An AP Physics 1 lesson focusing on the advanced graphical, algebraic, and vector representations of velocity and acceleration. Includes a comprehensive 12-page reading anthology and a matching 12-page student analysis packet.
C ✓Velocity is changing; acceleration is non-zero and perpendicular to velocity.
DVelocity is constant; acceleration is changing in direction.
Teacher Justification: Velocity direction changes continuously, so velocity changes. Since speed is constant, the net acceleration has no parallel component and is strictly perpendicular (centripetal).
AP Physics 1 Technical Anthology — ANSWER KEY Page 2 of 7
02. Vector Sign Matrix
Teacher Answer Key: Detailed solutions and signs diagnostic matrices.
2.1 Parallel vs. Antiparallel Vectors
To mathematically determine whether an object is speeding up or slowing down, you must evaluate the algebraic signs of both velocity and acceleration simultaneously. Never evaluate acceleration in isolation:
Same Signs (Speeding Up): When \(\vec{v}\) and \(\vec{a}\) possess identical mathematical signs (both \(+\) or both \(-\)), the vectors are parallel. Acceleration acts in the direction of travel, increasing velocity magnitude.
Opposite Signs (Slowing Down): When \(\vec{v}\) and \(\vec{a}\) possess opposite signs, the vectors are antiparallel. Acceleration acts as a brake, opposing the direction of motion, decreasing velocity magnitude.
Sign Diagnostic Matrix
Velocity Sign (\(v\))
Acceleration Sign (\(a\))
Geometric Alignment
Motion Result
Positive (+)
Positive (+)
Parallel (Pointing Right)
Speeding Up (+ Direction)
Positive (+)
Negative (-)
Antiparallel (Opposing)
Slowing Down (+ Direction)
Negative (-)
Negative (-)
Parallel (Pointing Left)
Speeding Up (- Direction)
Negative (-)
Positive (+)
Antiparallel (Opposing)
Slowing Down (- Direction)
This matrix represents a critical milestone in kinematic reasoning. Relying on the simplistic misconception that "negative acceleration always means slowing down" leads to massive errors on free-response exams when negative-direction motion is assessed.
AP Exam Edge
If a problem states "the speed of an object is decreasing," you can immediately write down an algebraic constraint: \(v(t) \cdot a(t) < 0\). If the speed is increasing, the constraint is \(v(t) \cdot a(t) > 0\). This mathematical relation is perfect for structured derivations.
Checkpoint 2.1 Solution Correct Answer: B
A ball rolls up a friction-laden ramp in the positive direction and slows down to a momentary stop. During this upward roll, what are the signs of its velocity and acceleration coordinates?
A\(v>0\), \(a>0\)
B ✓\(v>0\), \(a < 0\)
C\(v < 0\), \(a>0\)
D\(v < 0\), \(a < 0\)
Teacher Justification: Rolling up in the positive direction yields \(v > 0\). Since speed is decreasing, velocity and acceleration must have opposing signs, meaning acceleration is negative (\(a < 0\)).
AP Physics 1 Technical Anthology — ANSWER KEY Page 3 of 7
03. Positive Motion Maps
Teacher Answer Key: Detailed solutions and particle diagram maps.
3.1 Anatomy of Positive-Direction Maps (\(v > 0\))
A motion map represents an object's trajectory by plotting its position as a series of dots recorded at equal time intervals (typically \(\Delta t = 1.0\text{ s}\)). Velocity vectors (green) are drawn starting at each dot, indicating direction and speed. Acceleration vectors (red) are stacked below to show velocity changes.
A. Constant Speed (\(v > 0, a = 0\)) — Equal Dot Spacing t=0 t=1s t=2s t=3s
B. Speeding Up (\(v > 0, a > 0\)) — Spacing Increases t=0 t=1s t=2s t=3s
C. Slowing Down (\(v > 0, a < 0\)) — Spacing Decreases t=0 t=1s t=2s
Checkpoint 3.1 Solution Correct Answer: B
On horizontal motion map, if the green velocity vectors drawn from sequential dots grow progressively shorter, what does this tell us about the direction of red acceleration vectors?
