocket flight. Translate each stage of the narrative into a corresponding path on the blank position-time grid. Use a straightedge for straight lines and draw smooth, clear curves for intervals where speed changes!
Stage 1 (0s to 4s): The model rocket is launched vertically from rest at the launchpad (\(x = 0\text{ m}\)). Its engine fires, providing constant thrust and causing the rocket to accelerate upwards, speeding up in the positive direction until it reaches coordinate position \(x = 12\text{ m}\) at clock reading \(t = 4\text{ s}\).
Stage 2 (4s to 8s): The engine runs out of fuel at \(t = 4\text{ s}\). The rocket continues to coast upwards under momentum but is **slowing down in the positive direction** as gravity pulls on it, finally coming to a temporary split-second stop at its peak altitude of \(x = 18\text{ m}\) at clock reading \(t = 8\text{ s}\).
Stage 3 (8s to 12s): From its peak altitude, the rocket starts to fall back toward Earth. Gravity accelerates it downward, causing it to fall faster and faster—**speeding up in the negative direction**—until it reaches coordinate position \(x = 10\text{ m}\) at clock reading \(t = 12\text{ s}\).
Stage 4 (12s to 16s): At \(t = 12\text{ s}\), a recovery parachute deploys. This slows the fall, meaning the rocket continues downward but is **slowing down in the negative direction**, decelerating until it gently lands on the ground at the origin coordinate (\(x = 0\text{ m}\)) at clock reading \(t = 16\text{ s}\).
Conceptual Drawing Grid (Student Work Area) t (s) x (m) 0 5 10 15 20 0 4 8 12 16
Unit: Tracking Position and Time Page 5 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name:
Date: Period:
Problem 1: Moving at a Constant Rate
An object starts at an initial position of \(x_0 = +25\text{ m}\) and moves such that its position changes at a constant rate (slope) of \(-6.5\text{ meters per second}\). Write its motion equation \(x(t)\), and calculate its position at \(t = 12.0\text{ s}\).
Show setup and calculate coordinates here...
Problem 2: Finding Clock Readings
An object's motion is modeled by the equation: \(x(t) = -15 + 4.0t\). Identify its position at \(t = 0\text{ s}\), and calculate the clock reading \(t\) when the object reaches the coordinate \(x = 0\text{ m}\).
Show calculation steps here...
1. On a position-time graph, what does a flat, completely horizontal line tell you about the position of the object as the clock continues to run? Explain.
2. If you are comparing two lines on a position-time graph, and Line 1 is steeper than Line 2, what does this tell you about how their positions are changing over time? Which is changing position more rapidly?
Unit: Tracking Position and Time Page 6 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
An observer records the coordinate positions of a rolling ball at specific clock readings. The data is presented in the table below.
| Clock Reading \(t\) (s) | Position \(x\) (m) |
|---|---|
| 0.0 | -10.0 |
| 2.0 | -5.0 |
| 4.0 | 0.0 (Origin) |
| 6.0 | +5.0 |
| 8.0 | +10.0 |
| 10.0 | +15.0 |
3. Describe how the ball's position is changing. Does it change by equal amounts during each 2.0-second interval?
Yes. During each 2.0s clock interval, the coordinate position increases by exactly +5.0 meters, showing constant forward movement.
4. Calculate the constant rate of change of position (the slope) for the entire 10-second trip using any two rows from the table.
slope = (15.0 - 0.0) / (10.0 - 4.0) = 15.0m / 6.0s = +2.5 m/s
5. What is the total change in position (\(\Delta x = x_f - x_i\)) from \(t = 0\text{ s}\) to \(t = 10\text{ s}\)? Show your calculation.
Δx = +15.0m - (-10.0m) = +25.0 meters
6. Based on the pattern in the table, predict the ball's exact coordinate position at a clock reading of \(t = 12.0\text{ s}\).
x = +15.0m + 5.0m = +20.0 meters
Dot diagrams map an object's position at equal intervals of time (e.g., a camera flash recording position every 1.0 second).
Diagram 1: Constant Spacing
t=0s
t=4s
Constant Position Change
Diagram 2: Growing Spacing
t=0s
t=4s
Increasing Position Change
Diagram 3: Shrinking Spacing
t=0s
t=4s
Decreasing Position Change
7. Explain how the spacing between the dots in Diagram 2 reveals how the object's position changes over time compared to Diagram 1.
The increasing gap between dots in Diagram 2 shows that the object covers a larger distance in each successive second, meaning its position coordinate changes faster and faster compared to the steady, constant spacing of Diagram 1.
Unit: Tracking Position and Time Page 2 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
An object undergoes a complex journey along a straight line over a period of 20 seconds. The position-time graph below represents this entire motion divided into five distinct segments labeled A, B, C, D, and E.
t (s) x (m) -15 -10 -5 0 5 10 15 0 4 8 12 16 20 Segment A Segment B Segment C Segment D Segment E
8. Read the coordinate values directly from the graph above and describe the qualitative features of each segment, noting direction of motion and whether the speed is constant, speeding up, or slowing down:
Segment A:
Moves forward (positive direction) at a constant speed of +3.0 m/s from starting coordinate 0m to 12m over the first 4 seconds (t = 0s to 4s). Graph line is straight.
Segment B:
Object is stationary (at rest) at coordinate position x = 12m from t = 4s to 8s. Graph line is completely horizontal with zero slope.
Segment C:
Moves backward (negative direction), speeding up from coordinate 12m to 6m from t = 8s to 12s. Graph slopes down, getting progressively steeper (curving down).
Segment D:
Moves backward (negative direction), slowing down from coordinate 6m to -6m from t = 12s to 16s. Graph curves downward, flattening out to zero slope at t = 16s.
Segment E:
Moves forward (positive direction), speeding up from coordinate position -6m back to the origin (0m) from t = 16s to 20s. Graph curves upward, getting steeper.
Unit: Tracking Position and Time Page 3 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Refer back to the mixed linear-curved graph on Page 3. Complete a precise mathematical average slope calculation (\(\text{slope}_{\text{avg}} = \frac{x_f - x_i}{t_f - t_i}\)) for each of the five segments. For curved segments, this represents the average rate of change of position (secant line slope). Show your formula substitution and units.
Segment A Average Slope:
slope = (12m - 0m) / (4s - 0s)
slope_avg = +3.0 m/s
Segment B Average Slope:
slope = (12m - 12m) / (8s - 4s)
slope_avg = 0 m/s (stationary)
Segment C Average Slope:
slope = (6m - 12m) / (12s - 8s) = -6m / 4s
slope_avg = -1.5 m/s
Segment D Average Slope:
slope = (-6m - 6m) / (16s - 12s) = -12m / 4s
slope_avg = -3.0 m/s
Segment E Average Slope:
slope = (0m - (-6m)) / (20s - 16s) = +6m / 4s
slope_avg = +1.5 m/s
9. Calculate the total net change in position (\(\Delta x_{\text{net}} = x_{\text{final}} - x_{\text{initial}}\)) of the object for the entire 20-second trip.
Δx = x_final - x_initial
Δx = 0m - 0m = 0.0 meters (ended exactly where it started).
10. Calculate the total cumulative path distance traveled by the object during these 20 seconds. (Add up the absolute sizes of each change in position).
d = |+12m| + |0m| + |-6m| + |-12m| + |+6m|
d = 12 + 0 + 6 + 12 + 6 = 36.0 meters of total path length.
Unit: Tracking Position and Time Page 4 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Narrative Summary: Stage 1: Speeding up forward (curve gets steeper up to x=12m). Stage 2: Slowing down forward (curves to flat at peak x=18m). Stage 3: Speeding up backward (curves steeper down to x=10m). Stage 4: Slowing down backward (curves flatter down to x=0m, soft landing).
Plotted Curved Path (Red Ink Grading Guide) t (s) x (m) 0 5 10 15 20 0 4 8 12 16 Stage 1 Stage 2 Stage 3 Stage 4
Stage 1: Speed Up (+)
Starts flat at (40,140) and bends steeper upward to (105,68).
Stage 2: Slow Down (+)
Starts steep at (105,68) and bends flat at its peak (170,32).
Stage 3: Speed Up (-)
Starts flat at peak (170,32) and bends steeper downward to (235,80).
Stage 4: Slow Down (-)
Starts steep at (235,80) and bends flat at touchdown (300,140).
Unit: Tracking Position and Time Page 5 of 6 Reading-to-Practice Connections
Physics • Position and Time Analysis
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Problem 1: Moving at a Constant Rate
An object starts at an initial position of \(x_0 = +25\text{ m}\) and moves such that its position changes at a constant rate (slope) of \(-6.5\text{ meters per second}\). Write its motion equation \(x(t)\), and calculate its position at \(t = 12.0\text{ s}\).
Equation: x(t) = 25 - 6.5t
At t = 12.0s: x(12) = 25 - 6.5(12) = 25 - 78 = -53.0 meters (53 meters backward from origin).
Problem 2: Finding Clock Readings
An object's motion is modeled by the equation: \(x(t) = -15 + 4.0t\). Identify its position at \(t = 0\text{ s}\), and calculate the clock reading \(t\) when the object reaches the coordinate \(x = 0\text{ m}\).
Position at t = 0s: x(0) = -15.0 meters (starting location).
Reaches origin (x = 0) when: 0 = -15 + 4.0t -> 4.0t = 15 -> t = 3.75 seconds.
1. On a position-time graph, what does a flat, completely horizontal line tell you about the position of the object as the clock continues to run? Explain.
A horizontal line tells us that the coordinate position remains constant as the clock ticks forward. Since the slope is zero, the rate of change of position is zero, indicating that the object is completely stationary (at rest).
2. If you are comparing two lines on a position-time graph, and Line 1 is steeper than Line 2, what does this tell you about how their positions are changing over time? Which is changing position more rapidly?
A steeper line has a larger mathematical slope. This tells us that the position of Object 1 changes by a larger distance in any given time interval compared to Object 2. Therefore, Object 1 is changing position more rapidly (moving faster).
Unit: Tracking Position and Time Page 6 of 6 Reading-to-Practice Connections
Question 5: Multi-Stage Average (Equal Times) 1 Point
A vehicle drives down a long straight road at a constant speed of \(20.0\text{ m/s}\) for exactly \(30.0\text{ seconds}\), and then accelerates to a constant speed of \(40.0\text{ m/s}\) for the next \(30.0\text{ seconds}\). What is its average speed for this entire \(60.0\text{-second}\) trip?
A Average Speed = 26.7 m/s
B Average Speed = 30.0 m/s
C Average Speed = 33.3 m/s
D Average Speed = 35.0 m/s
Question 6: Multi-Stage Average (Equal Distances) 1 Point
An athlete runs exactly \(100\text{ m}\) due East at a constant speed of \(4.00\text{ m/s}\), and immediately runs another \(100\text{ m}\) due East at a constant speed of \(6.00\text{ m/s}\). What is their average speed for the entire \(200\text{-meter}\) run?
A Average Speed = 4.80 m/s
B Average Speed = 5.00 m/s
C Average Speed = 5.20 m/s
D Average Speed = 5.50 m/s
Speed and Velocity MCQ • MCQ Worksheet Page 2 of 7 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name:
Date: Period:
Question 7: Velocity Sign & Vector Averaging 1 Point
A toy train moves West at a constant speed of \(3.0\text{ m/s}\) for \(10.0\text{ s}\), stops for \(5.0\text{ s}\), and then moves East at \(3.0\text{ m/s}\) for \(10.0\text{ s}\). Taking East as the positive coordinate direction, what are the average speed and average velocity vector for this \(25.0\text{-second}\) interval?
