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2021 AP Calculus AB FRQ Question 2: Two Particles Moving on the x-axis

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 2

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Two Particles Moving on the x-axis

Particle Motion

Hard
A particle, PP, is moving along the xx-axis. The velocity of particle PP at time tt is given by vP(t)=sin⁡(t1.5)v_P(t)=\sin(t^{1.5}) for 0≤t≤π0\leq t\leq\pi. At time t=0t=0, particle PP is at position x=5x=5. A second particle, QQ, also moves along the xx-axis. The velocity of particle QQ at time tt is given by vQ(t)=(t−1.8)⋅1.25tv_Q(t)=(t-1.8)\cdot1.25^t for 0≤t≤π0\leq t\leq\pi. At time t=0t=0, QQ is at position x=10x=10.
Part AMedium2 points
Find the positions of particles PP and QQ at time t=1t=1.

Answer

xP(1)≈5.371x_P(1)\approx5.371; xQ(1)≈8.564x_Q(1)\approx8.564
Full Solution & Work

Set up each position as initial position plus accumulated change

xP(1)=5+∫01vP(t) dt,xQ(1)=10+∫01vQ(t) dtx_P(1) = 5+\int_0^1 v_P(t)\,dt, \qquad x_Q(1)=10+\int_0^1 v_Q(t)\,dt

Evaluate with a calculator

xP(1)≈5+0.371=5.371,xQ(1)≈10−1.436=8.564x_P(1) \approx 5+0.371 = 5.371, \qquad x_Q(1) \approx 10-1.436 = 8.564

AP Scoring — 2 Points

**P1**: Correct value, xP(1)≈5.371x_P(1)\approx5.371.
**P2**: Correct value, xQ(1)≈8.564x_Q(1)\approx8.564.
Part BMedium2 points
Are particles PP and QQ moving toward each other or away from each other at time t=1t=1? Explain your reasoning.

Answer

Moving toward each other.
Full Solution & Work

Evaluate velocities at t = 1

vP(1)=sin⁡(1)≈0.841>0,vQ(1)=(1−1.8)(1.25)≈−1<0v_P(1)=\sin(1)\approx0.841>0, \qquad v_Q(1)=(1-1.8)(1.25)\approx-1<0

Interpret using relative positions

From part (a), QQ is ahead of PP (xQ(1)≈8.564>xP(1)≈5.371x_Q(1)\approx8.564 > x_P(1)\approx5.371). Since vP(1)>0v_P(1)>0, PP is moving right (toward QQ); since vQ(1)<0v_Q(1)<0, QQ is moving left (toward PP).

Conclude

Since PP moves toward QQ and QQ moves toward PP, the two particles are **moving toward each other** at t=1t=1.

AP Scoring — 2 Points

**P1**: Correctly evaluates the signs of vP(1)v_P(1) and vQ(1)v_Q(1).
**P2**: Correct conclusion ("toward each other"), tied to the relative positions of PP and QQ.
Part CMedium2 points
Find the acceleration of particle QQ at time t=1t=1. Is the speed of particle QQ increasing or decreasing at time t=1t=1? Give a reason for your answer.

Answer

aQ(1)≈1.027a_Q(1)\approx1.027; speed is decreasing.
Full Solution & Work

Differentiate v_Q(t) with the Product Rule

aQ(t)=vQ′(t)=1.25t+(t−1.8)⋅1.25tln⁡(1.25)=1.25t[1+(t−1.8)ln⁡1.25]a_Q(t) = v_Q'(t) = 1.25^t + (t-1.8)\cdot1.25^t\ln(1.25) = 1.25^t\big[1+(t-1.8)\ln1.25\big]

Evaluate at t = 1

aQ(1)≈1.027a_Q(1) \approx 1.027

Compare signs of v_Q and a_Q

vQ(1)≈−1<0v_Q(1)\approx-1<0 and aQ(1)≈1.027>0a_Q(1)\approx1.027>0 — opposite signs, so the speed of particle QQ is **decreasing** at t=1t=1.

AP Scoring — 2 Points

**P1**: Correct value, aQ(1)≈1.027a_Q(1)\approx1.027, via the Product Rule.
**P2**: Correct conclusion ("decreasing") tied to the opposite signs of vQ(1)v_Q(1) and aQ(1)a_Q(1).
Part DHard3 points
Find the total distance traveled by particle PP over the time interval 0≤t≤π0\leq t\leq\pi.

Answer

≈ 1.931
Full Solution & Work

Set up the total-distance integral

Distance=∫0π∣vP(t)∣ dt=∫0π∣sin⁡(t1.5)∣ dt\text{Distance} = \int_0^\pi |v_P(t)|\,dt = \int_0^\pi \big|\sin(t^{1.5})\big|\,dt

Note the sign change

sin⁡(t1.5)=0\sin(t^{1.5})=0 when t1.5=πt^{1.5}=\pi, i.e. t=π2/3≈2.145t=\pi^{2/3}\approx2.145. vP(t)>0v_P(t)>0 on (0,2.145)(0,2.145) and vP(t)<0v_P(t)<0 on (2.145,π)(2.145,\pi).

Evaluate

∫0π∣vP(t)∣ dt≈1.931\int_0^\pi |v_P(t)|\,dt \approx 1.931

AP Scoring — 3 Points

**P1**: Correct setup, ∫0π∣vP(t)∣ dt\int_0^\pi |v_P(t)|\,dt, with reference to the sign change at t=π2/3≈2.145t=\pi^{2/3}\approx2.145.
**P2**: Splits the integral correctly at the sign change (or uses a calculator's absolute-value integration directly).
**P3**: Correct final answer, 1.931.

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