Where acceleration is zero (\(a(t) = 0\)), the position graph \(x(t)\) has an inflection point (where concavity changes).
3. Concavity Markers
Where \(a(t) < 0\), position is concave down. Where \(a(t) > 0\), position is concave up.
UNIT 1: CALC-BASED KINEMATICS GRAPH TRANSLATION GUIDE COHESIVE LESSON REFERENCE MATERIAL
An electric bike starts from rest (\(v(0) = 0\)). Its motor provides a linearly decreasing acceleration modeled by \(a(t) = 4 - t \, \text{m/s}^2\) for the first 4 seconds.
Step 1: Geometry
At \(t = 0\), \(a = 4\). At \(t = 4\), \(a = 0\). This forms a right triangle with base \(4 \, \text{s}\) and height \(4 \, \text{m/s}^2\).
\(\Delta v = \text{Area} = \frac{1}{2} (4) (4) = 8 \, \text{m/s}\).
Step 2: Calculus Integral
We integrate using the reverse power rule: \[\Delta v = \int_{0}^{4} (4 - t) dt = \left[ 4t - \frac{1}{2}t^2 \right]_{0}^{4} = 8 \, \text{m/s}\]
Formulate your thoughts individually, then discuss with a partner. Use the sentence stems below to explain why both approaches yielded the exact same result:
Discussion Sentence Stems: - "Both methods agreed because the acceleration function is _______________, which forms a simple _______________ shape on a graph, allowing..."
- "If the acceleration function had been \(a(t) = 4 - t^2\) instead of \(4-t\), geometry would fail because..."
UNIT 1: KINEMATICS ACCUMULATION PAGE 2 OF 11
Core Theoretical Guide
Page 3
When an object's velocity is modeled by a curved function (e.g., quadratic or cubic), its position graph cannot be solved by simply calculating triangles or rectangles. We must utilize the power rule for integration to find the exact area under the curved velocity path.
THE INTEGRATION POWER RULE: For any term \(t^n\) where \(n \neq -1\): \[\int t^n dt = \frac{t^{n+1}}{n+1} + C\] To find position change from velocity \(v(t) = At^2 + Bt + C\): \[\Delta x = \int_{t_1}^{t_2} (At^2 + Bt + C) dt = \left[ \frac{A}{3}t^3 + \frac{B}{2}t^2 + Ct \right]_{t_1}^{t_2}\]
A high-performance drone has a curved velocity profile described by the quadratic function: \(v(t) = t^2 - 4t + 3\). Assuming the drone starts at position \(x(0) = 0\), find its position at \(t = 3\) seconds.
Analytical Calculus Solution: To find position, integrate the velocity function from 0 to 3 seconds: \[x(3) - x(0) = \int_{0}^{3} (t^2 - 4t + 3) dt = \left[ \frac{1}{3}t^3 - 2t^2 + 3t \right]_{0}^{3} = 0 \, \text{m}\]
The drone flew the entire time from \(t = 0 \to 3\) s, yet its displacement at \(t = 3\) is exactly 0 meters. How is this physically possible? Use area concepts to explain.
Explain the positive and negative area balance here...
UNIT 1: KINEMATICS ACCUMULATION PAGE 3 OF 11
Core Theoretical Guide
Page 4
A sailboat is cruising forward. A variable wind starts gusting, providing a linearly increasing acceleration modeled by: \(a(t) = 0.5t \, \text{m/s}^2\). The boat's initial velocity is \(v(0) = 2 \, \text{m/s}\). Find the boat's velocity at \(t = 4\) seconds.
The Algebraic Dilemma
The student tries to apply the standard constant-acceleration equation:
\(v_f = v_i + at\) But what value of \(a\) should they use?
If they use the average acceleration \(a_{\text{avg}} = 1 \, \text{m/s}^2\):
\(v(4) = 2 + (1)(4) = 6 \, \text{m/s}\)
Limitation: While taking the average happened to work here because the acceleration change was perfectly linear, this method fails completely if \(a(t)\) is quadratic, cubic, or sinusoidal.
The Calculus Solution
The calculus student integrates the acceleration function to find the precise velocity: \[v(t) = v(0) + \int_{0}^{t} a(t) dt\] \[v(t) = 2 + \int_{0}^{t} 0.5t \, dt = 2 + 0.25t^2\] Now evaluate at exactly \(t=4\) s: \[v(4) = 2 + 0.25(4)^2 = 6 \, \text{m/s}\]
Strength: This exact same calculus process handles any function of \(a(t)\) with 100% precision.
Why is using average rates a dangerous trap when analyzing non-constant physical forces?
Discussion Stem: "Relying on average values is risky because if acceleration changes non-linearly over time, the average value will _______________, causing our predictions of final speed to be _______________."
UNIT 1: KINEMATICS ACCUMULATION PAGE 4 OF 11
Core Theoretical Guide
Page 5
In physics, Work (\(W\)) is the measure of energy transfer. The algebraic formula taught in basic physics is: \(W = F \cdot d\). But this formula assumes the force is completely constant as the object moves. Let's look at what happens when the force varies, like stretching a spring.
