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
PROBLEM 06 Two-Stage Trip: Equal Times (2 Points)
A sports car travels along a straight highway at a constant speed of \(30.0\text{ m/s}\) for exactly \(15.0\text{ s}\), and then accelerates to a constant speed of \(45.0\text{ m/s}\) for the next \(15.0\text{ s}\). (A) Calculate the overall average speed of the car. (B) Explain why the average speed in this specific case is exactly equal to the arithmetic average of the two speeds.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 3 OF 8
Honors Physics // Motion Unit
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PROBLEM 07 Two-Stage Trip: Equal Distances (3 Points)
A swimmer completes a \(100.0\text{-meter}\) race in a \(50.0\text{-meter}\) pool (one lap down and back). On the way down, they swim at a steady speed of \(2.00\text{ m/s}\). On the return lap, they tire and swim at a steady speed of \(1.00\text{ m/s}\). Calculate: (A) the average speed for the entire race, and (B) explain conceptually why this is NOT equal to the simple arithmetic average of \(1.50\text{ m/s}\).
PROBLEM 08 Sprinting Cheetah & Dimensional Analysis (2 Points)
A cheetah sprints a distance of \(150.0\text{ m}\) in exactly \(5.00\text{ s}\) along a straight vector. (A) Find its average speed in \(\text{m/s}\). (B) Convert this speed to miles per hour (\(\text{mph}\)) using dimensional analysis, showing your conversion steps. (Take \(1.00\text{ m/s} = 2.237\text{ mph}\)).
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 4 OF 8
Honors Physics // Motion Unit
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PROBLEM 09 Interpreting Horizontal Motion (2 Points)
The motion diagram below represents a skateboarder moving along a horizontal path, captured at equal time intervals (\(\Delta t = 1.0\text{ s}\)).
Start
End
t=0st=1st=2st=3st=4st=5st=6s
State the exact time interval(s) during which the skateboarder is: (A) decelerating, (B) traveling at a constant speed, and (C) accelerating. Justify each using dot spacing.
PROBLEM 10 Position vs. Time Graph: Slope & Motion (3 Points)
A remote-controlled car runs on a linear track. The graph below displays its position \(x\) (meters) as a function of time \(t\) (seconds).
20 15 10 5 0 0 2 4 6 8 10 x (m) t (s)
Tasks:
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 5 OF 8
Honors Physics // Motion Unit
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PROBLEM 11 Ramp Roll and Turning Point (3 Points)
A small steel marble is rolled up a smooth incline. It slows down to a stop, turns around at its peak, and rolls back down the ramp, speeding up. Sketch a single, clean motion diagram for the complete trip. Explain the direction of the velocity vector compared to the acceleration vector during: (A) the upward trip, (B) the instant of stopping at the peak, and (C) the downward trip.
PROBLEM 12 Two-Object Meeting Point (3 Points)
Car A is stationary at coordinate \(x = 0.0\text{ m}\). At \(t = 0.0\text{ s}\), it begins to move East at a constant velocity of \(+12.0\text{ m/s}\). At that same split-second, Car B is at coordinate \(x = +200.0\text{ m}\) and is traveling West toward Car A at a constant velocity of \(-8.0\text{ m/s}\). Calculate: (A) the linear position-time functions \(x_A(t)\) and \(x_B(t)\) for both cars, and (B) the exact elapsed time \(t\) and coordinate position \(x\) where the two cars pass each other.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 6 OF 8
Honors Physics // Motion Unit
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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)
GRAPH DETAILS: The graph above shows the position \(x\) over a total time interval of \(12.0\text{ seconds}\) for a toy cart traveling along a straight track. It consists of three distinct linear intervals:
PROBLEM 13 Segment Velocity Calculations, Distance & Displacement (6 Points)
Using the cart's graph from above, calculate: (A) the cart's constant instantaneous velocity during Intervals A, B, and C (show slope equations with units); (B) the total physical distance covered and net displacement (\(\Delta x\)) for the full \(12.0\text{-second}\) trip; and (C) the overall average speed and average velocity vector of the cart for the entire trip.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 7 OF 8
Honors Physics // Motion Unit
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PROBLEM 14 Curved Graph Slope Tangents (3 Points)
Suppose a position-time graph displays a smooth curve starting at coordinate \(x = 0\text{ m}\) at time \(t = 0\text{ s}\). The slope of the curve becomes progressively steeper as time increases (concave upward). (A) Explain what this upward curvature implies about the speed of the object. (B) Describe step-by-step how an honors physics student can visually estimate the cart's instantaneous velocity at exactly \(t = 3.0\text{ s}\) using a straightedge (ruler) on this curved plot without using calculus limits.