A Speed = 2.40 m/s, Velocity = 0 m/s
B Speed = 2.40 m/s, Velocity = +2.40 m/s
C Speed = 3.00 m/s, Velocity = 0 m/s
D Speed = 0 m/s, Velocity = -2.40 m/s
Question 8: Analyzing Tabular Motion Data 1 Point
The coordinate position of an object is recorded at specific clock readings in the table below. Calculate the average velocity vector of the object between clock readings \(t = 2.0\text{ s}\) and \(t = 8.0\text{ s}\).
| Clock Reading \(t\) (s) | 0.0 | 2.0 | 5.0 | 8.0 |
|---|---|---|---|---|
| Coordinate \(x\) (m) | -12.0 | -6.0 | +9.0 | +12.0 |
A \(v_{\text{avg}} = +1.50\text{ m/s}\)
B \(v_{\text{avg}} = +3.00\text{ m/s}\)
C \(v_{\text{avg}} = +2.00\text{ m/s}\)
D \(v_{\text{avg}} = +2.40\text{ m/s}\)
Question 9: Slope Calculation (Positive direction) 1 Point
Determine the instantaneous velocity of the object during the first stage of motion (\(0\text{ s}\) to \(4\text{ s}\)) shown on the position-time graph below.
0m 12m 0s 4s t (s) x (m)
A \(v = +3.00\text{ m/s}\)
B \(v = +12.0\text{ m/s}\)
C \(v = +4.00\text{ m/s}\)
D \(v = +1.50\text{ m/s}\)
Speed and Velocity MCQ • MCQ Worksheet Page 3 of 7 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
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Question 10: Slope of Stationary Segment 1 Point
An object's motion is tracked below. What is the calculated slope of the object during the interval from \(t = 4\text{ s}\) to \(t = 10\text{ s}\)?
15m 4s 10s t (s) x (m)
A Object is moving, slope = +1.50 m/s
B Object is stationary, slope = 15.0 m/s
C Object is stationary, slope = 0 m/s
D Object is reversing, slope = -1.50 m/s
Question 11: Slope Calculation (Negative direction) 1 Point
An object moves backward toward the origin. Determine the object's constant instantaneous velocity vector from the position-time graph below.
18m 0m 5s 11s t (s) x (m)
A v = -3.00 m/s
B v = +1.50 m/s
C v = -1.50 m/s
D v = -1.64 m/s
Question 12: Graphical Comparison of Speeds 1 Point
Two distinct objects (labeled Line 1 and Line 2) are plotted on the same position-time graph below. Which of the following statements is mathematically true regarding their speeds?
Line 1 Line 2 t (s) x (m)
A Line 2 is faster because its line is longer.
B Line 1 is faster because its slope is steeper.
C Line 2 is faster because its angle is wider.
D Both objects travel with equal average speeds.
Speed and Velocity MCQ • MCQ Worksheet Page 4 of 7 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
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Question 13: Curved Path Analysis (Speeding Up Forward) 1 Point
Analyze the curved position-time graph below. Which of the following physical behaviors is depicted?
t (s) x (m)
A Moving backward, slowing down
B Moving forward, speeding up
C Moving forward, slowing down
D Moving backward, speeding up
Question 14: Curved Path Analysis (Slowing Down Forward) 1 Point
Analyze the curved position-time graph below. Which physical description represents this system?
t (s) x (m)
A Moving forward, slowing down
B Moving forward, speeding up
C Moving backward, slowing down
D Moving backward, speeding up
Question 15: Curved Path Analysis (Speeding Up Backward) 1 Point
Identify the motion profile of an object moving according to the following position-time graph:
t (s) x (m)
A Moving forward, speeding up
B Moving backward, slowing down
C Moving backward, speeding up
D Moving forward, slowing down
Speed and Velocity MCQ • MCQ Worksheet Page 5 of 7 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
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Question 16: Curved Path Analysis (Slowing Down Backward) 1 Point
Determine the correct physical behavior plotted on the position-time curve below:
t (s) x (m)
A Moving backward, slowing down
B Moving backward, speeding up
C Moving forward, slowing down
D Moving forward, speeding up
Question 17: Dot-to-Graph Translation 1 Point
Which type of position-time graph slope matches the collinear motion diagram shown below? (Arrows represent velocity vectors).
A A completely flat, horizontal line
B A curving line that gets progressively steeper
C A straight, upward-sloping line
D A straight, downward-sloping line
Question 18: Translating Variable Spacing 1 Point
Identify the motion behavior represented by the following collinear motion diagram where dot spacing increases dynamically over time:
A Moving forward, constant speed
B Moving forward, speeding up
C Moving forward, slowing down
D Moving backward, speeding up
Speed and Velocity MCQ • MCQ Worksheet Page 6 of 7 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
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Date: Period:
Question 19: Translating Multi-Stage Motion 1 Point
A cart moves according to a position-time graph consisting of: (1) a steep rising linear segment, (2) a horizontal flat segment, and (3) a shallow falling linear segment. What represents this entire sequence of motion?
A Fast forward, at rest, slow backward
B Slow forward, at rest, fast backward
C Fast forward, speeding up, slow backward
D Fast backward, at rest, slow forward
Question 20: Colliding Vectors & Meeting Coordinates 1 Point
Car A is stationary at coordinate \(x = 0.0\text{ m}\). At \(t = 0\text{ s}\), it moves East at a constant rate of \(+15.0\text{ m/s}\). At that same moment, Car B is at coordinate \(x = +240\text{ m}\) and travels West at a constant rate of \(-10.0\text{ m/s}\). At what elapsed clock reading \(t\) do they pass each other?
A \(t = 24.0\text{ seconds}\)
B \(t = 16.0\text{ seconds}\)
C \(t = 9.60\text{ seconds}\)
D \(t = 48.0\text{ seconds}\)
Speed and Velocity MCQ • MCQ Worksheet Page 7 of 7 Collinear Physics Mechanics
A lab cart is pushed such that it rolls \(5.0\text{ m}\) North, then slams into a bumper and rebounds \(12.0\text{ m}\) South. Which statement accurately compares the total physical path distance and coordinate shift (displacement)?
A Distance > |Displacement| by 10.0 m
B Distance > |Displacement| by 5.0 m
C Distance > |Displacement| by 12.0 m
D Distance > |Displacement| by 17.0 m
Solution Setup & Physical Reasoning:
• Path distance \(d = d_1 + d_2 = 5.0\text{ m} + 12.0\text{ m} = 17.0\text{ m}\).
• Displacement vector \(\Delta x = +5.0 - 12.0 = -7.0\text{ m}\) (its absolute magnitude \(= 7.0\text{ m}\export \)).
• The path distance exceeds the magnitude of displacement by: \(17.0\text{ m} - 7.0\text{ m} = \mathbf{10.0\text{ m}}\).
Question 4: Round-Trip Theorem 1 Point
A physical system undergoes a complex, multi-stage journey starting at origin \(x = 0\text{ m}\) and eventually returning to origin \(x = 0\text{ m}\). What can be mathematically proven about its scalar distance \(d\) and net displacement vector \(\Delta x\)?
A \(d = 0\text{ m}\) and \(\Delta x = 0\text{ m}\)
B \(d > 0\text{ m}\) and \(\Delta x > 0\text{ m}\)
C \(d > 0\text{ m}\) and \(\Delta x = 0\text{ m}\)
D \(d = 0\text{ m}\) and \(\Delta x > 0\text{ m}\)
Solution Setup & Physical Reasoning:
• Since the final coordinate is equal to the initial coordinate (\(x_f = x_i = 0\text{ m}\)), the net displacement is exactly \(\Delta x = x_f - x_i = \mathbf{0\text{ m}}\).
• Because the object actively moved along a path, its accumulated scalar path distance \(d\) must be strictly positive (\(\mathbf{d > 0\text{ m}}\)).
Speed and Velocity MCQ • MCQ Answer Key Page 2 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 5: Multi-Stage Average (Equal Times) 1 Point
A vehicle drives down a long straight road at a constant speed of \(20.0\text{ m/s}\) for exactly \(30.0\text{ seconds}\), and then accelerates to a constant speed of \(40.0\text{ m/s}\) for the next \(30.0\text{ seconds}\). What is its average speed for this entire \(60.0\text{-second}\) trip?
A Average Speed = 26.7 m/s
B Average Speed = 30.0 m/s
C Average Speed = 33.3 m/s
D Average Speed = 35.0 m/s
Solution Setup & Physical Reasoning:
• Distance 1: \(d_1 = s_1 \cdot t_1 = 20.0\text{ m/s} \times 30.0\text{ s} = 600\text{ m}\).
• Distance 2: \(d_2 = s_2 \cdot t_2 = 40.0\text{ m/s} \times 30.0\text{ s} = 1200\text{ m}\).
• Average speed: \(s_{\text{avg}} = \frac{d_1 + d_2}{t_1 + t_2} = \frac{600 + 1200}{30 + 30} = \frac{1800\text{ m}}{60.0\text{ s}} = \mathbf{30.0\text{ m/s}}\).
• (Note: The average speed equals the simple arithmetic mean \(\frac{20 + 40}{2}\) only because the time intervals are equal).
Question 6: Multi-Stage Average (Equal Distances) 1 Point
An athlete runs exactly \(100\text{ m}\) due East at a constant speed of \(4.00\text{ m/s}\), and immediately runs another \(100\text{ m}\) due East at a constant speed of \(6.00\text{ m/s}\). What is their average speed for the entire \(200\text{-meter}\) run?
A Average Speed = 4.80 m/s
B Average Speed = 5.00 m/s
C Average Speed = 5.20 m/s
D Average Speed = 5.50 m/s
Solution Setup & Physical Reasoning:
• Time 1: \(t_1 = 100 / 4.00 = 25.0\text{ s}\). Time 2: \(t_2 = 100 / 6.00 = 16.67\text{ s}\). Total time \(= 41.67\text{ s}\).
• Average speed: \(s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{200\text{ m}}{41.67\text{ s}} = \mathbf{4.80\text{ m/s}}\).
• (Note: The average is lower than \(5.0\text{ m/s}\) because the runner spent more time traveling at the slower speed of \(4.0\text{ m/s}\)).
Speed and Velocity MCQ • MCQ Answer Key Page 3 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 7: Velocity Sign & Vector Averaging 1 Point
A toy train moves West at a constant speed of \(3.0\text{ m/s}\) for \(10.0\text{ s}\), stops for \(5.0\text{ s}\), and then moves East at \(3.0\text{ m/s}\) for \(10.0\text{ s}\). Taking East as the positive coordinate direction, what are the average speed and average velocity vector for this \(25.0\text{-second}\) interval?
A Speed = 2.40 m/s, Velocity = 0 m/s
B Speed = 2.40 m/s, Velocity = +2.40 m/s
C Speed = 3.00 m/s, Velocity = 0 m/s
D Speed = 0 m/s, Velocity = -2.40 m/s
Solution Setup & Physical Reasoning:
• Stage 1: West displacement \(\Delta x_1 = -3.0 \times 10 = -30\text{ m}\). Distance \(d_1 = 30\text{ m}\).