The force required to stretch a spring increases linearly with the displacement \(x\). The constant \(k\) is the spring constant. Because the force constantly changes as you stretch the spring, we cannot simply multiply the final force by the total displacement.
The Algebraic Formula Memorizer
Because \(W = F \cdot d\) fails, the algebra student is forced to memorize a separate, specialized equation for spring work: \[W = \frac{1}{2}kx^2\]
"I have to memorize different work formulas for springs, gravity, and electrical charges."
The Calculus Integrator
The calculus student knows work is the integral of force over displacement: \[W = \int_{0}^{x} F(x) dx\] Substitute Hooke's law (\(F = kx\)): \[W = \int_{0}^{x} kx \, dx = \frac{1}{2}kx^2\]
The "Aha!" Moment: The spring work formula is just the integral of a linear force function!
The Spring Integration Synthesis
Explain in your own words how definite integration unifies different force/work formulas:
Synthesis Stem: "Integrating a variable force \(F(x)\) with respect to \(x\) allows us to derive the specific work equation __________, bypassing the need to memorize arbitrary equations because..."
UNIT 1: KINEMATICS ACCUMULATION PAGE 5 OF 11
Core Practice Set
Page 6
Solve the integration problems below. Use the ample workspace to write out antiderivatives, upper and lower limits of evaluation, and final numeric values with units.
Problem 1: Definite Integration for Velocity
A maglev train undergoes variable acceleration described by \(a(t) = 0.6t^2 \, \text{m/s}^2\). If the train starts with an initial velocity of \(v(0) = 5 \, \text{m/s}\), calculate its exact velocity at \(t = 5\) seconds using definite integration.
Show integration and boundary steps here:
Problem 2: Curved Area Accumulation
An object's velocity is modeled by the function \(v(t) = 3t^2 - 6t \, \text{m/s}\). Find the exact change in position (\(\Delta x\)) from \(t = 0\) to \(t = 4\) seconds. Be careful with signs when integrating!
Show integration and boundary steps here:
Problem 3: Variable Force & Spring Work
A variable force acts on a block modeled by the function \(F(x) = 3x^2 \, \text{N}\). Find the exact work done (\(W = \int F(x) dx\)) in stretching the object from position \(x = 1 \, \text{m}\) to \(x = 3 \, \text{m}\).
Show integration and boundary steps here:
UNIT 1: KINEMATICS ACCUMULATION PAGE 6 OF 11
Core Practice Set
Page 7
Continue applying integration methods to these variable rates of change. Be sure to include physical units in your final answers.
Problem 4: Electric Vehicle Launch
An electric vehicle launches with a time-dependent acceleration function \(a(t) = 2t + 1 \, \text{m/s}^2\). If it starts from rest (\(v(0) = 0\)), calculate its exact speed at \(t = 4\) seconds.
Show integration and boundary steps here:
Problem 5: Parabolic Speed Deceleration
A motorboat's engine cuts out, and it decelerates with \(a(t) = -0.3t^2 \, \text{m/s}^2\). If its initial velocity at \(t=0\) was \(v(0) = 12 \, \text{m/s}\), calculate its exact velocity at \(t = 3\) seconds.
Show integration and boundary steps here:
Problem 6: Rocket Booster Acceleration
A solid fuel booster accelerative load increases linearly according to the function \(a(t) = 1.2t \, \text{m/s}^2\). Starting from rest, calculate the exact change in velocity during the interval from \(t = 1\) to \(t = 5\) seconds.
Show integration and boundary steps here:
UNIT 1: KINEMATICS ACCUMULATION PAGE 7 OF 11
Core Practice Set
Page 8
Finalize your practice set by resolving position and force integration problems. Take careful note of any boundary offsets.
Problem 7: Harmonic Velocity Translation
An oscillating mechanical piston's velocity function is described by \(v(t) = 6t - 2 \, \text{m/s}\). If the initial position is \(x(0) = 4\) meters, solve the definite integral to write the complete algebraic function for position \(x(t)\).
Show integration and boundary steps here:
Problem 8: Variable Pull Tension Work
A winch cable pulls a heavy block with a tension force that increases linearly with distance: \(F(x) = 4x + 2 \, \text{N}\). Integrate this force function over displacement to calculate the exact work done during the first \(x = 3\) meters of displacement.
Show integration and boundary steps here:
Problem 9: Falling Object Drag Resistance
A skydiver's net downward acceleration decreases due to variable drag according to \(a(t) = 9.8 - 0.8t \, \text{m/s}^2\) during their first few seconds. Compute the change in downward velocity during the first 5 seconds of the jump.