PROBLEM 15 Graph-to-Motion Translation Challenge (3 Points)
Synthesize: Translate the entire \(12.0\text{-second}\) position-time graph from Page 7 into a single, cohesive horizontal motion diagram. In the box below: (1) draw the series of dots representing the cart's positions at each of the intervals, (2) include velocity vectors representing speed and direction, and (3) include labels to mark the turning points or states of rest.
Student Motion Diagram Drawing Area
TRANSLATION GUIDE:
Be precise with vector sizes and arrow directions.
SPEED AND VELOCITY PROBLEMS // HONORS PHYSICS PAGE 8 OF 8
12 9 6 3 0 0 2 4 6 8 x (m) t (s)
SOLVED SLOPES:
Step-by-step Solution:
(A) Slopes (Velocity calculations):
• Interval 1 (\(0-2\text{ s}\)): \(v_1 = \frac{6.0\text{ m} - 0.0\text{ m}}{2.0\text{ s} - 0.0\text{ s}} = \mathbf{+3.00\text{ m/s}}\) (or \(3.00\text{ m/s East}\)).
• Interval 2 (\(2-6\text{ s}\)): \(v_2 = \frac{6.0\text{ m} - 6.0\text{ m}}{6.0\text{ s} - 2.0\text{ s}} = \mathbf{0.00\text{ m/s}}\) (at rest / stationary).
• Interval 3 (\(6-8\text{ s}\)): \(v_3 = \frac{12.0\text{ m} - 6.0\text{ m}}{8.0\text{ s} - 6.0\text{ s}} = \frac{6.0}{2.0} = \mathbf{+3.00\text{ m/s}}\) (or \(3.00\text{ m/s East}\)).
(B) Interval 2 physical state: The train is **completely at rest** (stationary) at coordinate position \(x = +6.0\text{ m}\) for a duration of \(4.0\text{ seconds}\).
PROBLEM 04 (SOLUTION) Key
An ant crawls along a meter stick. It starts at coordinate \(x = +2.0\text{ cm}\), crawls to \(x = -8.0\text{ cm}\), and then reverses direction to crawl back to \(x = -2.0\text{ cm}\). Calculate: (A) total distance, and (B) net displacement.
Step-by-step Solution:
(A) Distance \(d\) (sum of absolute stage increments):
Stage 1 (from +2.0 to -8.0): \(|d_1| = |-8.0\text{ cm} - 2.0\text{ cm}| = 10.0\text{ cm}\).
Stage 2 (from -8.0 to -2.0): \(|d_2| = |-2.0\text{ cm} - (-8.0\text{ cm})| = 6.0\text{ cm}\).
Total Distance \(d = |d_1| + |d_2| = 10.0\text{ cm} + 6.0\text{ cm} = \mathbf{16.0\text{ cm}}\).
(B) Displacement \(\Delta x\) (only depends on final and initial states):
\(\Delta x = x_f - x_i = -2.0\text{ cm} - (+2.0\text{ cm}) = \mathbf{-4.0\text{ cm}}\) (or \(4.0\text{ cm Left}\)).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 2 OF 8
Honors Motion Key // Teacher Copy
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PROBLEM 05 (SOLUTION) Key
A dog runs \(24.0\text{ m}\) North in \(6.0\text{ s}\), stops completely for \(2.0\text{ s}\) to sniff a flower, and then walks \(12.0\text{ m}\) South in \(4.0\text{ s}\). Taking North as positive, calculate: (A) average speed, and (B) average velocity vector.