• Stage 3: East displacement \(\Delta x_3 = +3.0 \times 10 = +30\text{ m}\). Distance \(d_3 = 30\text{ m}\).
• Total path distance \(d = 30 + 0 + 30 = 60\text{ m}\) \(\implies\) average speed \(= 60 / 25.0 = \mathbf{2.40\text{ m/s}}\).
• Net coordinate shift \(\Delta x = -30 + 0 + 30 = 0\text{ m}\) \(\implies\) average velocity \(= \mathbf{0\text{ m/s}}\).
Question 8: Analyzing Tabular Motion Data 1 Point
The coordinate position of an object is recorded at specific clock readings in the table below. Calculate the average velocity vector of the object between clock readings \(t = 2.0\text{ s}\) and \(t = 8.0\text{ s}\).
| Clock Reading \(t\) (s) | 0.0 | 2.0 | 5.0 | 8.0 |
|---|---|---|---|---|
| Coordinate \(x\) (m) | -12.0 | -6.0 | +9.0 | +12.0 |
A \(v_{\text{avg}} = +1.50\text{ m/s}\)
B \(v_{\text{avg}} = +3.00\text{ m/s}\)
C \(v_{\text{avg}} = +2.00\text{ m/s}\)
D \(v_{\text{avg}} = +2.40\text{ m/s}\)
Solution Setup & Physical Reasoning:
• Coordinate at starting clock reading \(t_i = 2.0\text{ s}\) is \(x_i = -6.0\text{ m}\). Coordinate at final reading \(t_f = 8.0\text{ s}\) is \(x_f = +12.0\text{ m}\).
• Net displacement: \(\Delta x = x_f - x_i = +12.0 - (-6.0) = +18.0\text{ m}\). Time interval \(\Delta t = 8.0 - 2.0 = 6.0\text{ s}\).
• Average velocity vector: \(v_{\text{avg}} = \Delta x / \Delta t = +18.0\text{ m} / 6.0\text{ s} = \mathbf{+3.00\text{ m/s}}\).
Speed and Velocity MCQ • MCQ Answer Key Page 4 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 9: Slope Calculation (Positive direction) 1 Point
Determine the instantaneous velocity of the object during the first stage of motion (\(0\text{ s}\) to \(4\text{ s}\)) shown on the position-time graph below.
0m 12m 0s 4s t (s) x (m)
A \(v = +3.00\text{ m/s}\)
B \(v = +12.0\text{ m/s}\)
C \(v = +4.00\text{ m/s}\)
D \(v = +1.50\text{ m/s}\)
Solution Setup & Physical Reasoning:
• The line has a constant positive slope, pointing Eastward (forward). Coordinate goes from \(0\text{ m}\) to \(12.0\text{ m}\).
• slope \(= \text{rise} / \text{run} = (12.0 - 0) / (4.0 - 0) = \mathbf{+3.00\text{ m/s}}\).
Question 10: Slope of Stationary Segment 1 Point
An object's motion is tracked below. What is the calculated slope of the object during the interval from \(t = 4\text{ s}\) to \(t = 10\text{ s}\)?
15m 4s 10s t (s) x (m)
A Object is moving, slope = +1.50 m/s
B Object is stationary, slope = 15.0 m/s
C Object is stationary, slope = 0 m/s
D Object is reversing, slope = -1.50 m/s
Solution Setup & Physical Reasoning:
• During the interval from clock reading \(4\text{ s}\) to \(10\text{ s}\), the coordinate position remains constant at \(x = 15\text{ m}\).
• slope \(= \text{rise} / \text{run} = (15 - 15) / (10 - 4) = 0 / 6 = \mathbf{0\text{ m/s}}\) (representing an object completely at rest).
Speed and Velocity MCQ • MCQ Answer Key Page 5 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 11: Slope Calculation (Negative direction) 1 Point
An object moves backward toward the origin. Determine the object's constant instantaneous velocity vector from the position-time graph below.
18m 0m 5s 11s t (s) x (m)
A v = -3.00 m/s
B v = +1.50 m/s
C v = -1.50 m/s
D v = -1.64 m/s
Solution Setup & Physical Reasoning:
• Initial coordinate \(x_i = 18.0\text{ m}\) at clock reading \(t_i = 5.0\text{ s}\). Final position \(x_f = 0.0\text{ m}\) at \(t_f = 11.0\text{ s}\).
• Rise \(= x_f - x_i = 0.0 - 18.0 = -18.0\text{ m}\). Run \(= t_f - t_i = 11.0 - 5.0 = 6.0\text{ s}\).
• slope \(= \text{rise} / \text{run} = -18.0\text{ m} / 6.0\text{ s} = \mathbf{-3.00\text{ m/s}}\) (indicating Westward motion).
Question 12: Graphical Comparison of Speeds 1 Point
Two distinct objects (labeled Line 1 and Line 2) are plotted on the same position-time graph below. Which of the following statements is mathematically true regarding their speeds?
Line 1 Line 2 t (s) x (m)
A Line 2 is faster because its line is longer.
B Line 1 is faster because its slope is steeper.
C Line 2 is faster because its angle is wider.
D Both objects travel with equal average speeds.
Solution Setup & Physical Reasoning:
• The slope of a position-time graph is mathematically defined as the velocity (\(\Delta x / \Delta t\)).
• Visual steepness indicates speed magnitude. Since Line 1 is steeper than Line 2, the speed of Object 1 is greater.
Speed and Velocity MCQ • MCQ Answer Key Page 6 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 13: Curved Path Analysis (Speeding Up Forward) 1 Point
Analyze the curved position-time graph below. Which of the following physical behaviors is depicted?
t (s) x (m)
A Moving backward, slowing down
B Moving forward, speeding up
C Moving forward, slowing down
D Moving backward, speeding up
Solution Setup & Physical Reasoning:
• The curve slopes upwards, indicating that the position coordinate increases over time (Forward direction).
• The curve starts flat at the bottom and becomes increasingly steep (slanted vertically) as time increments, proving the speed is increasing (speeding up).
Question 14: Curved Path Analysis (Slowing Down Forward) 1 Point
Analyze the curved position-time graph below. Which physical description represents this system?
t (s) x (m)
A Moving forward, slowing down
B Moving forward, speeding up
C Moving backward, slowing down
D Moving backward, speeding up
Solution Setup & Physical Reasoning:
• The curve slopes upwards, indicating forward motion (positive displacement).
• The curve starts highly steep (fast speed) and bends to become horizontal (completely flat, zero speed). Thus, speed is decreasing (slowing down).
Speed and Velocity MCQ • MCQ Answer Key Page 7 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 15: Curved Path Analysis (Speeding Up Backward) 1 Point
Identify the motion profile of an object moving according to the following position-time graph:
t (s) x (m)
A Moving forward, speeding up
B Moving backward, slowing down
C Moving backward, speeding up
D Moving forward, slowing down
Solution Setup & Physical Reasoning:
• The curve slopes downwards, indicating the coordinate decreases over time (Backward direction).
• The curve starts completely horizontal (flat at the peak altitude) and arches steeper downward (more vertical). Thus, speed is increasing (speeding up backward).
Question 16: Curved Path Analysis (Slowing Down Backward) 1 Point
Determine the correct physical behavior plotted on the position-time curve below:
t (s) x (m)
A Moving backward, slowing down
B Moving backward, speeding up
C Moving forward, slowing down
D Moving forward, speeding up
Solution Setup & Physical Reasoning:
• The curve slopes downward, showing backward motion (coordinate decreases).
• The curve starts highly steep (fast speed downward) and curves to become horizontal at the bottom (flat line, stopped). Thus, it is slowing down backward.
Speed and Velocity MCQ • MCQ Answer Key Page 8 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 17: Dot-to-Graph Translation 1 Point
Which type of position-time graph slope matches the collinear motion diagram shown below? (Arrows represent velocity vectors).
A A completely flat, horizontal line
B A curving line that gets progressively steeper
C A straight, upward-sloping line
D A straight, downward-sloping line
Solution Setup & Physical Reasoning:
• The motion diagram exhibits completely uniform spatial intervals between dots at equal time increments, which signifies constant velocity.
• Green velocity vectors point East (positive). Hence, this translates to a straight, upward-sloping line (constant positive slope).
Question 18: Translating Variable Spacing 1 Point
Identify the motion behavior represented by the following collinear motion diagram where dot spacing increases dynamically over time:
A Moving forward, constant speed
B Moving forward, speeding up
C Moving forward, slowing down
D Moving backward, speeding up
Solution Setup & Physical Reasoning:
• The gap between dots expands sequentially, which reveals that the object traverses larger distances in each successive second (speeding up).
• Green velocity vector arrows point Eastwards (positive), indicating forward movement. Option B is correct.
Speed and Velocity MCQ • MCQ Answer Key Page 9 of 10 Collinear Physics Mechanics
Physics • 20 honors multiple choice problems
Name: TEACHER COPY
Date: ANSWER KEY Period: ALL
Question 19: Translating Multi-Stage Motion 1 Point
A cart moves according to a position-time graph consisting of: (1) a steep rising linear segment, (2) a horizontal flat segment, and (3) a shallow falling linear segment. What represents this entire sequence of motion?
A Fast forward, at rest, slow backward
B Slow forward, at rest, fast backward
C Fast forward, speeding up, slow backward
D Fast backward, at rest, slow forward
Solution Setup & Physical Reasoning:
• Stage 1: Steep slope upward = high speed Eastward (fast forward).
• Stage 2: Flat slope = coordinate does not shift over time (at rest).
• Stage 3: Shallow slope downward = low speed Westward (slow backward). Option A fits the description perfectly.
Question 20: Colliding Vectors & Meeting Coordinates 1 Point
Car A is stationary at coordinate \(x = 0.0\text{ m}\). At \(t = 0\text{ s}\), it moves East at a constant rate of \(+15.0\text{ m/s}\). At that same moment, Car B is at coordinate \(x = +240\text{ m}\) and travels West at a constant rate of \(-10.0\text{ m/s}\). At what elapsed clock reading \(t\) do they pass each other?
A \(t = 24.0\text{ seconds}\)
B \(t = 16.0\text{ seconds}\)
C \(t = 9.60\text{ seconds}\)
D \(t = 48.0\text{ seconds}\)
Solution Setup & Physical Reasoning:
• Let East be positive. Car A position: \(x_A(t) = +15.0t\). Car B position: \(x_B(t) = 240.0 - 10.0t\).
• Meeting coordinate occurs when position coordinates match: \(x_A(t) = x_B(t)\).
• \(15.0t = 240.0 - 10.0t \implies 25.0t = 240.0 \implies t = 240.0 / 25.0 = \mathbf{9.60\text{ s}}\).
Speed and Velocity MCQ • MCQ Answer Key Page 10 of 10 Collinear Physics Mechanics
Question 5
A test vehicle travels along a straight track at speed \(v_1 = 15.0\text{ m/s}\) for \(\Delta t_1 = 20.0\text{ s}\), and then at \(v_2 = 35.0\text{ m/s}\) for \(\Delta t_2 = 20.0\text{ s}\). What is its average speed for the entire \(40.0\text{-second}\) interval?