Show integration and boundary steps here:
UNIT 1: KINEMATICS ACCUMULATION PAGE 8 OF 11
Core Theoretical Guide
Page 9
Use this quick reference card to recall relationships when converting motion components using calculus, contrasted with their constant algebraic counterparts.
| Physical Parameter | Algebra Formula (Constant) | Calculus Formula (Variable) |
|---|---|---|
| Displacement | \(\Delta x = v \cdot t\) | \(\Delta x = \int v(t) dt\) |
| Velocity Change | \(\Delta v = a \cdot t\) | \(\Delta v = \int a(t) dt\) |
| Work Done | \(W = F \cdot d\) | \(W = \int F(x) dx\) |
| Power | \(P = \frac{W}{t}\) | \(P = \frac{dW}{dt}\) |
| Impulse | \(J = F \cdot t\) | \(J = \int F(t) dt\) |
Integration Level-Up Checklist:
Key Integral Strategies Checklist:
UNIT 1: KINEMATICS ACCUMULATION PAGE 9 OF 11
Interactive Workshop
Page 10
Use these structured discussion frameworks and sentence stems to collaborate with peer groups and summarize kinematic transformations.
Task 1: Stop & Jot — Graph Transformation Summary
How does the definite integral of a function behave differently when crossing above versus below the horizontal axis? Formulate your written explanation.
Write your individual explanation here...
Task 2: Think-Pair-Share — Concavity and Curvature Connection
Discuss with your partner the direct physical bridge between the sign of acceleration and the curvature of position.
Discussion Sentence Stems: - "If the acceleration value \(a(t)\) is completely negative, the position graph \(x(t)\) must exhibit a _______________ curvature (concave _______________) because the velocity's slope is..."
- "An inflection point on the position curve happens at exactly the time \(t\) where acceleration equals _______________ and changes sign because..."
Peer Evaluation Stem
"I reviewed my partner's graph translations and agreed with their slope behaviors because _______________. However, we corrected the position graph curvature at interval _______________ since..."
UNIT 1: KINEMATICS ACCUMULATION PAGE 10 OF 11
Core Theoretical Guide
Page 11
Use this dedicated, structured notebook section to summarize your personal strategies, formulate questions for study groups, or document critical calculus derivations.
NOTES / INQUIRIES: __________________________________________________________________________
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UNIT 1: KINEMATICS ACCUMULATION PAGE 11 OF 11
UNIT 1: KINEMATICS & CALCULUS — ANSWER KEY PAGE 1 OF 2
TEACHER ANSWER KEY — RED EDITION
Part 2 Key
4
The quadratic velocity curve you sketched on Page 1 is modeled by: \(v(t) = t^2 - 4t + 3\). To determine position \(x(t)\), integrate velocity: \(x(t) - x(0) = \int_{0}^{t} v(t) dt\). Given initial position: \(x(0) = 0 \, \text{m}\).
Step 4.1: Analytic Integration & Exact Coordinates SOLUTIONS
Derive the general cubic position equation, then calculate coordinates at specific times:
\(x(t) = \int (t^2 - 4t + 3) dt = \mathbf{\frac{1}{3}t^3 - 2t^2 + 3t + C}\)
Cubic function with \(x(0) = 0\): \(x(t) = \mathbf{\frac{1}{3}t^3 - 2t^2 + 3t}\)
At \(t = 1\) s (Max)
\(x(1) = \mathbf{\frac{4}{3} \approx 1.33 \, m}\)
At \(t = 3\) s (Min)
\(x(3) = \mathbf{0 \, m}\)
At \(t = 4\) s (Final)
\(x(4) = \mathbf{\frac{4}{3} \approx 1.33 \, m}\)
Step 4.2: Inflection Points & Concavity SOLUTIONS
Inflection occurs when \(x''(t) = a(t) = 0\):
\(t = \mathbf{2}\) seconds
For \(t \in (0, 2)\), \(a(t) < 0\), so \(x(t)\) is:
Concave Down
For \(t \in (2, 4)\), \(a(t) > 0\), so \(x(t)\) is:
Concave Up
5
Plot the calculated points, mark the inflection point, and trace a smooth cubic curve.
COMPLETED POSITION GRAPH: PLOT \(x(t)\) 0.0 -0.5 0.5 1.0 1.5 2.0 x (m) 0 1.0 2.0 3.0 4.0 t (s)
Explain the physical meaning of what occurs at \(t = 2\) s across all three graphs. Focus specifically on the velocity minimum, acceleration sign change, and position inflection.
At t = 2 s, the acceleration changes sign from negative to positive. This crossing point (a = 0) corresponds to the minimum value of velocity (v = -1 m/s), where the object momentarily stops slowing down in reverse and begins speeding up forward. On the position graph, this is a distinct inflection point where the curvature changes from concave down (since a < 0) to concave up (since a > 0).