Step-by-step Solution:
Stage 1: \(\Delta x_1 = +24.0\text{ m}\), \(\Delta t_1 = 6.0\text{ s}\). Stage 2: \(\Delta x_2 = 0.0\text{ m}\), \(\Delta t_2 = 2.0\text{ s}\). Stage 3: \(\Delta x_3 = -12.0\text{ m}\), \(\Delta t_3 = 4.0\text{ s}\).
Total Time: \(t_{\text{total}} = 6.0\text{ s} + 2.0\text{ s} + 4.0\text{ s} = 12.0\text{ s}\).
(A) Average Speed (scalar rates over total time):
\(s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{|+24.0| + 0 + |-12.0|}{12.0\text{ s}} = \frac{36.0\text{ m}}{12.0\text{ s}} = \mathbf{3.00\text{ m/s}}\).
(B) Average Velocity (vector displacement over total time):
\(v_{\text{avg}} = \frac{\Delta x_{\text{total}}}{t_{\text{total}}} = \frac{+24.0\text{ m} + 0\text{ m} - 12.0\text{ m}}{12.0\text{ s}} = \frac{+12.0\text{ m}}{12.0\text{ s}} = \mathbf{+1.00\text{ m/s}}\) (or \(1.00\text{ m/s North}\)).
PROBLEM 06 (SOLUTION) Key
A sports car travels along a straight highway at a constant speed of \(30.0\text{ m/s}\) for exactly \(15.0\text{ s}\), and then accelerates to a constant speed of \(45.0\text{ m/s}\) for the next \(15.0\text{ s}\). (A) Calculate overall average speed. (B) Explain why it equals arithmetic average.
Step-by-step Solution:
(A) Calculate distance in each stage: \(d_1 = v_1 \cdot t_1 = 30.0\text{ m/s} \cdot 15.0\text{ s} = 450.0\text{ m}\).
\(d_2 = v_2 \cdot t_2 = 45.0\text{ m/s} \cdot 15.0\text{ s} = 675.0\text{ m}\).
Total Distance: \(d_{\text{total}} = 450.0\text{ m} + 675.0\text{ m} = 1125.0\text{ m}\). Total Time: \(t = 30.0\text{ s}\).
Average Speed: \(s_{\text{avg}} = \frac{1125.0\text{ m}}{30.0\text{ s}} = \mathbf{37.5\text{ m/s}}\).
(B) Proof/Explanation: If \(t_1 = t_2 = t\), then:
\(v_{\text{avg}} = \frac{d_1 + d_2}{2t} = \frac{v_1 t + v_2 t}{2t} = \frac{(v_1 + v_2)t}{2t} = \frac{v_1 + v_2}{2}\).
Since the time interval for both segments is identical, they are weighted equally in the time-average, making it equal to the simple arithmetic mean: \(\frac{30.0 + 45.0}{2} = 37.5\text{ m/s}\).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 3 OF 8
Honors Motion Key // Teacher Copy
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PROBLEM 07 (SOLUTION) Key
A swimmer completes a \(100.0\text{-meter}\) race in a \(50.0\text{-meter}\) pool (one lap down and back). On the way down, they swim at a steady speed of \(2.00\text{ m/s}\). On the return lap, they tire and swim at a steady speed of \(1.00\text{ m/s}\). Calculate: (A) the average speed for the entire race, and (B) explain conceptually why this is NOT equal to the simple arithmetic average of \(1.50\text{ m/s}\).
Step-by-step Solution:
(A) Calculate elapsed time for each stage:
Lap 1 (down): \(t_1 = \frac{d_1}{v_1} = \frac{50.0\text{ m}}{2.00\text{ m/s}} = 25.0\text{ s}\). Lap 2 (back): \(t_2 = \frac{d_2}{v_2} = \frac{50.0\text{ m}}{1.00\text{ m/s}} = 50.0\text{ s}\).
