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2022 AP Calculus AB FRQ Question 6: Two Particles in Motion

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 6

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Two Particles in Motion

Particle Motion

Hard
Particle PP moves along the xx-axis such that, for time t>0t>0, its position is given by xP(t)=6−4e−tx_P(t)=6-4e^{-t}. Particle QQ moves along the yy-axis such that, for time t>0t>0, its velocity is given by vQ(t)=1t2v_Q(t)=\dfrac{1}{t^2}. At time t=1t=1, the position of particle QQ is yQ(1)=2y_Q(1)=2.
Part AEasy1 point
Find vP(t)v_P(t), the velocity of particle PP at time tt.

Answer

vP(t)=4e−tv_P(t)=4e^{-t}
Full Solution & Work

Differentiate x_P(t)

vP(t)=xP′(t)=−4(−e−t)=4e−tv_P(t) = x_P'(t) = -4(-e^{-t}) = 4e^{-t}

AP Scoring — 1 Point

**P1**: Correct answer, vP(t)=4e−tv_P(t)=4e^{-t}.
Part BHard3 points
Find aQ(t)a_Q(t), the acceleration of particle QQ at time tt. Find all times tt, for t>0t>0, when the speed of particle QQ is decreasing. Justify your answer.

Answer

aQ(t)=−2t3a_Q(t)=-\dfrac{2}{t^3}; speed is decreasing for all t>0t>0.
Full Solution & Work

Differentiate v_Q(t)

aQ(t)=vQ′(t)=−2t3a_Q(t) = v_Q'(t) = -\frac{2}{t^3}

Compare signs of v_Q and a_Q for t > 0

vQ(t)=1t2>0v_Q(t)=\frac{1}{t^2}>0 for all t>0t>0. aQ(t)=−2t3<0a_Q(t)=-\frac{2}{t^3}<0 for all t>0t>0. So vQv_Q and aQa_Q have **opposite signs** for every t>0t>0.

Conclude

The speed of particle QQ is **decreasing for all t>0t>0** — the particle continually slows as it moves.

AP Scoring — 3 Points

**P1**: Correct aQ(t)=−2t3a_Q(t)=-\frac{2}{t^3}.
**P2**: Correct sign analysis showing vQ(t)v_Q(t) and aQ(t)a_Q(t) have opposite signs.
**P3**: Correct conclusion — for **all** t>0t>0, with justification.
Part CMedium2 points
Find yQ(t)y_Q(t), the position of particle QQ at time tt.

Answer

yQ(t)=3−1ty_Q(t)=3-\dfrac{1}{t}
Full Solution & Work

Set up position as initial position plus accumulated change

yQ(t)=yQ(1)+∫1tvQ(s) ds=2+∫1ts−2 dsy_Q(t) = y_Q(1)+\int_1^t v_Q(s)\,ds = 2+\int_1^t s^{-2}\,ds

Evaluate the integral

=2+[−s−1]1t=2+(−1t+1)=3−1t= 2+\left[-s^{-1}\right]_1^t = 2+\left(-\frac1t+1\right) = 3-\frac1t

AP Scoring — 2 Points

**P1**: Correct setup yQ(1)+∫1tvQ(s) dsy_Q(1)+\int_1^t v_Q(s)\,ds.
**P2**: Correct final answer, yQ(t)=3−1ty_Q(t)=3-\frac1t.
Part DMedium2 points
As t→∞t\to\infty, which particle will eventually be farther from the origin? Give a reason for your answer.

Answer

Particle P.
Full Solution & Work

Find the limiting positions

lim⁡t→∞xP(t)=lim⁡t→∞(6−4e−t)=6,lim⁡t→∞yQ(t)=lim⁡t→∞(3−1t)=3\lim_{t\to\infty} x_P(t) = \lim_{t\to\infty}\big(6-4e^{-t}\big) = 6, \qquad \lim_{t\to\infty} y_Q(t) = \lim_{t\to\infty}\left(3-\frac1t\right)=3

Compare

Both xP(t)x_P(t) and yQ(t)y_Q(t) increase monotonically toward their respective limits, staying below them for all finite tt. Since lim⁡xP=6>lim⁡yQ=3\lim x_P = 6 > \lim y_Q=3, particle PP's distance from the origin approaches a strictly larger value.

Conclude

As t→∞t\to\infty, particle **PP** will eventually be farther from the origin.

AP Scoring — 2 Points

**P1**: Correctly computes both limits, 6 and 3.
**P2**: Correct conclusion (particle PP) with a reason comparing the two limits.

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