UNIT 1: KINEMATICS & CALCULUS — ANSWER KEY PAGE 2 OF 2
DEFINITE INTEGRATION AS AREA: \[\text{Area} = \int_{t_1}^{t_2} f(t) dt\] In Kinematics, accumulating acceleration over time gives change in velocity: \[\Delta v = v(t_2) - v(t_1) = \int_{t_1}^{t_2} a(t) dt\]
An electric bike starts from rest (\(v(0) = 0\)). Its motor provides a linearly decreasing acceleration modeled by \(a(t) = 4 - t \, \text{m/s}^2\) for the first 4 seconds.
Step 1: Geometry
At \(t = 0\), \(a = 4\). At \(t = 4\), \(a = 0\). This forms a right triangle with base \(4 \, \text{s}\) and height \(4 \, \text{m/s}^2\).
\(\Delta v = \text{Area} = \frac{1}{2} (4) (4) = 8 \, \text{m/s}\).
Step 2: Calculus Integral
We integrate using the reverse power rule: \[\Delta v = \int_{0}^{4} (4 - t) dt = \left[ 4t - \frac{1}{2}t^2 \right]_{0}^{4} = 8 \, \text{m/s}\]
Formulate your thoughts individually, then discuss with a partner. Use the sentence stems below to explain why both approaches yielded the exact same result:
Completed Sentence Stems: - "Both methods agreed because the acceleration function is LINEAR AND CONTINUOUS, which forms a simple RIGHT TRIANGULAR shape on a graph, allowing GEOMETRIC FORMULAS TO COMPUTE EXACT AREA BOUNDARIES."
- "If the acceleration function had been \(a(t) = 4 - t^2\) instead of \(4-t\), geometry would fail because THE UPPER REGION IS PARABOLIC, REQUIRING CALCULUS TO COMPUTE CURVED AREA EXACTLY."
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 2 OF 11
TEACHER EDITION — KEY EDITION
Page 3 Key
When an object's velocity is modeled by a curved function (e.g., quadratic or cubic), its position graph cannot be solved by simply calculating triangles or rectangles. We must utilize the power rule for integration to find the exact area under the curved velocity path.
THE INTEGRATION POWER RULE: For any term \(t^n\) where \(n \neq -1\): \[\int t^n dt = \frac{t^{n+1}}{n+1} + C\] To find position change from velocity \(v(t) = At^2 + Bt + C\): \[\Delta x = \int_{t_1}^{t_2} (At^2 + Bt + C) dt = \left[ \frac{A}{3}t^3 + \frac{B}{2}t^2 + Ct \right]_{t_1}^{t_2}\]
A high-performance drone has a curved velocity profile described by the quadratic function: \(v(t) = t^2 - 4t + 3\). Assuming the drone starts at position \(x(0) = 0\), find its position at \(t = 3\) seconds.
Analytical Calculus Solution: To find position, integrate the velocity function from 0 to 3 seconds: \[x(3) - x(0) = \int_{0}^{3} (t^2 - 4t + 3) dt = \left[ \frac{1}{3}t^3 - 2t^2 + 3t \right]_{0}^{3} = 0 \, \text{m}\]
The drone flew the entire time from \(t = 0 \to 3\) s, yet its displacement at \(t = 3\) is exactly 0 meters. How is this physically possible? Use area concepts to explain.
This occurs because velocity is directional (vector). From t = 0 to 1 s, velocity is positive, meaning the drone moves forward (+4/3 m displacement). From t = 1 to 3 s, velocity is negative, meaning the drone flies backward (-4/3 m displacement). The accumulated negative area perfectly cancels the positive area, yielding exactly 0 net displacement.
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 3 OF 11
TEACHER EDITION — KEY EDITION
Page 4 Key
A sailboat is cruising forward. A variable wind starts gusting, providing a linearly increasing acceleration modeled by: \(a(t) = 0.5t \, \text{m/s}^2\). The boat's initial velocity is \(v(0) = 2 \, \text{m/s}\). Find the boat's velocity at \(t = 4\) seconds.
The Algebraic Dilemma
The student tries to apply the standard constant-acceleration equation:
\(v_f = v_i + at\) But what value of \(a\) should they use?
If they use the average acceleration \(a_{\text{avg}} = 1 \, \text{m/s}^2\):
\(v(4) = 2 + (1)(4) = 6 \, \text{m/s}\)
Limitation: While taking the average happened to work here because the acceleration change was perfectly linear, this method fails completely if \(a(t)\) is quadratic, cubic, or sinusoidal.
The Calculus Solution
The calculus student integrates the acceleration function to find the precise velocity: \[v(t) = v(0) + \int_{0}^{t} a(t) dt\] \[v(t) = 2 + \int_{0}^{t} 0.5t \, dt = 2 + 0.25t^2\] Now evaluate at exactly \(t=4\) s: \[v(4) = 2 + 0.25(4)^2 = 6 \, \text{m/s}\]
Strength: This exact same calculus process handles any function of \(a(t)\) with 100% precision.
Why is using average rates a dangerous trap when analyzing non-constant physical forces?