AThey must point in the same direction as velocity.
B ✓They must point in the opposite direction of velocity.
CThey must point strictly perpendicular to velocity.
DThey must be zero, because spacing changes require no acceleration.
AP Physics 1 Technical Anthology — ANSWER KEY Page 4 of 7
04. Negative Motion Maps
Teacher Answer Key: Detailed solutions and negative particle maps.
4.1 Negative Direction Motion Dynamics (\(v < 0\))
Moving in the negative direction is mechanically identical to moving in the positive direction; it simply reflects our coordinate definition. When \(v < 0\), the velocity vectors point to the left. Dots are recorded sequentially as time counts forward.
A. Constant Speed (\(v < 0, a = 0\)) — Equal Dot Spacing Leftward t=0 t=1s t=2s t=3s
B. Speeding Up (\(v < 0, a < 0\)) — Spacing Increases Leftward t=0 t=1s t=2s t=3s
C. Slowing Down (\(v < 0, a > 0\)) — Spacing Decreases Leftward t=0 t=1s t=2s
In negative speeding up (Case B), acceleration points left. In negative slowing down (Case C), acceleration points right to oppose leftward motion.
Think-Pair-Share
Question: For Case C, the acceleration vector points in the positive horizontal direction (\(a > 0\)), yet the object is slowing down. Explain this counter-intuitive result. Why is it physically incorrect to define "negative acceleration" as deceleration?
AP Physics 1 Technical Anthology — ANSWER KEY Page 5 of 7
05. Positive Graph Suites
Teacher Answer Key: Correct velocity and acceleration solutions drawn in Red.
5.1 Cohesive Interpretation of Positive Motion Suites (Solved)
Below are the teacher keys. Position curves feature a single bottom axis (`y=35`), while correct velocity-time (\(v-t\)) and acceleration-time (\(a-t\)) curves are plotted in bold red over the centered axes (`y=20`).
1. Constant Speed in Positive Direction (\(v > 0, a = 0\)) Linear Position
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since the slope of \(x-t\) is constant and positive, \(v-t\) is a horizontal line at a positive value (above `y=20`). Since \(v-t\) has zero slope, \(a-t\) is a horizontal line exactly at zero (`y=20`).
2. Speeding Up (\(v > 0, a > 0\)) — Concave Up Parabola Concave Up Parabola
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since \(x-t\) tangent slope starts low and grows increasingly positive, \(v-t\) is a sloped line rising upwards from zero. Since \(v-t\) rises linearly, \(a-t\) is a constant positive horizontal line (above `y=20`).
3. Slowing Down (\(v > 0, a < 0\)) — Concave Down parabola Concave Down Parabola
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since \(x-t\) tangent slope starts steep positive and flattens out to horizontal, \(v-t\) is a downward sloped line heading from high positive towards zero. Since \(v-t\) has a constant negative slope, \(a-t\) is constant negative (below `y=20`).
Think-Pair-Share
Question: Examine Suite 3 (Slowing Down). At the final timestamp shown, the position graph becomes perfectly horizontal, and the velocity graph touches the horizontal axis (\(v = 0\)). Is the acceleration of the object at this instant also zero? Explain.
AP Physics 1 Technical Anthology — ANSWER KEY Page 6 of 7
06. Negative Graph Suites
Teacher Answer Key: Correct velocity and acceleration solutions drawn in Red.
6.1 Cohesive Interpretation of Negative Motion Suites (Solved)
Below are the teacher keys. Position curves feature a single bottom axis (`y=35`), while correct velocity-time (\(v-t\)) and acceleration-time (\(a-t\)) curves are plotted in bold red over the centered axes (`y=20`).
1. Constant Speed in Negative Direction (\(v < 0, a = 0\)) Linear Decrease
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since the slope of \(x-t\) is constant and negative, \(v-t\) is a horizontal line at a constant negative value (below `y=20`). Since \(v-t\) has zero slope, \(a-t\) is a horizontal line exactly at zero (`y=20`).