Total Time: \(t_{\text{total}} = t_1 + t_2 = 25.0\text{ s} + 50.0\text{ s} = 75.0\text{ s}\). Total Distance \(d_{\text{total}} = 100.0\text{ m}\).
Average Speed: \(s_{\text{avg}} = \frac{100.0\text{ m}}{75.0\text{ s}} = \mathbf{1.33\text{ m/s}}\) (or \(\frac{4}{3}\text{ m/s}\)).
(B) Conceptual Explanation:
The swimmer spends twice as much time traveling at the slower speed (\(50.0\text{ s}\) at \(1.00\text{ m/s}\)) than at the faster speed (\(25.0\text{ s}\) at \(2.00\text{ m/s}\)). Because average speed is a time-weighted average, the overall rate is pulled heavily toward the slower rate. This conforms to the harmonic mean of rates: \(v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2} = \frac{2(2)(1)}{2+1} = 1.33\text{ m/s}\).
PROBLEM 08 (SOLUTION) Key
A cheetah sprints a distance of \(150.0\text{ m}\) in exactly \(5.00\text{ s}\) along a straight vector. (A) Find its average speed in \(\text{m/s}\). (B) Convert this speed to miles per hour (\(\text{mph}\)) using dimensional analysis, showing your conversion steps. (Take \(1.00\text{ m/s} = 2.237\text{ mph}\)).
Step-by-step Solution:
(A) Find average speed in m/s:
\(s_{\text{avg}} = \frac{\text{Distance}}{\text{Time}} = \frac{150.0\text{ m}}{5.00\text{ s}} = \mathbf{30.0\text{ m/s}}\).
(B) Dimensional analysis conversion to mph:
Using conversion factor \(1.00\text{ m/s} = 2.237\text{ mph}\):
\(v = 30.0\text{ m/s} \times \left( \frac{2.237\text{ mph}}{1.00\text{ m/s}} \right) = \mathbf{67.11\text{ mph}}\) (or \(67.1\text{ mph}\)).
Alternative formal method (optional): \(\frac{30\text{ m}}{1\text{ s}} \times \frac{3600\text{ s}}{1\text{ hr}} \times \frac{1\text{ mile}}{1609\text{ m}} = \frac{108000}{1609} \approx 67.12\text{ mph}\).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 4 OF 8
Honors Motion Key // Teacher Copy
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PROBLEM 09 (SOLUTION) Key
The motion diagram represents a skateboarder moving along a horizontal path, captured at equal time intervals (\(\Delta t = 1.0\text{ s}\)).
Start
End
t=0st=1st=2st=3st=4st=5st=6s
State the exact time interval(s) during which the skateboarder is: (A) decelerating, (B) traveling at constant speed, and (C) accelerating.
(A) Decelerating: \(\mathbf{t = 0\text{ s} \text{ to } t = 3\text{ s}}\). The spacing decreases (12px \(\to\) 8px \(\to\) 6px), proving they slowed down.
(B) Constant Speed: \(\mathbf{t = 3\text{ s} \text{ to } t = 4\text{ s}}\). Spacing remains uniform at 6px.
(C) Accelerating: \(\mathbf{t = 4\text{ s} \text{ to } t = 6\text{ s}}\). Spacing increases (8px \(\to\) 12px), proving they sped up.
PROBLEM 10 (SOLUTION) Key
A remote-controlled car runs on a linear track. The graph displays its position \(x\) (meters) over time \(t\) (seconds). Calculate slopes and describe motion.
20 15 10 5 0 0 2 4 6 8 10 x (m) t (s)
COMPUTED SLOPES:
(1) Slopes: Interval 1 slope: \(v_1 = \frac{15.0\text{ m} - 0.0\text{ m}}{4.0\text{ s} - 0.0\text{ s}} = \mathbf{+3.75\text{ m/s}}\). Interval 2 slope: \(v_2 = \frac{1.0\text{ m} - 15.0\text{ m}}{10.0\text{ s} - 4.0\text{ s}} = \frac{-14.0\text{ m}}{6.0\text{ s}} \approx \mathbf{-2.33\text{ m/s}}\).