Completed Stem: "Relying on average values is risky because if acceleration changes non-linearly over time, the average value will UNDERESTIMATE OR OVERESTIMATE THE ACTUAL ACCUMULATION PATH, causing our predictions of final speed to be MATHEMATICALLY INACCURATE AND PHYSICALLY UNRELIABLE IN COMPLEX ENGINEERING CONTEXTS."
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 4 OF 11
TEACHER EDITION — KEY EDITION
Page 5 Key
In physics, Work (\(W\)) is the measure of energy transfer. The algebraic formula taught in basic physics is: \(W = F \cdot d\). But this formula assumes the force is completely constant as the object moves. Let's look at what happens when the force varies, like stretching a spring.
The force required to stretch a spring increases linearly with the displacement \(x\). The constant \(k\) is the spring constant. Because the force constantly changes as you stretch the spring, we cannot simply multiply the final force by the total displacement.
The Algebraic Formula Memorizer
Because \(W = F \cdot d\) fails, the algebra student is forced to memorize a separate, specialized equation for spring work: \[W = \frac{1}{2}kx^2\]
"I have to memorize different work formulas for springs, gravity, and electrical charges."
The Calculus Integrator
The calculus student knows work is the integral of force over displacement: \[W = \int_{0}^{x} F(x) dx\] Substitute Hooke's law (\(F = kx\)): \[W = \int_{0}^{x} kx \, dx = \frac{1}{2}kx^2\]
The "Aha!" Moment: The spring work formula is just the integral of a linear force function!
The Spring Integration Synthesis Solution
Explain in your own words how definite integration unifies different force/work formulas:
Synthesis Stem Completed: "Integrating a variable force \(F(x)\) with respect to \(x\) allows us to derive the specific work equation FOR ANY FORCE PROFILE (SUCH AS SPRINGS, GRAVITY, OR COULOMB FORCES) ON THE FLY, bypassing the need to memorize arbitrary equations because DEFINITE INTEGRATION IS THE SINGLE MATHEMATICAL FOUNDATION FROM WHICH ALL ENERGY EQUATIONS DESCEND."
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 5 OF 11
TEACHER EDITION — KEY EDITION
Page 6 Key
Solutions for problems 1–3 mathematically completed in bold red ink:
Problem 1: Definite Integration for Velocity
A maglev train undergoes variable acceleration described by \(a(t) = 0.6t^2 \, \text{m/s}^2\). If the train starts with an initial velocity of \(v(0) = 5 \, \text{m/s}\), calculate its exact velocity at \(t = 5\) seconds using definite integration.
\[ \color{red}{\Delta v = \int_{0}^{5} 0.6t^2 dt = \left[ 0.2t^3 \right]_{0}^{5} = 0.2(125) - 0 = 25 \, \text{m/s}} \] \[ \color{red}{v(5) = v(0) + \Delta v = 5 + 25 = \mathbf{30 \, \text{m/s}}} \]
Problem 2: Curved Area Accumulation
An object's velocity is modeled by the function \(v(t) = 3t^2 - 6t \, \text{m/s}\). Find the exact change in position (\(\Delta x\)) from \(t = 0\) to \(t = 4\) seconds. Be careful with signs when integrating!
\[ \color{red}{\Delta x = \int_{0}^{4} (3t^2 - 6t) dt = \left[ t^3 - 3t^2 \right]_{0}^{4}} \] \[ \color{red}{\Delta x = (4^3 - 3(4)^2) - (0) = 64 - 48 = \mathbf{16 \, \text{meters}}} \]
Problem 3: Variable Force & Spring Work
A variable force acts on a block modeled by the function \(F(x) = 3x^2 \, \text{N}\). Find the exact work done (\(W = \int F(x) dx\)) in stretching the object from position \(x = 1 \, \text{m}\) to \(x = 3 \, \text{m}\).
\[ \color{red}{W = \int_{1}^{3} 3x^2 dx = \left[ x^3 \right]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = \mathbf{26 \, \text{Joules}}} \]
(Algebraic \(W = Fd\) fails because force changes continuously from 3 N to 27 N as distance increases.)
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 6 OF 11
TEACHER EDITION — KEY EDITION
Page 7 Key
Solutions for problems 4–6 mathematically completed in bold red ink:
Problem 4: Electric Vehicle Launch
An electric vehicle launches with a time-dependent acceleration function \(a(t) = 2t + 1 \, \text{m/s}^2\). If it starts from rest (\(v(0) = 0\)), calculate its exact speed at \(t = 4\) seconds.
\[ \color{red}{\Delta v = \int_{0}^{4} (2t + 1) dt = \left[ t^2 + t \right]_{0}^{4} = (16 + 4) - 0 = 20 \, \text{m/s}} \] \[ \color{red}{v(4) = v(0) + \Delta v = 0 + 20 = \mathbf{20 \, \text{m/s}}} \]
Problem 5: Parabolic Speed Deceleration
A motorboat's engine cuts out, and it decelerates with \(a(t) = -0.3t^2 \, \text{m/s}^2\). If its initial velocity at \(t=0\) was \(v(0) = 12 \, \text{m/s}\), calculate its exact velocity at \(t = 3\) seconds.