2. Speeding Up (\(v < 0, a < 0\)) — Concave Down parabola Concave Down Parabola
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since \(x-t\) tangent slope starts near zero and drops increasingly negative, \(v-t\) is a downward sloped line starting at zero (`y=20`) and heading negative. Since \(v-t\) drops linearly, \(a-t\) is a constant negative line (below `y=20`).
3. Slowing Down (\(v < 0, a > 0\)) — Concave Up parabola Concave Up Parabola
x vs t
v vs t (Solved) + -
a vs t (Solved) + -
Solution Profile: Since \(x-t\) tangent slope starts steep negative and flattens out to horizontal, \(v-t\) is an upward sloped line rising from negative (`y=32`) to zero (`y=20`). Since \(v-t\) has a constant positive slope, \(a-t\) is constant positive (above `y=20`).
Kinematic Analysis Scaffolding — ANSWER KEY
Sentence Stem Solved: Because the velocity vector points unidirectionally and the acceleration vector points opposingly, the vectors are opposed, and the object is slowing down.
Stop & Jot Solved
Because velocity is negative (\(v = -8\text{ m/s}\)) and acceleration is positive (\(a = +2\text{ m/s}^2\)), the signs are opposite (opposed), which means the skateboarder is slowing down.
Below are the complete algebraic and graphical solutions. Correct velocity and acceleration curves are plotted in bold red lines on the coordinate grids.
Given: Position vs. Time Graph (Parabolic curved motion)
Position (x) Time (t) t = 4 s Interval A (t = 0 to 4 s) Interval B (t = 4 to 8 s)
Solved 1: Velocity vs. Time Graph (Straight linear line crossing zero)
Velocity (v) Time (t) 4 s + -
Solved 2: Acceleration vs. Time Graph (Constant positive flat line)
Acceleration (a) Time (t) 4 s + -
Structured Sentence Stems Solved — ANSWER KEY
For Interval A (t = 0 to 4 s): The slope of the position-time graph is negative (positive / negative / zero), which means velocity is negative. The curve is concave up (up / down), which means acceleration is positive. Because the velocity and acceleration coordinates have opposite (same / opposite) signs, the car is slowing down (speeding up / slowing down).
For Interval B (t = 4 to 8 s): The slope of the position-time graph is positive (positive / negative / zero), which means velocity is positive. The curve is concave up (up / down), which means acceleration is positive. Because the velocity and acceleration coordinates have same (same / opposite) signs, the car is speeding up (speeding up / slowing down).
Positive Direction Motion \(\vec{v}_{avg} = +5\text{ m/s}\)
Motion in the positive direction results in positive average velocity and displacement.
Negative Direction Motion \(\vec{v}_{avg} = -5\text{ m/s}\)
Motion in the negative direction results in negative average velocity and displacement.
While average velocity provides a general overview of a trip, it conceals the moment-to-moment behaviors. For example, if a runner completes a full lap around a circular \(400\text{-m}\) track in \(100\text{ s}\), their displacement is zero, yielding an average velocity of \(0\text{ m/s}\). However, their average speed is a brisk \(4\text{ m/s}\). To analyze physical motion with precision, we must narrow down our measurement windows.
Think-Pair-Share
Can an object experience zero average velocity over an extended interval while maintaining a non-zero instantaneous velocity at every intermediate second? Discuss.
Interpreting slopes of position-time graphs as exact instantaneous states.
3.1 Conceptualizing Instantaneous Rate
To find the instantaneous velocity of an object at an exact moment, we calculate the average velocity over a time interval that is infinitesimally small. On a Position vs. Time (\(x-t\)) graph, this translates geometrically to calculating the slope of the line tangent to the curve at that specific point.
Figure 3.1: Tangents and Slopes on Position-Time Curve
Point A (Low positive slope, low speed) Point B (Steep slope, high speed) Time (t) Pos (x)
Geometrically, as the curve in Figure 3.1 bends upwards, the slope of the tangent lines becomes increasingly steep. This demonstrates that the object's instantaneous velocity is rising over time. The sign of the tangent slope indicates the direction of travel: a positive slope denotes motion in the positive direction; a negative slope denotes motion in the negative direction.