A \(s_{\text{avg}} = 21.0\text{ m/s}\)
B \(s_{\text{avg}} = 25.0\text{ m/s}\)
C \(s_{\text{avg}} = 27.5\text{ m/s}\)
D \(s_{\text{avg}} = 30.0\text{ m/s}\)
Question 6
A runner sprints the first \(d_1 = 120.0\text{ m}\) at speed \(v_1 = 6.0\text{ m/s}\) and the second \(d_2 = 120.0\text{ m}\) at speed \(v_2 = 4.0\text{ m/s}\). What is the runner's average speed over the full \(240.0\text{-meter}\) distance?
A \(s_{\text{avg}} = 4.80\text{ m/s}\)
B \(s_{\text{avg}} = 5.00\text{ m/s}\)
C \(s_{\text{avg}} = 5.20\text{ m/s}\)
D \(s_{\text{avg}} = 4.50\text{ m/s}\)
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 2 OF 12 PART 1: QUESTIONS 4–6
Speed & Velocity Unit Test • Honors Physics Part 1: Multiple Choice
Question 7
A remote-control car moves East (+ direction) at \(4.0\text{ m/s}\) for \(10.0\text{ s}\), stops for \(5.0\text{ s}\), and reverses West (- direction) at \(4.0\text{ m/s}\) for \(10.0\text{ s}\). What are the car's average speed and average velocity for the entire \(25.0\text{-second}\) period?
A Speed = 3.20 m/s, Velocity = 0 m/s
B Speed = 3.20 m/s, Velocity = +3.20 m/s
C Speed = 4.00 m/s, Velocity = 0 m/s
D Speed = 0 m/s, Velocity = 0 m/s
Question 8
The coordinate position of a dynamics cart is recorded at specific clock readings in the table below. Calculate the average velocity vector of the cart between \(t = 2.0\text{ s}\) and \(t = 8.0\text{ s}\).
| Clock Reading \(t\) (s) | 0.0 | 2.0 | 5.0 | 8.0 |
|---|---|---|---|---|
| Position Coordinate \(x\) (m) | -15.0 | -9.0 | +6.0 | +9.0 |
A \(v_{\text{avg}} = +1.50\text{ m/s}\)
B \(v_{\text{avg}} = +3.00\text{ m/s}\)
C \(v_{\text{avg}} = +2.25\text{ m/s}\)
D \(v_{\text{avg}} = -3.00\text{ m/s}\)
Question 9
Determine the constant velocity of the moving object whose position-time graph is shown below:
0 m 16 m 0 s 4.0 s t (s) x (m)
A \(v = +4.00\text{ m/s}\)
B \(v = +16.0\text{ m/s}\)
C \(v = +0.25\text{ m/s}\)
D \(v = +2.00\text{ m/s}\)
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 3 OF 12 PART 1: QUESTIONS 7–9
Speed & Velocity Unit Test • Honors Physics Part 1: Multiple Choice
Question 10
An object's motion is recorded on the graph below. What is its calculated velocity between \(t = 3.0\text{ s}\) and \(t = 9.0\text{ s}\)?
+10 m 3.0 s 9.0 s t (s) x (m)
A Object is moving forward at \(+1.67\text{ m/s}\)
B Object is at rest, velocity = \(0\text{ m/s}\)
C Object has constant velocity of \(+10.0\text{ m/s}\)
D Object is accelerating at \(+9.8\text{ m/s}^2\)
Question 11
An object moves along the straight path plotted below. Determine the magnitude and sign of its constant velocity:
+24 m 0 m 2.0 s 8.0 s t (s) x (m)
A \(v = -4.00\text{ m/s}\)
B \(v = +4.00\text{ m/s}\)
C \(v = -3.00\text{ m/s}\)
D \(v = -24.0\text{ m/s}\)
Question 12
The positions of Cart A and Cart B are plotted below. Which statement is physically valid regarding their speeds?
Cart A Cart B t (s) x (m)
A Cart B is faster because its line is drawn longer.
B Cart A is faster because its rate of change of position (\(\Delta x/\Delta t\)) is steeper.
C Both carts travel with identical constant speeds.
D Cart B is accelerating while Cart A has zero velocity.
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 4 OF 12 PART 1: QUESTIONS 10–12
Speed & Velocity Unit Test • Honors Physics Part 1: Multiple Choice
Question 13
An object moves along a straight track in two distinct stages shown on the position-time graph below: from \(t = 0\) to \(2.0\text{ s}\) (Segment 1), and from \(t = 2.0\) to \(6.0\text{ s}\) (Segment 2). In which segment does the object travel with greater speed, and in which direction is it moving during Segment 2?
+12 m 0 m 0 s 2.0 s 6.0 s t (s) x (m)
A Segment 1 has greater speed; moving in negative direction in Segment 2
B Segment 2 has greater speed; moving in positive direction in Segment 2
C Both segments have equal speed; moving in negative direction in Segment 2
D Segment 1 has greater speed; moving in positive direction in Segment 2
Question 14
An object moves along the \(x\)-axis as plotted below, starting at \(x = +4.0\text{ m}\) at \(t = 0\text{ s}\), reaching \(x = +16.0\text{ m}\) at \(t = 4.0\text{ s}\), and finishing at \(x = +20.0\text{ m}\) at \(t = 8.0\text{ s}\). What is the object's average velocity for the entire \(8.0\text{-second}\) period?
+20 m +16 m +4 m 0 s 4.0 s 8.0 s t (s) x (m)
A \(v_{\text{avg}} = +2.50\text{ m/s}\)
B \(v_{\text{avg}} = +2.00\text{ m/s}\)
C \(v_{\text{avg}} = +3.00\text{ m/s}\)
D \(v_{\text{avg}} = +1.00\text{ m/s}\)
Question 15
The position-time graphs for three independent objects (1, 2, and 3) are shown below. Which object travels with a negative velocity, and which object remains at rest?
Object 1 Object 2 Object 3 t (s) x (m)
A Object 3 has negative velocity; Object 2 is at rest
B Object 1 has negative velocity; Object 3 is at rest
C Object 2 has negative velocity; Object 1 is at rest
D Object 3 has negative velocity; Object 1 is at rest
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 5 OF 12 PART 1: QUESTIONS 13–15
Speed & Velocity Unit Test • Honors Physics Part 1: Multiple Choice
Question 16
Two runners travel along parallel linear lanes: Runner A starts at \(x = 0\text{ m}\) at constant velocity \(v_A = +4.0\text{ m/s}\). Runner B starts with a head start at \(x = +20.0\text{ m}\) at constant velocity \(v_B = +2.0\text{ m/s}\). At what clock reading \(t\) does Runner A catch up to Runner B?
Runner A Runner B Catch-up +20 m t (s) x (m)
A \(t = 10.0\text{ s}\) at position \(x = +40.0\text{ m}\)
B \(t = 5.0\text{ s}\) at position \(x = +20.0\text{ m}\)
C \(t = 15.0\text{ s}\) at position \(x = +60.0\text{ m}\)
D \(t = 20.0\text{ s}\) at position \(x = +80.0\text{ m}\)
Question 17
Which type of position-time graph matches the motion diagram below (equal time spacing between recorded position dots)?
A A straight, linear line with constant positive slope
B A parabolic curve bending progressively steeper
C A flat, horizontal line at zero slope
D A vertical line with undefined infinite slope
Question 18
An object moves toward the right (+ direction) producing the ticker tape motion diagram below:
A Moving forward, constant speed
B Moving forward, increasing speed
C Moving forward, decreasing speed
D Moving backward, increasing speed
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 6 OF 12 PART 1: QUESTIONS 16–18
Speed & Velocity Unit Test • Honors Physics Part 1: Multiple Choice
Question 19
An object moves toward the right (+ direction) producing the ticker tape motion diagram below:
A Moving forward, increasing speed
B Moving forward, decreasing speed
C Moving backward, decreasing speed
D Moving backward, constant velocity
Question 20
A robotic cart executes the three distinct motion segments plotted below. Which description accurately reflects this full motion?
(1) (2) (3) t (s) x (m)
A Fast forward, at rest, slow backward
B Slow forward, at rest, fast backward
C Fast forward, speeding up, slow backward
D Constant speed the entire trip
Question 21
Cart A starts at \(x_A = 0\text{ m}\) moving East at \(+12.0\text{ m/s}\). Cart B starts at \(x_B = +180\text{ m}\) moving West at \(-8.0\text{ m/s}\). At what clock reading \(t\) do they pass each other?
A +12 m/s B -8 m/s 0 m +180 m
A \(t = 9.0\text{ s}\)
B \(t = 15.0\text{ s}\)
C \(t = 45.0\text{ s}\)
D \(t = 12.0\text{ s}\)
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 7 OF 12 PART 1: QUESTIONS 19–21
Speed & Velocity Unit Test • Honors Physics Part 1: Questions 22–25
Question 22
An object moves with constant velocity as shown on the position-time graph below, passing through \((t = 2.0\text{ s}, x = +4.0\text{ m})\) and \((t = 6.0\text{ s}, x = +16.0\text{ m})\). What is the velocity of the object?
+16 m +4 m 2.0 s 6.0 s t (s) x (m)
A \(v = +3.00\text{ m/s}\)
B \(v = +2.50\text{ m/s}\)
C \(v = +10.0\text{ m/s}\)
D \(v = +1.50\text{ m/s}\)
Question 23
An object moves back and forth according to the position-time plot below. What is the net displacement \(\Delta x\) between \(t = 0\text{ s}\) and \(t = 12\text{ s}\)?
+8 m 0 m -6 m 0 s 12 s xf = -2 m
A \(\Delta x = +4.0\text{ m}\)
B \(\Delta x = -8.0\text{ m}\)
C \(\Delta x = +14.0\text{ m}\)
D \(\Delta x = -2.0\text{ m}\)
Question 24
A student walks exactly once around a circular 400.0-meter track, completing the full lap in 80.0 seconds and returning to the exact starting marker. Which pairing accurately characterizes the student's average speed and average velocity for the entire walk?
A Avg Speed = 5.0 m/s, Avg Velocity = 0 m/s
B Avg Speed = 0 m/s, Avg Velocity = 5.0 m/s
C Avg Speed = 5.0 m/s, Avg Velocity = 5.0 m/s
D Avg Speed = 0 m/s, Avg Velocity = 0 m/s
Question 25
The two-tier motion diagram below tracks an object that travels to the right, momentarily stops, and reverses direction toward the left. At which labelled dot does the turning point occur?
1 2 3 4 5 6
A Dot 2
B Dot 4
C Dot 6
D Dot 1
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 8 OF 12 PART 1 CONCLUSION (QUESTIONS 22–25)
Speed & Velocity Unit Test • Honors Physics Section 2: Free Response Questions
PART 2: FREE RESPONSE
6 Complete Problems
Instructions: Show all governing physics equations, complete algebraic substitutions with correct units, and clear final numerical statements with appropriate directional signs and vector labels in the designated columns below.
Word Problem 1
An autonomous exploration rover begins at coordinate waypoint \(x_0 = -24.0\text{ m}\) along a straight scientific test grid. It drives in the positive direction to sample rocks at \(x_1 = +36.0\text{ m}\), and then reverses direction to navigate back to its base station landing zone at \(x_2 = -8.0\text{ m}\).
