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2024 AP Calculus AB FRQ Question 2: Particle Velocity — Logarithmic Model

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2024.

Question 2

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Particle Velocity — Logarithmic Model

Particle Motion

Hard
A particle moves along the xx-axis so that its velocity at time t≥0t \geq 0 is given by v(t)=ln⁡(t2−4t+5)−0.2tv(t) = \ln(t^2-4t+5) - 0.2t.
Part AEasy2 points
There is one time, t=tRt=t_R, in the interval 0<t<20<t<2 when the particle is at rest (not moving). Find tRt_R. For 0<t<tR0<t<t_R, is the particle moving to the right or to the left? Give a reason for your answer.

Answer

tR=1.426t_R = 1.426 (or 1.4251.425); moving right for 0<t<tR0<t<t_R.
Full Solution & Work

Solve v(t) = 0 on (0,2)

v(t)=0  ⟹  t=1.425610v(t) = 0 \implies t = 1.425610

So tR=1.426t_R = 1.426 (or 1.425).

Determine direction on (0, t_R)

For 0<t<tR0<t<t_R, v(t)>0v(t)>0. Therefore, the particle is moving to the **right** on that interval.

AP Scoring — 2 Points

**P1**: The value tR=1.426t_R=1.426 (or 1.425) — a response that finds no value or an incorrect value of tRt_R is not eligible for the second point.
**P2**: Correct direction (right) with the explanation that v(t)>0v(t)>0 on (0,tR)(0,t_R).
Part BMedium2 points
Find the acceleration of the particle at time t=1.5t=1.5. Show the setup for your calculations. Is the speed of the particle increasing or decreasing at time t=1.5t=1.5? Explain your reasoning.

Answer

a(1.5)=v′(1.5)=−1a(1.5)=v'(1.5)=-1; speed is increasing at t=1.5t=1.5.
Full Solution & Work

Relate acceleration to v'(t)

a(1.5)=v′(1.5)=−1 (or −0.999)a(1.5) = v'(1.5) = -1 \text{ (or } -0.999\text{)}

Compare signs of v and a

v(1.5)=−0.076856<0v(1.5) = -0.076856 < 0. Because a(1.5)a(1.5) and v(1.5)v(1.5) have the **same sign** (both negative), the speed of the particle is **increasing** at t=1.5t=1.5.

AP Scoring — 2 Points

**P1**: A response must demonstrate the relationship v′=av'=a to earn this point — just stating "a(1.5)=−1a(1.5)=-1" with no connection to v′v' is not sufficient.
**P2**: Earned only for a response consistent with a negative velocity at t=1.5t=1.5 and the presented value of a(1.5)a(1.5).
Part CMedium3 points
The position of the particle at time tt is x(t)x(t), and its position at time t=1t=1 is x(1)=−3x(1)=-3. Find the position of the particle at time t=4t=4. Show the setup for your calculations.

Answer

x(4)=x(1)+∫14v(t) dt=−3+0.197117=−2.803x(4) = x(1) + \displaystyle\int_1^4 v(t)\,dt = -3+0.197117=-2.803
Full Solution & Work

Set up x(4) using x(1) plus accumulated change

x(4)=x(1)+∫14v(t) dtx(4) = x(1) + \int_1^4 v(t)\,dt

Evaluate

=−3+0.197117=−2.802883= -3+0.197117 = -2.802883

The position of the particle at time t=4t=4 is **−2.803-2.803** (or −2.802-2.802).

AP Scoring — 3 Points

**P1**: A definite integral with integrand v(t)v(t). If the limits of integration are incorrect, this response is not eligible for the third point.
**P2**: Adding x(1)x(1) (or −3-3) to a definite integral with lower limit 1.
**P3**: Correct answer −3+0.197-3+0.197 with supporting work — an answer of just −2.803-2.803 with no supporting work earns no points.
Part DMedium2 points
Find the total distance traveled by the particle over the interval 1≤t≤41 \leq t \leq 4. Show the setup for your calculations.

Answer

∫14∣v(t)∣ dt=0.958\displaystyle\int_1^4 |v(t)|\,dt = 0.958
Full Solution & Work

Set up the distance integral

Distance=∫14∣v(t)∣ dt\text{Distance} = \int_1^4 |v(t)|\,dt

Evaluate with a calculator

=0.9581= 0.9581

The total distance traveled by the particle over 1≤t≤41 \leq t \leq 4 is **0.958**.

AP Scoring — 2 Points

**P1**: The integral ∫14∣v(t)∣ dt\int_1^4 |v(t)|\,dt, or an equivalent split at the sign changes of vv (at t≈1.425t\approx1.425 and t≈2.883t\approx2.883, the other root of v(t)=0v(t)=0).
**P2**: Due to variations in numerical integration across calculator models, answers of 0.958, 0.959, or 0.96 all earn this point.

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