(2) Motion Description: From \(t = 0\text{ s}\) to \(t = 4\text{ s}\), the car moves East (forward) at a constant velocity of \(+3.75\text{ m/s}\). At \(t = 4\text{ s}\), the car instantly turns around and moves West (backward) at a constant velocity of \(-2.33\text{ m/s}\) from \(t = 4\text{ s}\) to \(10\text{ s}\).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 5 OF 8
Honors Motion Key // Teacher Copy
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PROBLEM 11 (SOLUTION) Key
A small steel marble is rolled up a smooth incline. It slows down to a stop, turns around at its peak, and rolls back down the ramp, speeding up. Sketch a single, clean motion diagram for the complete trip. Explain the direction of the velocity vector compared to the acceleration vector during: (A) the upward trip, (B) the instant of stopping at the peak, and (C) the downward trip.
Step-by-step Solution:
(A) Upward Trip: The velocity vector points UP the ramp, whereas the acceleration vector (caused by the component of gravity parallel to the incline) points constant DOWN the ramp. Since they point in opposite directions, the marble decelerates.
(B) At the Peak: The instantaneous velocity is zero (\(\vec{v} = 0\)). However, the acceleration is still non-zero and points constant DOWN the ramp. (If acceleration were zero at the peak, the marble would stay suspended there at rest forever!).
(C) Downward Trip: The velocity vector now points DOWN the ramp, and the acceleration vector continues to point constant DOWN the ramp. Since they point in the same direction, the marble accelerates (speeds up).
PROBLEM 12 (SOLUTION) Key
Car A is stationary at coordinate \(x = 0.0\text{ m}\). At \(t = 0.0\text{ s}\), it begins to move East at a constant velocity of \(+12.0\text{ m/s}\). At that same split-second, Car B is at coordinate \(x = +200.0\text{ m}\) and is traveling West toward Car A at a constant velocity of \(-8.0\text{ m/s}\). Calculate passing time and coordinate.
Step-by-step Solution:
(A) Set up position equations as functions of time \(t\):
• Car A: starts at origin (\(x_{0A} = 0.0\text{ m}\)) moving East (\(v_A = +12.0\text{ m/s}\)): \(\mathbf{x_A(t) = 12.0 \cdot t}\).
• Car B: starts at \(x_{0B} = +200.0\text{ m}\) moving West (\(v_B = -8.0\text{ m/s}\)): \(\mathbf{x_B(t) = 200.0 - 8.0 \cdot t}\).
(B) Find meeting elapsed time and position coordinate:
Set the position functions equal: \(x_A(t) = x_B(t) \implies 12.0 \cdot t = 200.0 - 8.0 \cdot t\)
Combine like terms: \(20.0 \cdot t = 200.0 \implies \mathbf{t = 10.0\text{ seconds}}\).
Find coordinate \(x\) by substituting \(t = 10.0\text{ s}\) back into Car A's equation:
\(x = 12.0 \cdot (10.0) = \mathbf{120.0\text{ meters}}\) (Check Car B: \(200.0 - 8.0 \cdot 10.0 = 120.0\text{ m}\)).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 6 OF 8
Honors Motion Key // Teacher Copy
PAGE 7 OF 8
+8 +6 +4 +2 0 -2 -4 -6 -8 0 2 4 6 8 10 12 Position x (m) Time t (seconds)
KEY SCHEMA ANALYSIS: Continuous piecewise functions on \(x\text{-}t\) grid.
PROBLEM 13 (SOLUTION) Key
Using the cart's graph from above, calculate: (A) instantaneous velocity during Intervals A, B, and C; (B) total distance and net displacement; and (C) overall average speed and velocity.