\[ \color{red}{\Delta v = \int_{0}^{3} -0.3t^2 dt = \left[ -0.1t^3 \right]_{0}^{3} = -0.1(27) - 0 = -2.7 \, \text{m/s}} \] \[ \color{red}{v(3) = v(0) + \Delta v = 12 - 2.7 = \mathbf{9.3 \, \text{m/s}}} \]
Problem 6: Rocket Booster Acceleration
A solid fuel booster accelerative load increases linearly according to the function \(a(t) = 1.2t \, \text{m/s}^2\). Starting from rest, calculate the exact change in velocity during the interval from \(t = 1\) to \(t = 5\) seconds.
\[ \color{red}{\Delta v = \int_{1}^{5} 1.2t dt = \left[ 0.6t^2 \right]_{1}^{5} = 0.6(25) - 0.6(1) = 15 - 0.6 = \mathbf{14.4 \, \text{m/s}}} \]
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 7 OF 11
TEACHER EDITION — KEY EDITION
Page 8 Key
Solutions for problems 7–9 mathematically completed in bold red ink:
Problem 7: Harmonic Velocity Translation
An oscillating mechanical piston's velocity function is described by \(v(t) = 6t - 2 \, \text{m/s}\). If the initial position is \(x(0) = 4\) meters, solve the definite integral to write the complete algebraic function for position \(x(t)\).
\[ \color{red}{x(t) = x(0) + \int_{0}^{t} (6t - 2) dt = 4 + \left[ 3t^2 - 2t \right]_{0}^{t}} \] \[ \color{red}{x(t) = 4 + (3t^2 - 2t) - (0) = \mathbf{3t^2 - 2t + 4 \, \text{meters}}} \]
Problem 8: Variable Pull Tension Work
A winch cable pulls a heavy block with a tension force that increases linearly with distance: \(F(x) = 4x + 2 \, \text{N}\). Integrate this force function over displacement to calculate the exact work done during the first \(x = 3\) meters of displacement.
\[ \color{red}{W = \int_{0}^{3} (4x + 2) dx = \left[ 2x^2 + 2x \right]_{0}^{3}} \] \[ \color{red}{W = (2(9) + 2(3)) - 0 = (18 + 6) = \mathbf{24 \, \text{Joules}}} \]
Problem 9: Falling Object Drag Resistance
A skydiver's net downward acceleration decreases due to variable drag according to \(a(t) = 9.8 - 0.8t \, \text{m/s}^2\) during their first few seconds. Compute the change in downward velocity during the first 5 seconds of the jump.
\[ \color{red}{\Delta v = \int_{0}^{5} (9.8 - 0.8t) dt = \left[ 9.8t - 0.4t^2 \right]_{0}^{5}} \] \[ \color{red}{\Delta v = (9.8(5) - 0.4(25)) - 0 = 49 - 10 = \mathbf{39 \, \text{m/s}}} \]
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 8 OF 11
Core Theoretical Guide
Page 9
Use this quick reference card to recall relationships when converting motion components using calculus, contrasted with their constant algebraic counterparts.
| Physical Parameter | Algebra Formula (Constant) | Calculus Formula (Variable) |
|---|---|---|
| Displacement | \(\Delta x = v \cdot t\) | \(\Delta x = \int v(t) dt\) |
| Velocity Change | \(\Delta v = a \cdot t\) | \(\Delta v = \int a(t) dt\) |
| Work Done | \(W = F \cdot d\) | \(W = \int F(x) dx\) |
| Power | \(P = \frac{W}{t}\) | \(P = \frac{dW}{dt}\) |
| Impulse | \(J = F \cdot t\) | \(J = \int F(t) dt\) |
Integration Level-Up Checklist:
Key Integral Strategies Checklist:
UNIT 1: KINEMATICS ACCUMULATION PAGE 9 OF 11
TEACHER EDITION — KEY EDITION
Page 10 Key
Completed classroom discussion entries and stems completed in bold red:
Task 1: Stop & Jot Solution
How does the definite integral of a function behave differently when crossing above versus below the horizontal axis?
Definite integrals calculate NET SIGNED AREA. Any region above the horizontal axis represents positive accumulation (positive change in value, e.g. moving forward), while any region below represents negative accumulation (negative change, e.g. moving backward). They algebraically subtract from one another.
Task 2: Think-Pair-Share Solution
Discuss with your partner the direct physical bridge between the sign of acceleration and the curvature of position.
Completed Sentence Stems: - "If the acceleration value \(a(t)\) is completely negative, the position graph \(x(t)\) must exhibit a FROWNING curvature (concave DOWNWARD) because the velocity's slope is NEGATIVELY CHARGED (SLOPING DOWNWARD)."