Required Tasks:
(A) Determine the total scalar physical distance \(d\) traversed across both legs.
(B) Calculate the net displacement vector \(\Delta x\), stating its magnitude and directional sign.
(C) Compare \(d\) and \(|\Delta x|\), explaining conceptually why they are not equal in this journey.
Part (A): Total Distance
Part (B): Net Displacement
Part (C): Comparison & Reason
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 9 OF 12 PART 2: WORD PROBLEM 1
Speed & Velocity Unit Test • Honors Physics Part 2: Free Response
Word Problem 2
A medical courier travels along a straight North-South avenue (taking North as positive):
• Stage 1: Drives \(12.0\text{ km}\) North to Clinic A.
• Stage 2: Reverses and drives \(5.0\text{ km}\) South to Hospital B.
• Stage 3: Continues \(8.0\text{ km}\) North to Laboratory C.
• Stage 4: Drives \(15.0\text{ km}\) South to Central Pharmacy D.
Tasks: (A) Calculate total distance \(d\). (B) Calculate net displacement \(\Delta \vec{x}\). (C) Compute the numerical ratio \(\frac{|\Delta x|}{d}\) and state its physical significance.
Part (A): Distance
Part (B): Displacement
Part (C): Ratio & Meaning
Word Problem 3
+18m 0m -12m 0s 3s 7s 12s
Tasks:
(A) Calculate the instantaneous velocity for each of the three piecewise intervals: [0–3 s], [3–7 s], and [7–12 s].
(B) Determine the total scalar distance and net displacement for the entire \(12.0\text{-second}\) run.
Part (A): Velocities [0-3s], [3-7s], [7-12s]
Part (B): Distance & Net Displacement
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 10 OF 12 PART 2: WORD PROBLEMS 2 & 3
Speed & Velocity Unit Test • Honors Physics Part 2: Free Response
Word Problem 4
+40m +16m 0m 0s 4s 8s
Context & Tasks: A dynamic cart travels along a test track in two distinct constant-velocity stages: Stage 1 covers \(0\text{ to }16.0\text{ m}\) in \(4.0\text{ s}\), and Stage 2 covers \(16.0\text{ to }40.0\text{ m}\) in the next \(4.0\text{ s}\).
(A) Calculate the constant velocity in Stage 1 and Stage 2.
(B) Calculate the average velocity across the full \(8.0\text{-second}\) trip.
(C) Explain why the overall average velocity equals the simple average of the two stage velocities.
Part (A): Stage Velocities
Part (B): Average Velocity
Part (C): Explanation
Word Problem 5
A regional express train departs Station Alpha and travels \(90.0\text{ km}\) East at a constant speed of \(120.0\text{ km/h}\). The train stops at Station Beta for a scheduled layover of \(15.0\text{ minutes}\) (\(0.25\text{ h}\)). It then departs Station Beta and travels \(45.0\text{ km}\) West toward Station Gamma at a constant speed of \(90.0\text{ km/h}\).
Tasks:
(A) Calculate the elapsed travel time for Leg 1, the stop, Leg 2, and the total trip time \(\Delta t_{\text{total}}\) in hours.
(B) Determine the overall average speed \(s_{\text{avg}}\) of the train in \(\text{km/h}\).
(C) Determine the overall average velocity vector \(\vec{v}_{\text{avg}}\) in \(\text{km/h}\) (specify magnitude and direction).
Part (A): Travel Times
Part (B): Average Speed
Part (C): Average Velocity
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 11 OF 12 PART 2: WORD PROBLEMS 4 & 5
Speed & Velocity Unit Test • Honors Physics Part 2 Conclusion & Final Assessment
Word Problem 6
An honors athlete conducts two distinct comparative sprint conditioning protocols:
• Protocol 1 (Equal Distances): Sprints \(100.0\text{ m}\) East at \(8.00\text{ m/s}\), then immediately sprints another \(100.0\text{ m}\) East at \(5.00\text{ m/s}\).
• Protocol 2 (Equal Times): Sprints East at \(8.00\text{ m/s}\) for exactly \(20.0\text{ s}\), then continues East at \(5.00\text{ m/s}\) for exactly \(20.0\text{ s}\).
Tasks:
(A) Calculate the average speed for Protocol 1. (Show full calculation via total distance over total time).
(B) Calculate the average speed for Protocol 2.
(C) Honors Conceptual Derivation: Explain mathematically why the average speed in Protocol 1 is strictly less than the simple arithmetic mean of the two speeds (\(6.50\text{ m/s}\)), referencing time weighting.
Part (A): Protocol 1 (Equal d)
Part (B): Protocol 2 (Equal t)
Part (C): Time-Weighting Proof
Examination Section Completion Summary Teacher Evaluation Record
PART 1 (MCQ) Completed
WP 1–2 (DIST/DISP) Evaluated
WP 3–4 (GRAPHS) Evaluated
WP 5–6 (AVG RATES) Evaluated
SPEED & VELOCITY UNIT TEST • HONORS PHYSICS PAGE 12 OF 12 END OF EXAMINATION
Question 4
A swimmer starts at the edge of a pool (\(x = 0\text{ m}\)), swims \(50.0\text{ m}\) to the opposite end, executes a turn, and swims back to touch the starting wall. Which statement correctly evaluates distance \(d\) and displacement \(\Delta x\)?
A \(d = 100.0\text{ m}\) and \(\Delta x = 0\text{ m}\)
B \(d = 0\text{ m}\) and \(\Delta x = 100.0\text{ m}\)
C \(d = 100.0\text{ m}\) and \(\Delta x = +100.0\text{ m}\)
D \(d = 50.0\text{ m}\) and \(\Delta x = 0\text{ m}\)
Key Solution: Round trips always have zero displacement (\(x_f = x_i \implies \Delta x = 0\)). Distance accumulatively sums both laps: \(50.0 + 50.0 = 100.0\text{ m}\).
Question 5
A test vehicle travels along a straight track at speed \(v_1 = 15.0\text{ m/s}\) for \(\Delta t_1 = 20.0\text{ s}\), and then at \(v_2 = 35.0\text{ m/s}\) for \(\Delta t_2 = 20.0\text{ s}\). What is its average speed for the entire \(40.0\text{-second}\) interval?
A \(s_{\text{avg}} = 21.0\text{ m/s}\)
B \(s_{\text{avg}} = 25.0\text{ m/s}\)
C \(s_{\text{avg}} = 27.5\text{ m/s}\)
D \(s_{\text{avg}} = 30.0\text{ m/s}\)
Key Solution: For equal times, \(s_{\text{avg}} = \frac{v_1 + v_2}{2} = \frac{15.0 + 35.0}{2} = 25.0\text{ m/s}\). (Check: \(d_1 = 300\text{m}, d_2 = 700\text{m} \implies 1000/40 = 25.0\text{ m/s}\)).
Question 6
A runner sprints the first \(d_1 = 120.0\text{ m}\) at speed \(v_1 = 6.0\text{ m/s}\) and the second \(d_2 = 120.0\text{ m}\) at speed \(v_2 = 4.0\text{ m/s}\). What is the runner's average speed over the full \(240.0\text{-meter}\) distance?
A \(s_{\text{avg}} = 4.80\text{ m/s}\)
B \(s_{\text{avg}} = 5.00\text{ m/s}\)
C \(s_{\text{avg}} = 5.20\text{ m/s}\)
D \(s_{\text{avg}} = 4.50\text{ m/s}\)
Key Solution: \(t_1 = 120/6 = 20\text{ s}\); \(t_2 = 120/4 = 30\text{ s}\). \(t_{\text{total}} = 50\text{ s}\). \(s_{\text{avg}} = 240 / 50 = 4.80\text{ m/s}\) (harmonic mean).
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 2 OF 12 SOLUTIONS: QUESTIONS 4–6
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 7–9
Question 7
A remote-control car moves East (+ direction) at \(4.0\text{ m/s}\) for \(10.0\text{ s}\), stops for \(5.0\text{ s}\), and reverses West (- direction) at \(4.0\text{ m/s}\) for \(10.0\text{ s}\). What are the car's average speed and average velocity for the entire \(25.0\text{-second}\) period?
A Speed = 3.20 m/s, Velocity = 0 m/s
B Speed = 3.20 m/s, Velocity = +3.20 m/s
C Speed = 4.00 m/s, Velocity = 0 m/s
D Speed = 0 m/s, Velocity = 0 m/s
Key Solution: Distance \(d = 40 + 0 + 40 = 80.0\text{ m}\). Speed \(= 80.0 / 25.0 = 3.20\text{ m/s}\). Displacement \(\Delta x = +40 - 40 = 0\text{ m}\), so \(v_{\text{avg}} = 0\text{ m/s}\).
Question 8
The coordinate position of a dynamics cart is recorded at specific clock readings in the table below. Calculate the average velocity vector of the cart between \(t = 2.0\text{ s}\) and \(t = 8.0\text{ s}\).
A \(v_{\text{avg}} = +1.50\text{ m/s}\)
B \(v_{\text{avg}} = +3.00\text{ m/s}\)
C \(v_{\text{avg}} = +2.25\text{ m/s}\)
D \(v_{\text{avg}} = -3.00\text{ m/s}\)
Key Solution: \(\Delta x = x(8) - x(2) = +9.0 - (-9.0) = +18.0\text{ m}\). \(\Delta t = 8.0 - 2.0 = 6.0\text{ s}\). \(v_{\text{avg}} = \frac{+18.0}{6.0} = +3.00\text{ m/s}\).
Question 9
Determine the constant velocity of the moving object whose position-time graph is shown below:
A \(v = +4.00\text{ m/s}\)
B \(v = +16.0\text{ m/s}\)
C \(v = +0.25\text{ m/s}\)
D \(v = +2.00\text{ m/s}\)
Key Solution: Slope \(= \frac{\text{Rise}}{\text{Run}} = \frac{16.0 - 0}{4.0 - 0} = +4.00\text{ m/s}\).
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 3 OF 12 SOLUTIONS: QUESTIONS 7–9
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 10–12
Question 10
An object's motion is recorded on the graph below. What is its calculated velocity between \(t = 3.0\text{ s}\) and \(t = 9.0\text{ s}\)?
A Object is moving forward at \(+1.67\text{ m/s}\)
B Object is at rest, velocity = \(0\text{ m/s}\)
C Object has constant velocity of \(+10.0\text{ m/s}\)
D Object is accelerating at \(+9.8\text{ m/s}^2\)
Key Solution: Flat segment on \(x\)-vs-\(t\) means \(\Delta x = 0\text{ m} \implies \text{velocity} = 0\text{ m/s}\) (object remains stationary at \(x = +10.0\text{ m}\)).
Question 11
An object moves along the straight path plotted below. Determine the magnitude and sign of its constant velocity:
A \(v = -4.00\text{ m/s}\)
B \(v = +4.00\text{ m/s}\)
C \(v = -3.00\text{ m/s}\)
D \(v = -24.0\text{ m/s}\)
Key Solution: \(\Delta x = 0 - 24.0 = -24.0\text{ m}\). \(\Delta t = 8.0 - 2.0 = 6.0\text{ s}\). Slope \(= \frac{-24.0}{6.0} = -4.00\text{ m/s}\). Negative sign indicates motion in negative direction.
Question 12
The positions of Cart A and Cart B are plotted below. Which statement is physically valid regarding their speeds?