Stop & Jot
If an object's position-time graph has a horizontal tangent line at exactly \(t=3\text{ s}\), what is its instantaneous velocity at that moment? Is its average velocity over the preceding interval necessarily zero?
Exploring acceleration as a rate of change of a vector quantity.
4.1 Defining Acceleration
Acceleration (\(\vec{a}\)) is a vector defined as the rate at which velocity changes: \(\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}\). Because velocity includes both speed and direction, an object accelerates if it changes its speed, its direction, or both.
Figure 4.1: The Three Physical Modes of Acceleration
1. Speeding Up v a Velocity and acceleration point in the same direction.
2. Slowing Down v a Velocity and acceleration point in opposite directions.
3. Changing Direction v a Acceleration has a component perpendicular to velocity.
A turning pendulum, a car rounding a curve, and an apple falling under gravity are all systems undergoing acceleration. Recognizing direction changes is fundamental to success on advanced AP circular motion and orbital dynamics units.
Stop & Jot
A driver slams on the brakes of a car. Draw a quick mental schema showing the velocity and acceleration vectors during the braking interval. Are they in alignment or opposed?
Constructing logical rules for predicting changes in speed based on vector signs.
5.1 Same Sign vs. Opposite Sign Conventions
To predict whether an object is speeding up or slowing down, we compare the signs of its velocity and acceleration:
Same Sign (Speeding Up): If both \(\vec{v}\) and \(\vec{a}\) have the same sign (both positive or both negative), the acceleration acts in the direction of the velocity, causing the magnitude of the velocity to grow.
Opposite Sign (Slowing Down): If \(\vec{v}\) and \(\vec{a}\) have opposite signs (one positive, one negative), the acceleration opposes the velocity, acting as a brake. Speed decreases.
Sign Diagnostic Matrix
Velocity Sign (\(v\))
Acceleration Sign (\(a\))
Geometric Alignment
Motion Result
Positive (+)
Positive (+)
Aligned (pointing right)
Speeding Up (+ Direction)
Positive (+)
Negative (-)
Opposed (vel right, acc left)
Slowing Down (+ Direction)
Negative (-)
Negative (-)
Aligned (pointing left)
Speeding Up (- Direction)
Negative (-)
Positive (+)
Opposed (vel left, acc right)
Slowing Down (- Direction)
This matrix represents a cornerstone of AP Physics 1 mechanical reasoning. Students who rely on "negative acceleration always means slowing down" are highly susceptible to losing points on free response questions.
Stop & Jot
A ball is rolling along a floor in the negative direction and slows to a halt. What is the sign of its acceleration? Justify your answer.
Advanced particle diagrams for systems moving in the positive direction (\(v > 0\)).
6.1 Anatomy of Positive-Direction Maps
A motion map (or particle diagram) is a powerful tool to track position, velocity, and acceleration at uniform intervals of time (usually \(\Delta t = 1\text{ s}\)). Below are the key positive scenarios:
A. Constant Speed (\(v > 0, a = 0\)) t=0 t=1s t=2s t=3s
B. Speeding Up (\(v > 0, a > 0\)) t=0 t=1s t=2s t=3s
C. Slowing Down (\(v > 0, a < 0\)) t=0 t=1s t=2s
Notice that in Case B (speeding up), the velocity arrows (green) get longer and acceleration (red) points right. In Case C (slowing down), the velocity arrows get shorter and acceleration points left.
Advanced particle diagrams for systems moving in the negative direction (\(v < 0\)).
7.1 Anatomy of Negative-Direction Maps
When objects travel in the negative direction (\(v < 0\)), their velocity vectors point to the left (opposing the conventional horizontal axis direction):
A. Constant Speed (\(v < 0, a = 0\)) t=0 t=1s t=2s t=3s
B. Speeding Up (\(v < 0, a < 0\)) t=0 t=1s t=2s t=3s
C. Slowing Down (\(v < 0, a > 0\)) t=0 t=1s t=2s
In negative speeding up (Case B), acceleration points left (same direction as velocity, both negative). In negative slowing down (Case C), acceleration points right (opposite direction of velocity).