- "An inflection point on the position curve happens at exactly the time \(t\) where acceleration equals ZERO and changes sign because THAT REPRESENTS THE BOUNDARY WHERE CONCAVITY SHIFTS SIDES."
Peer Evaluation Completed Stem
"I reviewed my partner's graph translations and agreed with their slope behaviors because SLOPES OF VELOCITY DIRECTLY DICTATE ACCELERATION VALUES. However, we corrected the position graph curvature at interval \(t \in (0,2)\) since THEY HAD INVERTED CONCAVITY SHIFTS."
UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 10 OF 11
TEACHER EDITION — KEY EDITION
Page 11 Key
Key study and teaching strategies summarizing definite integration concepts:
KEY CONCEPTS TO REINFORCE WITH STUDENTS:
1. Definite integrals find NET signed area, unlike geometry which finds total absolute area.
2. Constant acceleration represents a HORIZONTAL line; velocity is then a LINEAR curve.
3. Non-constant acceleration (like a(t) = t) forces velocity to be QUADRATIC (degree 2).
4. Work done by a variable force is ALWAYS the area under the force-displacement graph.
5. Highlight that \(x''(t) = v'(t) = a(t)\) connects inflection points directly with \(a(t) = 0\).
6. Remind students that boundary conditions (\(x(0)\), \(v(0)\)) dictate the constant of integration C.
7. Stop & Jot questions can be graded based on conceptual completeness and vocab.
8. Stems like "fails because..." help surface typical student algebraic misconceptions.
9. Power rule: Raise power first, then divide. Avoid simple subtraction errors in bounds.
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UNIT 1: KINEMATICS ACCUMULATION — ANSWER KEY PAGE 11 OF 11
6 The Runway Takeoff Threshold (Constant Acceleration Limits)
A cargo airplane requires a minimum take-off speed of \(v_{\text{take-off}} = 80.0 \, \text{m/s}\). The plane starts from rest at the beginning of a straight runway of length \(L = 1000 \, \text{m}\). If the engines provide a constant acceleration of \(a = 3.00 \, \text{m/s}^2\), calculate the runway distance remaining when the plane reaches take-off speed, and check if the takeoff is successful.
7 Overtaking at a Stoplight (Linear Time Pursuit)
A motorcycle is stopped at a red light. An SUV passes the motorcycle at a constant speed of \(v_s = 20.0 \, \text{m/s}\). At that exact instant, the light turns green and the motorcycle accelerates from rest at a constant \(a_m = 4.00 \, \text{m/s}^2\) to chase the SUV. Find the time and distance from the intersection where the motorcycle overtakes the SUV.
8 Rocket Sled Deceleration Test (Sequential Constant Rates)
A rocket-propelled sled on a straight track accelerates from rest at \(a_1 = 30.0 \, \text{m/s}^2\) for \(t_1 = 4.00\) s. The rockets then cut off, and water-brakes deploy, decelerating the sled at a constant rate of \(a_2 = -15.0 \, \text{m/s}^2\) until its speed is reduced to \(v_2 = 20.0 \, \text{m/s}\). Calculate the total distance traveled during this entire test.
UNIT 1: KINEMATICS — AP PHYSICS C CONSTANT ACCELERATION PRACTICE PAGE 2 OF 2
SOLVED VIA QUADRATIC FORMULA FOR TIME
• Positions: \(x_{\text{truck}}(t) = 80.0 + 15.0t \quad x_{\text{car}}(t) = 25.0t + \frac{1}{2}a t^2 = 25.0t + 1.00t^2\)
• Passing: \(25.0t + t^2 = 80.0 + 15.0t \implies \mathbf{t^2 + 10.0t - 80.0 = 0}\)
• Solve: \(t = \frac{-10.0 \pm \sqrt{100 - 4(1)(-80.0)}}{2.00} = \frac{-10.0 + 20.49}{2.00} = \mathbf{5.25 \, \text{seconds}}\)
4 Braking Reaction Distance (1D Safety Bounds)
A motorist traveling at a constant speed of \(v_0 = 32.0 \, \text{m/s}\) suddenly notices a stalled delivery van \(D = 100.0 \, \text{m}\) ahead. After a reaction delay of \(\Delta t = 0.400\) s, the motorist applies the brakes, causing a constant deceleration of \(a = -6.00 \, \text{m/s}^2\). Calculate the stopping distance of the car from the moment the van was spotted and determine if a collision occurs.