A Cart B is faster because its line is drawn longer.
B Cart A is faster because its rate of change of position (\(\Delta x/\Delta t\)) is steeper.
C Both carts travel with identical constant speeds.
D Cart B is accelerating while Cart A has zero velocity.
Key Solution: On a position-time graph, speed is represented strictly by the magnitude of the slope (\(|\Delta x/\Delta t|\)). Steeper line = greater speed.
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 4 OF 12 SOLUTIONS: QUESTIONS 10–12
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 13–15 Solutions
Question 13
An object moves along a straight track in two distinct stages shown on the position-time graph below: from \(t = 0\) to \(2.0\text{ s}\) (Segment 1), and from \(t = 2.0\) to \(6.0\text{ s}\) (Segment 2). In which segment does the object travel with greater speed, and in which direction is it moving during Segment 2?
A Segment 1 has greater speed; moving in negative direction in Segment 2
B Segment 2 has greater speed; moving in positive direction in Segment 2
C Both segments have equal speed; moving in negative direction in Segment 2
D Segment 1 has greater speed; moving in positive direction in Segment 2
Key Solution: Speed in Segment 1: \(|\frac{12.0 - 0}{2.0 - 0}| = 6.00\text{ m/s}\). Speed in Segment 2: \(|\frac{0 - 12.0}{6.0 - 2.0}| = |\frac{-12.0}{4.0}| = 3.00\text{ m/s}\). Since \(6.00 > 3.00\), Segment 1 is faster. Negative slope (\(-3.00\text{ m/s}\)) means moving in the negative direction.
Question 14
An object moves along the \(x\)-axis as plotted below, starting at \(x = +4.0\text{ m}\) at \(t = 0\text{ s}\), reaching \(x = +16.0\text{ m}\) at \(t = 4.0\text{ s}\), and finishing at \(x = +20.0\text{ m}\) at \(t = 8.0\text{ s}\). What is the object's average velocity for the entire \(8.0\text{-second}\) period?
A \(v_{\text{avg}} = +2.50\text{ m/s}\)
B \(v_{\text{avg}} = +2.00\text{ m/s}\)
C \(v_{\text{avg}} = +3.00\text{ m/s}\)
D \(v_{\text{avg}} = +1.00\text{ m/s}\)
Key Solution: Average velocity depends only on total displacement over total elapsed time: \(v_{\text{avg}} = \frac{x(8) - x(0)}{\Delta t} = \frac{20.0 - 4.0}{8.0} = \frac{+16.0\text{ m}}{8.0\text{ s}} = \mathbf{+2.00\text{ m/s}}\).
Question 15
The position-time graphs for three independent objects (1, 2, and 3) are shown below. Which object travels with a negative velocity, and which object remains at rest?
A Object 3 has negative velocity; Object 2 is at rest
B Object 1 has negative velocity; Object 3 is at rest
C Object 2 has negative velocity; Object 1 is at rest
D Object 3 has negative velocity; Object 1 is at rest
Key Solution: Object 3 has a downward slope (\(\frac{\Delta x}{\Delta t} < 0 \implies v < 0\)). Object 2 has a horizontal slope (\(\Delta x = 0 \implies v = 0\), at rest). Object 1 has a positive slope (\(v > 0\)).
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 5 OF 12 SOLUTIONS: QUESTIONS 13–15
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 16–18 Solutions
Question 16
Two runners travel along parallel linear lanes: Runner A starts at \(x = 0\text{ m}\) at constant velocity \(v_A = +4.0\text{ m/s}\). Runner B starts with a head start at \(x = +20.0\text{ m}\) at constant velocity \(v_B = +2.0\text{ m/s}\). At what clock reading \(t\) does Runner A catch up to Runner B?
A \(t = 10.0\text{ s}\) at position \(x = +40.0\text{ m}\)
B \(t = 5.0\text{ s}\) at position \(x = +20.0\text{ m}\)
C \(t = 15.0\text{ s}\) at position \(x = +60.0\text{ m}\)
D \(t = 20.0\text{ s}\) at position \(x = +80.0\text{ m}\)
Key Solution: Equate position equations: \(x_A(t) = x_B(t) \implies 4.0t = 20.0 + 2.0t \implies 2.0t = 20.0 \implies \mathbf{t = 10.0\text{ s}}\). Meeting position: \(x = 4.0(10.0) = \mathbf{+40.0\text{ m}}\).
Question 17
Which type of position-time graph matches the motion diagram below (equal time spacing between recorded position dots)?
A A straight, linear line with constant positive slope
B A parabolic curve bending progressively steeper
C A flat, horizontal line at zero slope
D A vertical line with undefined infinite slope
Key Solution: Constant displacement \(\Delta x\) across equal \(\Delta t\) means constant velocity \(\implies\) linear, constant-slope position-time graph.
Question 18
An object moves toward the right (+ direction) producing the ticker tape motion diagram below:
A Moving forward, constant speed
B Moving forward, increasing speed
C Moving forward, decreasing speed
D Moving backward, increasing speed
Key Solution: Dots spread farther apart in equal time intervals \(\implies\) object covers increasing distance per second (increasing speed in the positive direction).
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 6 OF 12 SOLUTIONS: QUESTIONS 16–18
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 19–21
Question 19
An object moves toward the right (+ direction) producing the ticker tape motion diagram below:
A Moving forward, increasing speed
B Moving forward, decreasing speed
C Moving backward, decreasing speed
D Moving backward, constant velocity
Key Solution: Successive dots become closer together in equal time intervals \(\implies\) object covers decreasing distance per unit time (decreasing speed).
Question 20
A robotic cart executes the three distinct motion segments plotted below. Which description accurately reflects this full motion?
A Fast forward, at rest, slow backward
B Slow forward, at rest, fast backward
C Fast forward, speeding up, slow backward
D Constant speed the entire trip
Key Solution: Segment 1 has a steep positive slope (fast forward), Segment 2 has zero slope (at rest), and Segment 3 has a shallow negative slope (slow backward).
Question 21
Cart A starts at \(x_A = 0\text{ m}\) moving East at \(+12.0\text{ m/s}\). Cart B starts at \(x_B = +180\text{ m}\) moving West at \(-8.0\text{ m/s}\). At what clock reading \(t\) do they pass each other?
A \(t = 9.0\text{ s}\)
B \(t = 15.0\text{ s}\)
C \(t = 45.0\text{ s}\)
D \(t = 12.0\text{ s}\)
Key Solution: Equate positions: \(x_A(t) = x_B(t) \implies 0 + 12.0t = 180.0 - 8.0t \implies 20.0t = 180.0 \implies t = 9.0\text{ s}\).
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 7 OF 12 SOLUTIONS: QUESTIONS 19–21
Speed Velocity Test Answer Key • Teacher Edition Part 1: Questions 22–25
Question 22
An object moves with constant velocity as shown on the position-time graph below, passing through \((t = 2.0\text{ s}, x = +4.0\text{ m})\) and \((t = 6.0\text{ s}, x = +16.0\text{ m})\). What is the velocity of the object?
A \(v = +3.00\text{ m/s}\)
B \(v = +2.50\text{ m/s}\)
C \(v = +10.0\text{ m/s}\)
D \(v = +1.50\text{ m/s}\)
Key Solution: Velocity \(= \frac{\Delta x}{\Delta t} = \frac{16.0 - 4.0}{6.0 - 2.0} = \frac{12.0}{4.0} = \mathbf{+3.00\text{ m/s}}\).
Question 23
An object moves back and forth according to the position-time plot below. What is the net displacement \(\Delta x\) between \(t = 0\text{ s}\) and \(t = 12\text{ s}\)?
A \(\Delta x = +4.0\text{ m}\)
B \(\Delta x = -8.0\text{ m}\)
C \(\Delta x = +14.0\text{ m}\)
D \(\Delta x = -2.0\text{ m}\)
Key Solution: \(\Delta x = x(12) - x(0) = (-2.0\text{ m}) - (-6.0\text{ m}) = +4.0\text{ m}\).
Question 24
A student walks exactly once around a circular 400.0-meter track in 80.0 seconds, returning to the start. Which pairing accurately characterizes average speed and average velocity?
A Avg Speed = 5.0 m/s, Avg Velocity = 0 m/s
B Avg Speed = 0 m/s, Avg Velocity = 5.0 m/s
C Avg Speed = 5.0 m/s, Avg Velocity = 5.0 m/s
D Avg Speed = 0 m/s, Avg Velocity = 0 m/s
Key Solution: Distance \(d = 400\text{m} \implies s_{\text{avg}} = 400/80 = 5.0\text{ m/s}\). Displacement \(\Delta x = 0\text{m} \implies \vec{v}_{\text{avg}} = 0\text{ m/s}\).
Question 25
The two-tier motion diagram below tracks an object that travels to the right, momentarily stops, and reverses direction toward the left. At which labelled dot does the turning point occur?
A Dot 2
B Dot 4
C Dot 6
D Dot 1
Key Solution: Dot 4 marks the turning point where velocity reverses from positive to negative.
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 8 OF 12 SOLUTIONS: QUESTIONS 22–25
Speed Velocity Test Answer Key • Teacher Edition Section 2: Free Response Worked Solutions
SECTION 2 SOLUTIONS & RUBRIC
6 Complete Keys
Word Problem 1: Worked Solution
Rover Path: Starts at \(x_0 = -24.0\text{ m}\), moves to \(x_1 = +36.0\text{ m}\), reverses to \(x_2 = -8.0\text{ m}\).
Part (A): Total Distance
• Leg 1: \(d_1 = |+36 - (-24)| = 60.0\text{ m}\)
• Leg 2: \(d_2 = |-8 - 36| = 44.0\text{ m}\)
• Total Distance:
\(d = d_1 + d_2 = 60.0 + 44.0\)
\(\mathbf{d = 104.0\text{ m}}\)
Part (B): Net Displacement
• Formula:
\(\Delta x = x_{\text{final}} - x_{\text{initial}}\)
• Substitution:
\(\Delta x = -8.0\text{ m} - (-24.0\text{ m})\)
\(\mathbf{\Delta x = +16.0\text{ m}}\)
(16.0 m positive)
Part (C): Comparison
• \(d = 104.0\text{ m} \neq |\Delta x| = 16.0\text{ m}\).
Distance is a cumulative scalar of the total physical path. Displacement is a vector between endpoints. Because the rover reversed direction, coordinate gains cancel while path distance continues to accumulate.
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 9 OF 12 SOLUTIONS: WP 1
Speed Velocity Test Answer Key • Teacher Edition Part 2: Word Problems 2 & 3 Solutions
Word Problem 2 Solution
Part (A): Distance
\(d = d_1 + d_2 + d_3 + d_4\)
\(= 12.0 + 5.0 + 8.0 + 15.0\)
\(\mathbf{d = 40.0\text{ km}}\)
Part (B): Displacement
\(\Delta x = (+12) + (-5) + (+8) + (-15)\)
\(\mathbf{\Delta x = 0.0\text{ km}}\)
(Returned to origin)
Part (C): Ratio & Meaning
\(\frac{|\Delta x|}{d} = \frac{0.0\text{ km}}{40.0\text{ km}} = \mathbf{0.00}\).
0% net linear progress despite 40 km driven, demonstrating round-trip cancellation.