Step-by-step Solution:
(A) Instantaneous Velocity (Slope Calculations):
• Interval A (\(0-4\text{ s}\)): \(v_A = \frac{\Delta x}{\Delta t} = \frac{x(4) - x(0)}{4 - 0} = \frac{+6.0\text{ m} - (-2.0\text{ m})}{4.0\text{ s}} = \frac{+8.0\text{ m}}{4.0\text{ s}} = \mathbf{+2.00\text{ m/s}}\) (moving right).
• Interval B (\(4-8\text{ s}\)): \(v_B = \frac{x(8) - x(4)}{8 - 4} = \frac{+6.0\text{ m} - (+6.0\text{ m})}{4.0\text{ s}} = \frac{0.0\text{ m}}{4.0\text{ s}} = \mathbf{0.00\text{ m/s}}\) (stationary / at rest).
• Interval C (\(8-12\text{ s}\)): \(v_C = \frac{x(12) - x(8)}{12 - 8} = \frac{-6.0\text{ m} - (+6.0\text{ m})}{4.0\text{ s}} = \frac{-12.0\text{ m}}{4.0\text{ s}} = \mathbf{-3.00\text{ m/s}}\) (moving left).
(B) Total Distance & Net Displacement:
• Distance: sum of absolute path lengths = \(|8.0\text{ m}| + 0.0\text{ m} + |-12.0\text{ m}| = \mathbf{20.0\text{ meters}}\).
• Net Displacement: \(\Delta x = x_f - x_i = x(12) - x(0) = -6.0\text{ m} - (-2.0\text{ m}) = \mathbf{-4.0\text{ meters}}\) (or \(4.0\text{ m Left/Backward}\)).
(C) Average Speed & Average Velocity Vector:
• Average Speed: \(s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{20.0\text{ m}}{12.0\text{ s}} = \mathbf{1.67\text{ m/s}}\).
• Average Velocity Vector: \(v_{\text{avg}} = \frac{\Delta x_{\text{total}}}{t_{\text{total}}} = \frac{-4.0\text{ m}}{12.0\text{ s}} = \mathbf{-0.33\text{ m/s}}\) (or \(0.33\text{ m/s Left}\)).
SPEED AND VELOCITY PROBLEMS — KEY PAGE 7 OF 8
Honors Kinematics Key // Teacher Copy
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PROBLEM 14 (SOLUTION) Key
Suppose a position-time graph displays a smooth curve starting at coordinate \(x = 0\text{ m}\) at time \(t = 0\text{ s}\). The slope of the curve becomes progressively steeper as time increases (concave upward). (A) Curvature implication. (B) Instaneous velocity visual estimation.
(A) Curvature Implication: Since slope of a position-time graph equates to instantaneous velocity, a progressively steeper positive slope indicates the object is speeding up (accelerating in the positive coordinate direction).
(B) Tangent Estimation Step-by-Step:
1. Align a transparent ruler so it is tangent to the curve at exactly \(t = 3.0\text{ s}\), matching the local curve tilt.
2. Trace a straight line. 3. Read two coordinate coordinates on this straight tangent line, e.g., \((t_1, x_1)\) and \((t_2, x_2)\).
4. Calculate slope: \(v = \frac{x_2 - x_1}{t_2 - t_1}\). This slope equals the instantaneous velocity at \(t = 3.0\text{ s}\) algebraically without calculus.
PROBLEM 15 (SOLUTION) Key
Synthesize: Translate the entire \(12.0\text{-second}\) position-time graph from Page 7 into a single, cohesive horizontal motion diagram. In the box below, draw dots, velocity vectors, and labels.
SOLVED MOTION DIAGRAM:
x=-2 x=0 x=2 x=4 x=6 v = +2 m/s STATIONARY (t=4 to 8s) v = -3 m/s (Interval C, Leftward)
TRANSLATION VERIFICATION:
SPEED AND VELOCITY PROBLEMS — KEY PAGE 8 OF 8
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
PAGE 2 OF 2
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
PAGE 4 OF 5
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
PAGE 3 OF 5
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
PAGE 4 OF 5
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