• Reaction Stage: \(d_{\text{react}} = v_0 \cdot \Delta t = (32.0)(0.400) = 12.8 \, \text{m}\)
• Braking Stage: \(v_f^2 = v_i^2 + 2ad \implies 0 = 32.0^2 + 2(-6.00)d_{\text{brake}} \implies d_{\text{brake}} = 85.33 \, \text{m}\)
• Total stopping distance: \(d_{\text{total}} = d_{\text{react}} + d_{\text{brake}} = 12.8 + 85.33 = \mathbf{98.13 \, \text{meters}}\)
• Check: \(98.13 \, \text{m} < 100.0 \, \text{m} \implies \mathbf{\text{No collision occurs (stops 1.87 m before van).}}\)
UNIT 1: KINEMATICS — AP PHYSICS C MASTER KEY PAGE 1 OF 2
TEACHER ANSWER KEY — RED SOLUTIONS EDITION
Problem Set 2 Key Page 2 Solutions
5 Dual Head-On Train Intercept (1D Relational Passing)
Two model trains, Train A and Train B, are on parallel straight tracks facing each other, separated by a distance \(L = 40.0 \, \text{m}\) at \(t = 0\). Train A starts from rest and accelerates toward Train B at a constant \(a_A = 1.60 \, \text{m/s}^2\). Train B starts at the same instant from rest and accelerates toward Train A with constant \(a_B = 1.20 \, \text{m/s}^2\). Find the time and position when they pass each other.
• Train A Position: \(x_A(t) = \frac{1}{2}a_At^2 = 0.80t^2\)
• Train B Position: \(x_B(t) = 40.0 - \frac{1}{2}a_Bt^2 = 40.0 - 0.60t^2\)
• Passing Point: \(0.80t^2 = 40.0 - 0.60t^2 \implies 1.40t^2 = 40.0 \implies t^2 = \frac{40.0}{1.40} \approx 28.57\)
• Solve (Simple Square Root): \(t = \sqrt{28.57} = \mathbf{5.35 \, \text{seconds}}\)
• Position: \(x_A = 0.80(28.57) = \mathbf{22.86 \, \text{meters from Train A's initial position}}\)
6 The Runway Takeoff Threshold (Constant Acceleration Limits)
A cargo airplane requires a minimum take-off speed of \(v_{\text{take-off}} = 80.0 \, \text{m/s}\). The plane starts from rest at the beginning of a straight runway of length \(L = 1000 \, \text{m}\). If the engines provide a constant acceleration of \(a = 3.00 \, \text{m/s}^2\), calculate the runway distance remaining when the plane reaches take-off speed, and check if the takeoff is successful.
• Takeoff Distance: \(v_f^2 = v_i^2 + 2ad \implies 80.0^2 = 0 + 2(3.00)d_{\text{req}} \implies d_{\text{req}} = \frac{6400}{6.00} \approx 1066.67 \, \text{m}\)
• Runway distance remaining: \(L - d_{\text{req}} = 1000 - 1066.67 = \mathbf{-66.67 \, \text{meters}}\)
• Takeoff Status: \(\mathbf{\text{Takeoff is UNSUCCESSFUL. Plane overshoots runway by 66.67 m.}}\)
7 Overtaking at a Stoplight (Linear Time Pursuit)
A motorcycle is stopped at a red light. An SUV passes the motorcycle at a constant speed of \(v_s = 20.0 \, \text{m/s}\). At that exact instant, the light turns green and the motorcycle accelerates from rest at a constant \(a_m = 4.00 \, \text{m/s}^2\) to chase the SUV. Find the time and distance from the intersection where the motorcycle overtakes the SUV.
• Motorcycle Position: \(x_m(t) = \frac{1}{2}a_m t^2 = 2.00t^2\)
• SUV Position: \(x_s(t) = v_s t = 20.0t\)
• Overtake: \(x_m(t) = x_s(t) \implies 2.00t^2 = 20.0t \implies 2.00t = 20.0 \implies \mathbf{t = 10.0 \, \text{seconds}}\)
• Distance Covered: \(x = 2.00(10.0)^2 = \mathbf{200.0 \, \text{meters from the intersection.}}\)
8 Rocket Sled Deceleration Test (Sequential Constant Rates)
A rocket-propelled sled on a straight track accelerates from rest at \(a_1 = 30.0 \, \text{m/s}^2\) for \(t_1 = 4.00\) s. The rockets then cut off, and water-brakes deploy, decelerating the sled at a constant rate of \(a_2 = -15.0 \, \text{m/s}^2\) until its speed is reduced to \(v_2 = 20.0 \, \text{m/s}\). Calculate the total distance traveled during this entire test.
• Stage 1 (Accel): \(v_1 = a_1t_1 = (30.0)(4.00) = 120.0 \, \text{m/s} \quad d_1 = 0.5 a_1 t_1^2 = 0.5(30.0)(16.0) = 240.0 \, \text{m}\)
• Stage 2 (Decel): \(v_2^2 = v_1^2 + 2a_2d_2 \implies 20.0^2 = 120.0^2 + 2(-15.0)d_2\)
• Solve Stage 2: \(400 = 14400 - 30.0d_2 \implies 30.0d_2 = 14000 \implies d_2 \approx 466.67 \, \text{m}\)
• Total Distance: \(D = d_1 + d_2 = 240.0 + 466.67 = \mathbf{706.67 \, \text{meters}}\)
UNIT 1: KINEMATICS — AP PHYSICS C MASTER KEY PAGE 2 OF 2