Word Problem 3 Solution
Part (A): Piecewise Velocities
• [0–3 s]: \(v_1 = \frac{18.0 - 0.0}{3.0 - 0.0} = \mathbf{+6.0\text{ m/s}}\)
• [3–7 s]: \(v_2 = \frac{18.0 - 18.0}{7.0 - 3.0} = \mathbf{0.0\text{ m/s}}\) (At rest)
• [7–12 s]: \(v_3 = \frac{-12.0 - 18.0}{12.0 - 7.0} = \mathbf{-6.0\text{ m/s}}\)
Part (B): Distance & Displacement
• Total Distance:
\(d = 18.0 + 0.0 + 30.0 = \mathbf{48.0\text{ m}}\)
• Net Displacement:
\(\Delta x = x_f - x_i = -12.0 - 0.0 = \mathbf{-12.0\text{ m}}\)
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 10 OF 12 SOLUTIONS: WP 2 & 3
Speed Velocity Test Answer Key • Teacher Edition Part 2: Word Problems 4 & 5 Solutions
Word Problem 4 Solution
Part (A): Stage Velocities
• Stage 1 [0–4 s]:
\(v_1 = \frac{16.0 - 0.0}{4.0} = \mathbf{+4.00\text{ m/s}}\)
• Stage 2 [4–8 s]:
\(v_2 = \frac{40.0 - 16.0}{4.0} = \mathbf{+6.00\text{ m/s}}\)
Part (B): Average Velocity
\(v_{\text{avg}} = \frac{\Delta x_{\text{total}}}{\Delta t_{\text{total}}}\)
\(= \frac{40.0\text{ m} - 0.0\text{ m}}{8.0\text{ s}}\)
\(\mathbf{v_{\text{avg}} = +5.00\text{ m/s}}\)
Part (C): Explanation
Because the cart spent equal time (\(4.0\text{ s}\) each) in both stages, each velocity contributes equal 50% weight: \(\frac{4.00 + 6.00}{2} = 5.00\text{ m/s}\).
Word Problem 5 Solution
Part (A): Travel Times
• Leg 1: \(\frac{90.0}{120.0} = 0.75\text{ h}\)
• Stop: \(15\text{ min} = 0.25\text{ h}\)
• Leg 2: \(\frac{45.0}{90.0} = 0.50\text{ h}\)
\(\mathbf{t_{\text{total}} = 1.50\text{ h}}\) (90 min)
Part (B): Average Speed
\(s_{\text{avg}} = \frac{d_{\text{total}}}{\Delta t_{\text{total}}}\)
\(= \frac{90.0 + 45.0}{1.50\text{ h}}\)
\(\mathbf{s_{\text{avg}} = 90.0\text{ km/h}}\)
Part (C): Velocity Vector
\(\Delta x = +90 - 45 = +45\text{ km}\)
\(\vec{v}_{\text{avg}} = \frac{+45.0\text{ km}}{1.50\text{ h}}\)
\(\mathbf{= +30.0\text{ km/h [East]}}\)
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 11 OF 12 SOLUTIONS: WP 4 & 5
Speed Velocity Test Answer Key • Teacher Edition Part 2: Word Problem 6 & Final Solutions
Word Problem 6 Solution
Part (A): Protocol 1 (Equal d)
• \(t_1 = \frac{100.0}{8.00} = 12.50\text{ s}\)
• \(t_2 = \frac{100.0}{5.00} = 20.00\text{ s}\)
• \(t_{\text{total}} = 32.50\text{ s}\)
• \(s_{\text{avg}} = \frac{200.0\text{ m}}{32.50\text{ s}}\)
\(\mathbf{= 6.15\text{ m/s}}\)
(Harmonic mean)
Part (B): Protocol 2 (Equal t)
• \(d_1 = (8.00)(20) = 160.0\text{ m}\)
• \(d_2 = (5.00)(20) = 100.0\text{ m}\)
• \(d_{\text{total}} = 260.0\text{ m}\)
• \(s_{\text{avg}} = \frac{260.0\text{ m}}{40.0\text{ s}}\)
\(\mathbf{= 6.50\text{ m/s}}\)
(Arithmetic mean)
Part (C): Time Weighting
Average speed is time-weighted: \(s = \frac{\sum v t}{\sum t}\).
• In Protocol 1, covering equal distance at lower speed requires more time (\(20\text{s} > 12.5\text{s}\)), pulling the average closer to \(5\text{ m/s}\) (\(6.15 < 6.50\)).
• In Protocol 2, times are exactly equal (20s each), giving each speed 50% weight.
Teacher Master Assessment Record ALL PROBLEMS WORKED
PART 1 25 MCQs Completed
WP 1–2 Distance / Displacement
WP 3–4 Position-Time Graphs
WP 5–6 Average Speed & Velocity
SPEED VELOCITY TEST ANSWER KEY • TEACHER EDITION PAGE 12 OF 12 END OF ANSWER KEY
6
An object starts at position coordinate \(x_i = -15.0\text{ m}\) at clock reading \(t = 0\text{ s}\), and reaches \(x_f = -5.0\text{ m}\) at \(t = 2.0\text{ s}\). Which statement accurately characterizes its position and displacement?
(A) Position is negative, but displacement is positive (\(\Delta x = +10.0\text{ m}\))
(B) Both the position and displacement are negative (\(\Delta x = -10.0\text{ m}\))
(C) Displacement is \(-20.0\text{ m}\) because it moved further in \(-x\)
(D) Position is positive, but displacement is negative (\(\Delta x = -10.0\text{ m}\))
7
A runner moves \(+40\text{ m}\) east in \(10\text{ s}\), then reverses direction and walks \(-10\text{ m}\) west in \(10\text{ s}\). What are the runner's average speed and average velocity for the entire \(20\text{ s}\) trip?
(A) Average speed = \(1.5\text{ m/s}\), Average velocity = \(+2.5\text{ m/s}\)
(B) Average speed = \(2.5\text{ m/s}\), Average velocity = \(+1.5\text{ m/s}\)
(C) Average speed = \(2.5\text{ m/s}\), Average velocity = \(+2.5\text{ m/s}\)
(D) Average speed = \(5.0\text{ m/s}\), Average velocity = \(+3.0\text{ m/s}\)
8
On a position vs. time graph, the linear motion lines of two runners intersect at the coordinate point \((t = 6.0\text{ s}, x = +18\text{ m})\). What does this intersection point physically signify?
(A) Both runners have identical speeds at \(t = 6.0\text{ s}\)
(B) Both runners have covered identical total cumulative distances
(C) Both runners are at the exact same location along the track at \(t = 6.0\text{ s}\)
(D) Both runners have come to a complete stop at \(t = 6.0\text{ s}\)
9
A section of an object's position vs. time graph curves upward like a parabola (concave upward, becoming steeper over time). What does this curvature indicate about the object's motion?
(A) The object is moving with constant, unchanging velocity
(B) The object is speeding up in the positive direction as slope increases
(C) The object is slowing down because position coordinate is increasing
(D) The object has stopped moving completely
10
An automated cart travels along a 1-D coordinate line. At what graphical feature on a curved position-time (\(x-t\)) graph does the cart momentarily stop and reverse its direction of travel?
(A) Where the graph crosses the horizontal time axis (\(x = 0\text{ m}\))
(B) At a local peak or trough where the tangent slope momentarily equals zero
(C) Along any straight line segment with a constant non-zero slope
(D) At the initial vertical intercept (\(t = 0\text{ s}\))
Reading Mechanics Assignment Quiz Page 2 of 4 Part I: Multiple Choice Questions 6–10
Honors Physics // Motion Mechanics
Name:
Date: Period:
Show all formulas, algebraic substitutions, and appropriate units clearly in the work areas.
Questions 11–15
11
A toy cart rolls along a straight track calibrated in meters. At clock reading \(t_i = 2.0\text{ s}\), the cart is at position coordinate \(x_i = +8.0\text{ m}\). At a later clock reading \(t_f = 7.0\text{ s}\), the cart reaches position coordinate \(x_f = -12.0\text{ m}\).
Part A: Time Interval & Displacement
Calculate the elapsed time interval (\(\Delta t\)) and the object's net change in position (\(\Delta x\)). Include signs and units.
Part B: Conceptual Distinction
In your own words, explain the fundamental difference between a clock reading (\(t\)) and a time interval (\(\Delta t\)).
12
An observer records the coordinate positions of a glider on an air track at regular time intervals:
| Clock Reading \(t\) (s) | Position \(x\) (m) |
|---|---|
| 0.0 | +18.0 |
| 3.0 | +12.0 |
| 6.0 | +6.0 |
| 9.0 | 0.0 (Origin) |
| 12.0 | -6.0 |
Part A: Constant Rate of Change (Slope)
Calculate the slope (\(\frac{\Delta x}{\Delta t}\)) between \(t = 0\text{ s}\) and \(t = 6.0\text{ s}\). Show setup and units.
Part B: Extrapolation
Based on the table's pattern, determine the glider's position at clock reading \(t = 15.0\text{ s}\).
13
A stroboscopic camera flashes once every second (\(\Delta t = 1.0\text{ s}\)), recording the positions of two carts moving to the right:
Cart 1:
t=0s
t=4s
Cart 2:
t=0s
t=4s
Part A: Type of Motion
Compare the speed of Cart 1 and Cart 2. State whether each cart is moving at constant speed, speeding up, or slowing down.
Part B: Evidence from Dot Spacing
Explain how the changing distance between dots in Cart 2 provides physical proof of acceleration over equal time intervals.
Reading Mechanics Assignment Quiz Page 3 of 4 Part II: Problems 11–13 (Vocabulary, Tables & Dot Diagrams)
Honors Physics // Motion Mechanics
Name:
Date: Period:
14
The position-time graph below represents the one-dimensional motion of an automated shuttle over a 14-second interval with three distinct stages labeled A, B, and C.
t (s) x (m) -10 -5 0 +5 +10 0 4 8 11 14 Stage A Stage B Stage C
Part A: Calculate Stage Slopes
Calculate the slope (\(\frac{\Delta x}{\Delta t}\)) for Stage A and Stage C. Include proper signs and units.
Part B: Interpreting Horizontal Lines
What is the object doing during Stage B? What is the slope of the line, and what does it tell you about position?
Part C: Total Net Displacement
Calculate total net displacement (\(\Delta x_{\text{net}} = x_f - x_i\)) for the 14-second trip.
Part D: Total Cumulative Distance
Calculate total distance traveled by adding the absolute path lengths traveled in each stage.
15
A motorized rover moves along a straight exploration track. Its position as a function of time is modeled by the linear equation: \(x(t) = -24.0 + 6.0t\) (where position \(x\) is in meters and time \(t \ge 0\) is in seconds).
Part A: Initial State & Rate
Identify the rover's initial position \(x_0\) at \(t = 0\text{ s}\) and state its constant rate of change of position (slope).
Part B: Crossing the Origin
Calculate the exact clock reading \(t\) when the rover reaches the origin (\(x = 0.0\text{ m}\)). Show your algebraic steps.
Part C: Coordinate Evaluation
Calculate the rover's exact coordinate position at clock reading \(t = 8.5\text{ s}\). Show setup and include units.