Converting position curves into velocity and acceleration suites via slope analysis.
8.1 The Slope Connection
To translate graphs "forward" along the trajectory of Position (\(x\)) \(\to\) Velocity (\(v\)) \(\to\) Acceleration (\(a\)), you must evaluate the mathematical slope at each segment of time:
Figure 8.1: Translating a Segmented Motion Sequence via Slope
x (Position) Positive Slope = Constant +v Zero Slope = At Rest (v=0)
v (Velocity) Slope of v is Zero (a=0)
a (Acceleration) a = 0 throughout
This forward translation highlights that flat lines on position graphs result in zero velocity heights, and flat linear slopes on position graphs yield horizontal heights on velocity graphs.
Stop & Jot
If a position graph has a parabola opening downwards, what is the shape of the velocity graph? Is the acceleration value positive or negative?
Converting acceleration graphs back to velocity and displacement curves via area integration.
9.1 The Area Connection
To convert graphs "backward" along the trajectory of Acceleration (\(a\)) \(\to\) Velocity (\(v\)) \(\to\) Position (\(x\)), you must compute the geometric area under the curve:
Figure 9.1: Integrating Acceleration to find Delta-V
a (Acceleration) Shaded Area = a · t = \(\Delta v\)
v (Velocity) Sloped line grows from v_i to v_f
The calculated area on an acceleration graph defines the exact vertical height change (\(\Delta v\)) of the velocity curve. Similarly, the area on a velocity graph dictates the final displacement value (\(\Delta x\)) of the position curve.
Stop & Jot
An acceleration graph has a step area of \(-10\text{ m/s}\). If the object's initial velocity is \(+15\text{ m/s}\), what is its final velocity at the end of this step?
Formalizing the equations of uniformly accelerated motion through absolute geometric proof.
12.1 The Geometric Foundation of Calculus in Kinematics
In AP Physics 1, we analyze systems undergoing constant acceleration (\(a = \text{const}\)). This steady rate of change produces a linear, sloped line on a Velocity vs. Time (\(v-t\)) graph. From this single linear curve, we can geometrically derive the three master kinematic equations of motion by evaluating its slope (acceleration) and area under the curve (displacement).
The Slope is Acceleration
By definition, \(a = \frac{\Delta v}{\Delta t}\). On a \(v-t\) plot, this rise-over-run is exactly the slope of the velocity line.
The Area is Displacement
Since \(\Delta x = v \Delta t\), multiplying the vertical axis (velocity) by the horizontal axis (time) yields displacement.
Figure 12.1: Geometric Analysis of a Constant Acceleration Velocity-Time Curve
v_i v_f t 0 Velocity Curve: v(t) Rectangle Area = v_i · t Triangle Area = 1/2 · (v_f - v_i) · t Rise = v_f - v_i Run = t
Derivation I Velocity Equation
Premise: Constant acceleration is defined as the mathematical slope of the \(v-t\) line.
On an \(x(t)\) position-time plot, the sign of the second derivative dictates the graph's curvature (concavity), providing a direct visual identification of acceleration direction:
Concave Upward (\(a > 0\))
The tangent slope is getting more positive. The curve "holds water" like a cup.
Increasing Slope (Positive Acceleration)
Concave Downward (\(a < 0\))
The tangent slope is getting more negative. The curve "spills water" like an umbrella.
Decreasing Slope (Negative Acceleration)
An inflection point occurs where the curve changes concavity. At these precise inflection moments, the second derivative passes through zero, meaning \(a(t) = 0\).
Example: Finding the Acceleration Equation
Continuing with our particle from Page 1, where \(v(t) = 9t^2 - 16t + 4 \quad (\text{m/s})\): \[a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 - 16t + 4) = 18t - 16 \quad (\text{m/s}^2)\] To find when the acceleration is zero (the inflection point of position), solve: \[18t - 16 = 0 \implies t = \frac{16}{18} = \frac{8}{9} \approx 0.89\text{ s}\]
Stop & Jot
An object's position is given by \(x(t) = -t^2 + 6t\). Calculate its velocity and acceleration equations. Is the position graph concave up or concave down, and what does this tell you about the acceleration's direction?
