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2026 AP Calculus AB FRQ Question 5: Remote-Controlled Car Motion

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2026.

Question 5

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Remote-Controlled Car Motion

Particle Motion

Hard
A remote-controlled toy car moves back and forth along a straight path so that its velocity at time tt is given by the function vv, where v(t)v(t) is measured in feet per second and tt is measured in seconds. v(t)={t4−8t3+16t20≤t≤404<t<610cos⁡ ⁣(π3t)−106≤t≤12v(t) = \begin{cases} t^4 - 8t^3 + 16t^2 & 0 \leq t \leq 4 \\ 0 & 4 < t < 6 \\ 10\cos\!\left(\frac{\pi}{3}t\right) - 10 & 6 \leq t \leq 12 \end{cases} where v(t)v(t) is in ft/sec and tt in seconds.
 The graph of v(t)v(t) is shown.
2026 AP Calculus AB FRQ Question 5 Graph
Part AEasy2 points
Find the acceleration of the car at time t=1t = 1 second. Show the work that leads to your answer.

Answer

a(1)=v′(1)=4(1)3−24(1)2+32(1)=4−24+32=12a(1) = v'(1) = 4(1)^3 - 24(1)^2 + 32(1) = 4 - 24 + 32 = 12 ft/sec²
Full Solution & Work

Differentiate v(t) on [0, 4]

For 0≤t≤40 \leq t \leq 4:
v(t)=t4−8t3+16t2v(t) = t^4 - 8t^3 + 16t^2

a(t)=v′(t)=4t3−24t2+32ta(t) = v'(t) = 4t^3 - 24t^2 + 32t

Evaluate at t = 1

a(1)=4(1)3−24(1)2+32(1)=4−24+32=12 ft/sec2a(1) = 4(1)^3 - 24(1)^2 + 32(1) = 4 - 24 + 32 = 12 \text{ ft/sec}^2

AP Scoring — 2 Points

**P1**: Correct derivative 4t3−24t2+32t4t^3 - 24t^2 + 32t.
**P2**: Correct answer a(1)=12a(1) = 12 ft/sec².
Part BEasy1 point
Is the car speeding up or slowing down at time t=1t = 1 second? Give a reason for your answer.

Answer

The car is speeding up at t=1t = 1 because v(1)>0v(1) > 0 and a(1)>0a(1) > 0 (same sign).
Full Solution & Work

Find v(1)

v(1)=1−8+16=9>0v(1) = 1 - 8 + 16 = 9 > 0

Compare signs of v and a

Since v(1)=9>0v(1) = 9 > 0 and a(1)=12>0a(1) = 12 > 0, velocity and acceleration have the **same sign**. The car is **speeding up**.

AP Scoring — 1 Point

**P1**: 'Speeding up' with justification that v(1)v(1) and a(1)a(1) have the same sign.
Part CMedium3 points
Find the distance, in feet, that the car traveled over 0≤t≤40 \leq t \leq 4 seconds.Show the work that leads to your answer.

Answer

Distance=∫04∣v(t)∣ dt=∫04v(t) dt=[t55−2t4+16t33]04=10245−512+10243=51215≈34.133\text{Distance} = \int_0^4 |v(t)|\,dt = \int_0^4 v(t)\,dt = \left[\frac{t^5}{5} - 2t^4 + \frac{16t^3}{3}\right]_0^4 = \frac{1024}{5} - 512 + \frac{1024}{3} = \frac{512}{15} \approx 34.133 ft
Full Solution & Work

Check if v changes sign on [0, 4]

Factor: v(t)=t2(t2−8t+16)=t2(t−4)2≥0v(t) = t^2(t^2 - 8t + 16) = t^2(t-4)^2 \geq 0 for all tt. So ∣v(t)∣=v(t)|v(t)| = v(t) on [0,4][0,4] — no sign change.

Set up and evaluate the integral

Distance=∫04t2(t−4)2 dt=∫04(t4−8t3+16t2) dt\text{Distance} = \int_0^4 t^2(t-4)^2\,dt = \int_0^4 (t^4 - 8t^3 + 16t^2)\,dt

=[t55−2t4+16t33]04= \left[\frac{t^5}{5} - 2t^4 + \frac{16t^3}{3}\right]_0^4

=10245−2(256)+16(64)3=10245−512+10243= \frac{1024}{5} - 2(256) + \frac{16(64)}{3} = \frac{1024}{5} - 512 + \frac{1024}{3}

Simplify

Common denominator 15:
=307215−768015+512015=51215≈34.133 feet= \frac{3072}{15} - \frac{7680}{15} + \frac{5120}{15} = \frac{512}{15} \approx 34.133 \text{ feet}

AP Scoring — 3 Points

**P1**: Recognizes distance = ∫04∣v(t)∣ dt\int_0^4 |v(t)|\,dt and justifies v(t)≥0v(t) \geq 0.
**P2**: Correct antiderivative.
**P3**: Correct answer 51215\frac{512}{15} ft.
Part DMedium3 points
Find the average velocity of the car over the time interval 6≤t≤126 \leq t \leq 12 seconds. Show the work that leads to your answer.

Answer

vˉ=16∫612v(t) dt=16[10⋅3πsin⁡ ⁣(π3t)−10t]612=−10\bar{v} = \dfrac{1}{6}\int_6^{12} v(t)\,dt = \dfrac{1}{6}\left[10\cdot\dfrac{3}{\pi}\sin\!\left(\dfrac{\pi}{3}t\right) - 10t\right]_6^{12} = -10 ft/sec
Full Solution & Work

Set up the average velocity formula

vˉ=112−6∫612v(t) dt=16∫612[10cos⁡ ⁣(π3t)−10]dt\bar{v} = \frac{1}{12-6}\int_6^{12} v(t)\,dt = \frac{1}{6}\int_6^{12}\left[10\cos\!\left(\frac{\pi}{3}t\right) - 10\right]dt

Find the antiderivative

∫[10cos⁡ ⁣(π3t)−10]dt=30πsin⁡ ⁣(π3t)−10t+C\int \left[10\cos\!\left(\frac{\pi}{3}t\right) - 10\right]dt = \frac{30}{\pi}\sin\!\left(\frac{\pi}{3}t\right) - 10t + C

Evaluate from 6 to 12

[30πsin⁡ ⁣(π3t)−10t]612\left[\frac{30}{\pi}\sin\!\left(\frac{\pi}{3}t\right) - 10t\right]_6^{12}

At t=12t=12: 30πsin⁡(4π)−120=0−120=−120\frac{30}{\pi}\sin(4\pi) - 120 = 0 - 120 = -120.
At t=6t=6: 30πsin⁡(2π)−60=0−60=−60\frac{30}{\pi}\sin(2\pi) - 60 = 0 - 60 = -60.
=−120−(−60)=−60= -120 - (-60) = -60

Compute average velocity

vˉ=−606=−10 ft/sec\bar{v} = \frac{-60}{6} = -10 \text{ ft/sec}

The negative sign indicates net motion in the negative direction.

AP Scoring — 3 Points

**P1**: Correct average velocity setup 16∫612v(t) dt\frac{1}{6}\int_6^{12} v(t)\,dt.
**P2**: Correct antiderivative.
**P3**: Correct answer −10-10 ft/sec.

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