Reading Mechanics Assignment Quiz Page 4 of 4 Part II: Problems 14–15 (Position-Time Graphs & Algebraic Modeling)
(A) \(t = 3.0\text{ s}\)
(B) \(t = 5.0\text{ s}\)
(C) \(t = 12.0\text{ s}\)
(D) \(t = 15.0\text{ s}\)
Key Rationale: \(0 = 15.0 - 3.0t \implies 3.0t = 15.0 \implies t = \frac{15.0}{3.0} = 5.0\text{ s}\).
Reading Mechanics Assignment Quiz • Teacher Answer Key Page 1 of 4 Part I: Multiple Choice Questions 1–5
Reading Mechanics Quiz Key // Part I (Continued)
Questions 6–10 (10 pts)
2 Points Each
6
An object starts at position coordinate \(x_i = -15.0\text{ m}\) at clock reading \(t = 0\text{ s}\), and reaches \(x_f = -5.0\text{ m}\) at \(t = 2.0\text{ s}\). Which statement accurately characterizes its position and displacement?
(A) Position is negative, displacement is positive (\(\Delta x = +10.0\text{ m}\))
(B) Both the position and displacement are negative (\(\Delta x = -10.0\text{ m}\))
(C) Displacement is \(-20.0\text{ m}\) because it moved further in \(-x\)
(D) Position is positive, displacement is negative (\(\Delta x = -10.0\text{ m}\))
Key Rationale: \(\Delta x = x_f - x_i = -5.0\text{ m} - (-15.0\text{ m}) = +10.0\text{ m}\). Moving toward zero is positive.
7
A runner moves \(+40\text{ m}\) east in \(10\text{ s}\), then reverses direction and walks \(-10\text{ m}\) west in \(10\text{ s}\). What are the runner's average speed and average velocity for the entire \(20\text{ s}\) trip?
(A) Average speed = \(1.5\text{ m/s}\), Average velocity = \(+2.5\text{ m/s}\)
(B) Average speed = \(2.5\text{ m/s}\), Average velocity = \(+1.5\text{ m/s}\)
(C) Average speed = \(2.5\text{ m/s}\), Average velocity = \(+2.5\text{ m/s}\)
(D) Average speed = \(5.0\text{ m/s}\), Average velocity = \(+3.0\text{ m/s}\)
Key Rationale: Distance \(= 40+10 = 50\text{ m} \implies 50/20 = 2.5\text{ m/s}\). Displacement \(= +40-10 = +30\text{ m} \implies +30/20 = +1.5\text{ m/s}\).
8
On a position vs. time graph, the linear motion lines of two runners intersect at the coordinate point \((t = 6.0\text{ s}, x = +18\text{ m})\). What does this intersection point physically signify?
(A) Both runners have identical speeds at \(t = 6.0\text{ s}\)
(B) Both runners have covered identical total cumulative distances
(C) Both runners are at the exact same location along the track at \(t = 6.0\text{ s}\)
(D) Both runners have come to a complete stop at \(t = 6.0\text{ s}\)
Key Rationale: A graph of \(x\) vs \(t\) shows position. An intersection means sharing the exact same coordinate \(x\) at time \(t\).
9
A section of an object's position vs. time graph curves upward like a parabola (concave upward, becoming steeper over time). What does this curvature indicate about the object's motion?
(A) The object is moving with constant, unchanging velocity
(B) The object is speeding up in the positive direction as slope increases
(C) The object is slowing down because position coordinate is increasing
(D) The object has stopped moving completely
Key Rationale: The slope of the tangent line represents velocity. Becoming steeper upwards means velocity is increasing.
10
An automated cart travels along a 1-D coordinate line. At what graphical feature on a curved position-time (\(x-t\)) graph does the cart momentarily stop and reverse its direction of travel?
(A) Where the graph crosses the horizontal time axis (\(x = 0\text{ m}\))
(B) At a local peak or trough where the tangent slope momentarily equals zero
(C) Along any straight line segment with a constant non-zero slope
(D) At the initial vertical intercept (\(t = 0\text{ s}\))
Key Rationale: At peaks and troughs, the slope transitions from positive to negative (or vice versa), passing through zero instantaneous velocity.
Reading Mechanics Assignment Quiz • Teacher Answer Key Page 2 of 4 Part I: Multiple Choice Questions 6–10
Teacher Resource // Full Answer Key
TEACHER COPY
Total Points: 60
Worked mathematical setups, units, and conceptual justifications are pre-filled below in red ink.
20 Points Total
11
A toy cart rolls along a straight track calibrated in meters. At clock reading \(t_i = 2.0\text{ s}\), the cart is at position coordinate \(x_i = +8.0\text{ m}\). At a later clock reading \(t_f = 7.0\text{ s}\), the cart reaches position coordinate \(x_f = -12.0\text{ m}\).
Part A: Time Interval & Displacement
Calculate the elapsed time interval (\(\Delta t\)) and the object's net change in position (\(\Delta x\)). Include signs and units.
SOLUTION (3 pts): \(\Delta t = t_f - t_i = 7.0\text{ s} - 2.0\text{ s} = \mathbf{5.0\text{ s}}\)
\(\Delta x = x_f - x_i = -12.0\text{ m} - (+8.0\text{ m}) = \mathbf{-20.0\text{ m}}\)
Part B: Conceptual Distinction
In your own words, explain the fundamental difference between a clock reading (\(t\)) and a time interval (\(\Delta t\)).
SOLUTION (3 pts): A clock reading (\(t\)) is a single specific instant/timestamp in time (e.g., \(t = 2.0\text{ s}\)), whereas a time interval (\(\Delta t\)) is the duration or elapsed time between two clock readings (\(\Delta t = t_f - t_i\)).
12
An observer records the coordinate positions of a glider on an air track at regular time intervals:
| Clock Reading \(t\) (s) | Position \(x\) (m) |
|---|---|
| 0.0 | +18.0 |
| 3.0 | +12.0 |
| 6.0 | +6.0 |
| 9.0 | 0.0 (Origin) |
| 12.0 | -6.0 |
Part A: Constant Rate of Change (Slope)
Calculate the slope (\(\frac{\Delta x}{\Delta t}\)) between \(t = 0\text{ s}\) and \(t = 6.0\text{ s}\). Show setup and units.
\(\text{slope} = \frac{x(6) - x(0)}{6.0 - 0.0} = \frac{+6.0\text{ m} - (+18.0\text{ m})}{6.0\text{ s}} = \frac{-12.0\text{ m}}{6.0\text{ s}} = \mathbf{-2.0\text{ m/s}}\) (4 pts)
Part B: Extrapolation
Based on the table's pattern, determine the glider's position at clock reading \(t = 15.0\text{ s}\).
\(x(15) = x(12) + (-2.0\text{ m/s})(3.0\text{ s}) = -6.0\text{ m} - 6.0\text{ m} = \mathbf{-12.0\text{ m}}\) (3 pts)
13
A stroboscopic camera flashes once every second (\(\Delta t = 1.0\text{ s}\)), recording the positions of two carts moving to the right:
Cart 1:
t=0s
t=4s
Cart 2:
t=0s
t=4s
Part A: Type of Motion
Compare the speed of Cart 1 and Cart 2. State whether each cart is moving at constant speed, speeding up, or slowing down.
Cart 1: Constant speed (equal dot spacing).
Cart 2: Speeding up / accelerating (increasing dot spacing). (3 pts)
Part B: Evidence from Dot Spacing
Explain how the changing distance between dots in Cart 2 provides physical proof of acceleration over equal time intervals.
Since \(\Delta t = 1.0\text{ s}\) is constant, larger distances between consecutive dots indicate progressively greater displacement \(\Delta x\) each second (\(v = \frac{\Delta x}{\Delta t}\) is increasing). (4 pts)
Reading Mechanics Assignment Quiz • Teacher Answer Key Page 3 of 4 Part II: Problems 11–13 (Vocabulary, Tables & Dot Diagrams)
Teacher Resource // Full Answer Key
TEACHER COPY
Total Points: 60
14
The position-time graph below represents the one-dimensional motion of an automated shuttle over a 14-second interval with three distinct stages labeled A, B, and C.
t (s) x (m) -10 -5 0 +5 +10 0 4 8 11 14 Stage A Stage B Stage C
Part A: Calculate Stage Slopes
Calculate the slope (\(\frac{\Delta x}{\Delta t}\)) for Stage A and Stage C. Include proper signs and units.
\(\text{Slope}_A = \frac{+10 - (-10)\text{ m}}{4.0 - 0.0\text{ s}} = \frac{+20\text{ m}}{4\text{ s}} = \mathbf{+5.0\text{ m/s}}\)
\(\text{Slope}_C = \frac{-2.0 - (+10)\text{ m}}{14.0 - 8.0\text{ s}} = \frac{-12\text{ m}}{6\text{ s}} = \mathbf{-2.0\text{ m/s}}\) (3 pts)
Part B: Interpreting Horizontal Lines
What is the object doing during Stage B? What is the slope of the line, and what does it tell you about position?
The shuttle is at rest / stationary at \(x = +10.0\text{ m}\). Slope is \(0\text{ m/s}\); flat lines mean position is unchanged over time. (2 pts)
Part C: Total Net Displacement
Calculate total net displacement (\(\Delta x_{\text{net}} = x_f - x_i\)) for the 14-second trip.
\(\Delta x_{\text{net}} = x(14) - x(0) = -2.0\text{ m} - (-10.0\text{ m}) = \mathbf{+8.0\text{ m}}\) (2.5 pts)
Part D: Total Cumulative Distance
Calculate total distance traveled by adding the absolute path lengths traveled in each stage.
\(d = |\Delta x_A| + |\Delta x_B| + |\Delta x_C| = 20.0 + 0 + 12.0 = \mathbf{32.0\text{ m}}\) (2.5 pts)
15
A motorized rover moves along a straight exploration track. Its position as a function of time is modeled by the linear equation: \(x(t) = -24.0 + 6.0t\) (where position \(x\) is in meters and time \(t \ge 0\) is in seconds).
Part A: Initial State & Rate
Identify the rover's initial position \(x_0\) at \(t = 0\text{ s}\) and state its constant rate of change of position (slope).
SOLUTION (3 pts): • Initial position (\(t=0\)):
\(x_0 = \mathbf{-24.0\text{ m}}\)
• Rate of change (slope):
\(v = \mathbf{+6.0\text{ m/s}}\)
Part B: Crossing the Origin
Calculate the exact clock reading \(t\) when the rover reaches the origin (\(x = 0.0\text{ m}\)). Show your algebraic steps.
SOLUTION (3.5 pts): Set \(x(t) = 0.0\text{ m}\):
\(0 = -24.0 + 6.0t\)
\(6.0t = 24.0\)
\(t = \frac{24.0}{6.0} = \mathbf{4.0\text{ s}}\)
Part C: Coordinate Evaluation
Calculate the rover's exact coordinate position at clock reading \(t = 8.5\text{ s}\). Show setup and include units.
SOLUTION (3.5 pts): Substitute \(t = 8.5\text{ s}\):
\(x(8.5) = -24.0 + 6.0(8.5)\)
\(x(8.5) = -24.0 + 51.0\)
\(x(8.5) = \mathbf{+27.0\text{ m}}\)
Reading Mechanics Assignment Quiz • Teacher Answer Key Page 4 of 4 Part II: Problems 14–15 (Position-Time Graphs & Algebraic Modeling)