Reversing differentiation using definite and indefinite integrals to reconstruct motion histories.
3.1 The Antiderivative as a Reconstructive Tool
To reconstruct velocity from acceleration, or position from velocity, we must reverse the process of differentiation. This inverse operation is the antiderivative, represented by the mathematical integral.
Since \(a(t) = \frac{dv}{dt}\), separating variables yields \(dv = a(t) dt\). Integrating both sides gives: \[v(t) = \int a(t) \, dt + C_1 \quad \text{and similarly,} \quad x(t) = \int v(t) \, dt + C_2\] The integration constants \(C_1\) and \(C_2\) correspond physically to initial conditions: velocity \(v(0) = v_0\) and position \(x(0) = x_0\).
When boundaries are established, we utilize definite integrals. The definite integral calculates the exact net accumulated change (the area under the rate curve) between two times \(t_1\) and \(t_2\): \[\Delta x = x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t) \, dt\] Geometrically, area above the horizontal axis (\(v(t) > 0\)) represents positive displacement, while area below (\(v(t) < 0\)) represents negative displacement.
Stop & Jot
An object starts at rest at the origin. Its acceleration is given by \(a(t) = 6t\). Calculate its velocity and position equations. Why is the integration constant \(C_1 = 0\) here?
Applying differentiation and antiderivatives to solve advanced kinematics equations.
NAME: ____________________________________
DATE: ____________________
SCORE: ________ / 20
Problem 1: Analytically Differentiating Motion (6 Points)
A robot's position is tracked in 1D space according to the continuous equation: \[x(t) = 2t^3 - 9t^2 + 12t + 5 \quad \text{(where } x \text{ is in meters, } t \text{ is in seconds)}\]
Derive the analytical equation for the robot's instantaneous velocity \(v(t)\).
Find all values of \(t \ge 0\) where the robot is at rest (\(v(t) = 0\)).
Calculate the robot's exact acceleration \(a(t)\) at \(t = 3\text{ s}\).
Problem 2: Antiderivative and Initial Boundary Conditions (7 Points)
A particle's acceleration is model-mapped as a function of time: \(a(t) = 12t^2 - 4\). At \(t = 0\), the particle is at position \(x(0) = -3\text{ m}\) and possesses an initial velocity of \(v(0) = 8\text{ m/s}\).
Integrate the acceleration function to find the analytical velocity equation \(v(t)\). Show integration constant logic.
Integrate your \(v(t)\) equation to find the complete position equation \(x(t)\).
Problem 3: Area Accumulation vs. Net Displacement (7 Points)
A particle's velocity is given by \(v(t) = 2t - 4\) for the time interval from \(t = 0\text{ s}\) to \(t = 4\text{ s}\).
Write the definite integral representing the particle's net displacement \(\Delta x\) from \(t = 0\) to \(t = 4\). Calculate its value.
Explain why **total distance** traveled during this time interval is not equal to the displacement magnitude.
Constructing continuous piecewise linear velocity curves through segment-by-segment area accumulation.
2.1 Piecewise Constant Acceleration Challenge
An object moves in one dimension, starting from rest (\(v_0 = 0\text{ m/s}\)) at \(t = 0\text s\). Calculate the net velocity change \(\Delta v\) for each segment in the compact boxes below, then sketch the velocity trajectory on the massive grid.
Advanced challenge: Integrating your velocity trajectory to map continuous curved position graphs.
3.1 Integrating Your Velocity Profile
Just as area under acceleration yields change in velocity (\(\Delta v\)), the area under your constructed velocity graph yields the displacement or net change in position (\(\Delta x = x_f - x_i\)). Starting at an initial position of \(x_0 = 0\text{ m}\) at \(t = 0\text s\), evaluate the velocity areas segment-by-segment to draw the continuous position trajectory. Be extremely mindful of curve shapes: sloped velocity lines produce parabolic curves, and flat velocity lines produce straight, constant-slope lines!
AP Physics 1 Technical Library — ANSWER KEY Page 3 of 3
The Position Extension
Teacher Answer Key: Fully calculated displacement segments and curved position profiles.
