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2023 AP Calculus AB FRQ Question 2: Stephen's Swim — Particle Motion

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2023.

Question 2

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Stephen's Swim — Particle Motion

Particle Motion

Hard
Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen's velocity is modeled by v(t)=2.38e−0.02tsin⁡ ⁣(π56t)v(t) = 2.38e^{-0.02t}\sin\!\left(\dfrac{\pi}{56}t\right), where tt is measured in seconds and v(t)v(t) is measured in meters per second.
Part AEasy2 points
Find all times tt in the interval 0<t<900<t<90 at which Stephen changes direction. Give a reason for your answer.

Answer

t=56t=56 seconds
Full Solution & Work

Solve v(t) = 0 on (0, 90)

v(t)=0  ⟹  t=56v(t)=0 \implies t=56

Interpret the sign change

Stephen changes direction when his velocity changes sign. This occurs at t=56t=56 seconds.

AP Scoring — 2 Points

**P1**: Considers v(t)=0v(t)=0.
**P2**: The answer, t=56t=56, with reason — a response of just "t=56t=56" with no supporting work does not earn this point.
Part BMedium3 points
Find Stephen's acceleration at time t=60t=60 seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time t=60t=60 seconds? Give a reason for your answer.

Answer

a(60)≈−0.036a(60)\approx -0.036 m/sec²; Stephen is speeding up.
Full Solution & Work

Relate acceleration to v'(t)

v′(60)=a(60)=−0.0360162v'(60) = a(60) = -0.0360162

Stephen's acceleration at t=60t=60 seconds is −0.036-0.036 meter per second per second.

Compare signs of v and a

v(60)=−0.1595124<0v(60) = -0.1595124 < 0. Because Stephen's velocity and acceleration are both negative at t=60t=60, Stephen is **speeding up** at that time.

AP Scoring — 3 Points

**P1**: The minimum work needed is v′(60)=−0.036v'(60)=-0.036 — evaluating only a(60)=−0.0360162a(60)=-0.0360162 without explicitly connecting it to v′(t)v'(t) is not sufficient.
**P2**: A declared value for a(60)a(60) (units).
**P3**: Requires a presented value of v(60)v(60) correct to at least 1 decimal place, and consistent conclusion — "speeding up because a(60)a(60) and v(60)v(60) have the same sign" is sufficient without an explicit numeric v(60)v(60).
Part CMedium2 points
Find the distance between Stephen's position at time t=20t=20 seconds and his position at time t=80t=80 seconds. Show the setup for your calculations.

Answer

≈ 23.384 (or 23.383) meters
Full Solution & Work

Set up the definite integral

∫2080v(t) dt\int_{20}^{80} v(t)\,dt

Evaluate

=23.383997= 23.383997

The distance between Stephen's positions at t=20t=20 and t=80t=80 seconds is **23.384** (or 23.383) meters.

AP Scoring — 2 Points

**P1**: Only for ∫2080v(t) dt\int_{20}^{80} v(t)\,dt (with or without the differential).
**P2**: Only for the answer 23.384 (or 23.383), regardless of whether the first point was earned.
Part DMedium2 points
Find the total distance Stephen swims over the time interval 0≤t≤900 \leq t \leq 90 seconds. Show the setup for your calculations.

Answer

≈ 62.164 meters
Full Solution & Work

Set up the total-distance integral

∫090∣v(t)∣ dt\int_0^{90} |v(t)|\,dt

Evaluate

=62.164216= 62.164216

The total distance Stephen swims over 0≤t≤900\leq t\leq 90 seconds is **62.164** meters.

AP Scoring — 2 Points

**P1**: Only for ∫090∣v(t)∣ dt\int_0^{90}|v(t)|\,dt, or the equivalent ∫056v(t) dt−∫5690v(t) dt\int_0^{56}v(t)\,dt - \int_{56}^{90}v(t)\,dt, with or without differentials.
**P2**: Only for the answer 62.164, regardless of whether the first point was earned.

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