Calculus-Master

2025 AP Calculus AB FRQ Question 5: Two Particles in Motion

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2025.

Question 5

No Calculator

Two Particles in Motion

Particle Motion

Hard
Two particles, HH and JJ, are moving along the xx-axis. For 0≤t≤50 \leq t \leq 5, the position of particle HH at time tt is given by xH(t)=et2−4tx_H(t) = e^{t^2-4t} and the velocity of particle JJ at time tt is given by vJ(t)=2t(t2−1)3v_J(t) = 2t(t^2-1)^3.
Part AEasy2 points
Find the velocity of particle HH at time t=1t = 1. Show the work that leads to your answer.

Answer

vH(1)=xH′(1)=−2e−3v_H(1) = x_H'(1) = -2e^{-3}
Full Solution & Work

Differentiate x_H(t) with the Chain Rule

xH′(t)=vH(t)=(2t−4)et2−4tx_H'(t) = v_H(t) = (2t-4)e^{t^2-4t}

Evaluate at t = 1

vH(1)=(2−4)e1−4=−2e−3v_H(1) = (2-4)e^{1-4} = -2e^{-3}

AP Scoring — 2 Points

**P1**: Considers xH′(t)x_H'(t) — presenting xH′(t)x_H'(t), xH′(1)x_H'(1), or (2t−4)et2−4t(2t-4)e^{t^2-4t} earns this.
**P2**: Correct answer −2e−3-2e^{-3}. An unsupported answer of −2e−3-2e^{-3} alone earns P2 but not P1.
Part BHard3 points
During what open intervals of time tt, for 0<t<50 < t < 5, are particles HH and JJ moving in opposite directions? Give a reason for your answer.

Answer

1<t<21 < t < 2
Full Solution & Work

Analyze the sign of v_H(t)

vH(t)=(2t−4)et2−4t=0  ⟹  t=2v_H(t) = (2t-4)e^{t^2-4t} = 0 \implies t=2 (the exponential factor is never 0).
vH(t)<0v_H(t) < 0 for 0<t<20<t<2 (moving left); vH(t)>0v_H(t) > 0 for 2<t<52<t<5 (moving right).

Analyze the sign of v_J(t)

vJ(t)=2t(t2−1)3=0  ⟹  t=0v_J(t) = 2t(t^2-1)^3 = 0 \implies t=0 or t=1t=1 on (0,5)(0,5).
For 0<t<10<t<1: (t2−1)<0(t^2-1)<0, so (t2−1)3<0(t^2-1)^3<0, and 2t>02t>0, giving vJ(t)<0v_J(t)<0 (moving left).
For 1<t<51<t<5: (t2−1)>0(t^2-1)>0, so (t2−1)3>0(t^2-1)^3>0, giving vJ(t)>0v_J(t)>0 (moving right).

Compare directions

HH: left on (0,2)(0,2), right on (2,5)(2,5).
JJ: left on (0,1)(0,1), right on (1,5)(1,5).
The signs disagree only on (1,2)(1,2): HH is still moving left there while JJ has already turned right.

AP Scoring — 3 Points

**P3**: Sets vH(t)=0v_H(t)=0 or vJ(t)=0v_J(t)=0 (or identifies t=2t=2 for HH and no other value, or t=1t=1 for JJ and no other value, or identifies the interval 1<t<21<t<2 outright).
**P4**: A correct sign/direction analysis for at least one of the two particles on (0,5)(0,5).
**P5**: Correct analyses for **both** particles, leading to the answer 1<t<21<t<2.
Part CMedium1 point
It can be shown that vJ′(2)>0v_J'(2) > 0. Is the speed of particle JJ increasing, decreasing, or neither at time t=2t = 2? Give a reason for your answer.

Answer

Increasing, because vJ(2)>0v_J(2) > 0 and vJ′(2)>0v_J'(2) > 0 have the same sign.
Full Solution & Work

Find the sign of v_J(2)

vJ(2)=2(2)(22−1)3=4(27)=108>0v_J(2) = 2(2)\big(2^2-1\big)^3 = 4(27) = 108 > 0

Compare signs of velocity and acceleration

Since vJ(2)=108>0v_J(2)=108>0 and vJ′(2)>0v_J'(2)>0 (given), velocity and acceleration have the **same sign**, so the speed of particle JJ is **increasing** at t=2t=2.

AP Scoring — 1 Point

**P6**: "Increasing," with the reason that vJ(2)v_J(2) and vJ′(2)v_J'(2) have the same sign. (An explicit value for vJ(2)v_J(2) is not required, but if given it must be correct — vJ(2)=108v_J(2)=108.)
Part DMedium3 points
Particle JJ is at position x=7x = 7 at time t=0t = 0. Find the position of particle JJ at time t=2t = 2. Show the work that leads to your answer.

Answer

xJ(2)=7+14[(3)4−(−1)4]=27x_J(2) = 7 + \dfrac{1}{4}\big[(3)^4-(-1)^4\big] = 27
Full Solution & Work

Set up position as initial position + accumulated change

xJ(2)=xJ(0)+∫02vJ(t) dt=7+∫022t(t2−1)3 dtx_J(2) = x_J(0) + \int_0^2 v_J(t)\,dt = 7 + \int_0^2 2t(t^2-1)^3\,dt

Substitute u = t² − 1

Let u=t2−1u=t^2-1, du=2t dtdu=2t\,dt. When t=0t=0, u=−1u=-1; when t=2t=2, u=3u=3.
∫022t(t2−1)3 dt=∫−13u3 du=[u44]−13=81−14=20\int_0^2 2t(t^2-1)^3\,dt = \int_{-1}^3 u^3\,du = \left[\frac{u^4}{4}\right]_{-1}^3 = \frac{81-1}{4}=20

Combine

xJ(2)=7+20=27x_J(2) = 7+20 = 27

AP Scoring — 3 Points

**P7**: An indefinite or definite integral with integrand vJ(t)v_J(t) or 2t(t2−1)32t(t^2-1)^3.
**P8**: A correct antiderivative of the form k(t2−1)4k(t^2-1)^4 (equivalently) with k=14k=\frac{1}{4} — a different constant kk forfeits eligibility for P9.
**P9**: Correct final answer xJ(2)=27x_J(2)=27, requiring P8.

Study the concept

Learn the theory behind this question type with worked examples and strategy tips.

Practice More
Gary Chang

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

Discuss with Gary on Reddit

Finished reviewing?

Keep your momentum going

Practice multiple choice80 AP-style MCQs with full explanations, or a timed 45-question exam.Estimate your AP scorePlug in your MCQ and FRQ points to see your likely 1–5.Follow the 30-day planA day-by-day review of every AP Calc AB unit, from limits to volume.Practice FRQ by topicMaster each question type before the next exam.