3.1 Integrating Your Velocity Profile (Solved)
By finding geometric areas under the sloped velocity segments, we construct the continuous quadratic displacement profile. All answers are resolved below in bold red:
Determine the initial velocity \(v(0)\) and final velocity \(v(2)\) from slopes. Find the constant acceleration over this interval.
\(v(0) =\)
\(v(2) =\)
\(a_{0\to2} = \)
Interval 2: \(t = 2 \to 5\text{ s}\)
Calculate the constant slope (velocity) of this straight segment. Determine the acceleration of this constant-velocity phase.
\(v_{2\to5} = \)
\(a_{2\to5} = \)
Interval 3: \(t = 5 \to 7\text{ s}\)
Analyze the parabolic decay. State the final velocity at \(t=7\text{ s}\) where the curve flattens. Find the constant deceleration.
\(v(5) = \)
\(v(7) = \)
\(a_{5\to7} = \)
Graphing Suite 2: Aligned Reverse Translation
Given: Position \(x(t)\) +20.0 +16.0 +12.0 +8.0 +4.0 0 t (s) x (m) 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Sketch: Velocity \(v(t)\) +5.0 +4.0 +3.0 +2.0 +1.0 0 -1.0 -2.0 -3.0 t (s) v (m/s) 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Sketch: Acceleration \(a(t)\) +3.0 +2.0 +1.0 0 -1.0 -2.0 -3.0 t (s) a (m/s²) 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Socratic Seminar & Discussion Prompts
1. The Inertia & Continuity Paradox Explain why a physical velocity-time graph must always be continuous (no vertical gaps), while the acceleration-time graph is permitted to have instantaneous vertical steps. What would a vertical gap in velocity imply about forces?
2. Proving Smoothness (C¹ Continuity) At the transition points (\(t=2\text{ s}\) and \(t=5\text{ s}\) in Problem 2), what calculus/graphical criteria must be met to ensure the parabolic curves join the linear segments smoothly without sharp "kinks"? Prove this using your values.
3. The Area-Slope Fundamental Duality Explain the fundamental calculus duality in physics: why does the vertical displacement (\(\Delta x\)) map to a bounded *area* on the rate-of-change graph, whereas the vertical velocity (\(v\)) maps to a tangent *slope*?
A heavy elevator descends and ascends vertically. Its position \(x(t)\) starting from ground level (\(x_0 = 0\text{ m}\) at \(t = 0\text{ s}\)) is plotted in the graph on the right.
Interval 1: \(t = 0 \to 2\text{ s}\) ✔ Solved
Determine the initial velocity \(v(0)\) and final velocity \(v(2)\) from slopes. Find the constant acceleration over this interval.
1. The Inertia & Continuity Paradox Solution: Velocity vertical gaps represent infinite acceleration (\(a \to \infty\)), requiring infinite net force (\(F_{\text{net}} \to \infty\)) by Newton's Second Law (\(F=ma\)), which is physically impossible. Velocity must change continuously. Acceleration represents force over mass; instantaneous force changes (like sudden braking or changing gears) yield step-changes.
2. Proving Smoothness (C¹ Continuity) Solution: The derivative (velocity) must be continuous at boundaries. At \(t = 2\text{ s}\), the parabolic velocity is \(\lim_{t\to2^-} v(t) = +4.0\text{ m/s}\), matching the constant slope \(+4.0\text{ m/s}\) of segment 2. At \(t=5\text{ s}\), the constant slope is \(+4.0\text{ m/s}\), matching the initial slope of the quadratic decay. Since slopes match, \(x(t)\) is \(C^1\) smooth (no kinks).
3. The Area-Slope Fundamental Duality Solution: Slope is division (\(\Delta y / \Delta x\)) representing rate-of-change (differentiation: \(v = dx/dt\)). Area is multiplication (\(y \cdot x\)) representing accumulation (integration: \(\Delta x = \int v\,dt\)). Hence, displacement maps to bounded area under the velocity curve, while velocity represents the instantaneous slope of position.
