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[2026] AP Calculus Particle Motion FRQ: Step-by-Step SOP & Practice

A particle moves along a straight line. Given its velocity or position function, you will find displacement, total distance traveled, when the particle changes direction, and whether it is speeding up or slowing down at a given time.

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Key Formulashover to see usage
  • Average Velocity (Given Position s(t)s(t))Avg Vel=s(b)−s(a)b−a\text{Avg Vel} = \frac{s(b) - s(a)}{b - a}
    When to useUse the slope formula when you have position data. (Linked to Mean Value Theorem).
  • Average Velocity (Given Velocity v(t)v(t))Avg Vel=1b−a∫abv(t) dt\text{Avg Vel} = \frac{1}{b-a} \int_a^b v(t)\,dt
    When to useUse Mean Value Theorem for Integrals to find the 'average value' of the function v(t)v(t).
  • Average Acceleration (Given Velocity v(t)v(t))Avg Accel=v(b)−v(a)b−a\text{Avg Accel} = \frac{v(b) - v(a)}{b - a}
    When to useUse when asked for the average rate of change of velocity over [a,b][a, b].
  • Current Position s(t)s(t) (Fundamental Theorem of Calculus)s(t)=s(k)+∫ktv(x) dxs(t) = s(k) + \int_k^t v(x)\,dx
    When to useThe 'Accumulation' formula. Always start with the initial position s(k)s(k)!
  • Total Distance Traveled vs. Displacement∫ab∣v(t)∣ dt vs. ∫abv(t) dt\int_a^b |v(t)|\,dt \text{ vs. } \int_a^b v(t)\,dt
    When to useAbsolute value = Total distance (never negative). No absolute value = Net change (displacement).
  • Speeding Up or Slowing Down?v(t)⋅a(t)>0  ⟹  Speeding Upv(t) \cdot a(t) > 0 \implies \text{Speeding Up}
    When to useCompare signs of v(t)v(t) and a(t)a(t). Same sign: Speeding up; Opposite signs: Slowing down.
Step-by-Step SOP
  1. 1

    Establish the relationships

    Remember: s(t)→d/dtv(t)→d/dta(t)s(t) \xrightarrow{d/dt} v(t) \xrightarrow{d/dt} a(t). To go the other direction, integrate. Always track what you are given and what you need.
  2. 2

    Find when the particle changes direction

    Set v(t)=0v(t) = 0 and solve for t. Then check that v changes sign at that t — a zero velocity alone does not guarantee a direction change.
  3. 3

    Calculate displacement vs. total distance

    Displacement = ∫abv(t) dt\int_a^b v(t)\,dt (positive and negative areas cancel). Total distance = ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt (split the integral at every zero of v and add absolute values).
  4. 4

    Determine speeding up / slowing down

    The particle is speeding up when v(t) and a(t) have the same sign. It is slowing down when they have opposite signs.

Practice Questions

123

Question 1

2025 AP Calculus AB Exam — FRQ 5

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.
APart A
Easy
Find the velocity of particle HH at time t=1t = 1. Show the work that leads to your answer.
Need a Hint?
The velocity is the derivative of the position function. Use the chain rule to find xH′(t)x_H'(t), then evaluate at t=1t = 1.
Show Solution

Find the velocity function

vH(t)=xH′(t)=ddt(et2−4t)=(2t−4)et2−4tv_H(t) = x_H'(t) = \frac{d}{dt}(e^{t^2-4t}) = (2t - 4)e^{t^2-4t}

Evaluate at t = 1

vH(1)=(2(1)−4)e12−4(1)=−2e−3v_H(1) = (2(1) - 4)e^{1^2 - 4(1)} = -2e^{-3}

AP Scoring Note — P1 + P2

P1: Earned by presenting the derivative xH′(t)x_H'(t) or the specific expression (2(1)−4)e1−4(2(1)-4)e^{1-4}.

P2: Earned for the final answer −2e−3-2e^{-3}. An unsupported answer earns P2 but loses P1.
BPart B
Hard
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.
Need a Hint?
Particles move in opposite directions when their velocities have different signs. Find where vH(t)v_H(t) and vJ(t)v_J(t) are zero to determine their signs on the interval (0,5)(0, 5).
Show Solution

Analyze Particle H

vH(t)=(2t−4)et2−4t=0⇒t=2v_H(t) = (2t-4)e^{t^2-4t} = 0 \Rightarrow t = 2.
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 Particle J

vJ(t)=2t(t2−1)3=0⇒t=1v_J(t) = 2t(t^2-1)^3 = 0 \Rightarrow t = 1 (within 0<t<50 < t < 5).
vJ(t)<0v_J(t) < 0 for 0<t<10 < t < 1 (moving left).
vJ(t)>0v_J(t) > 0 for 1<t<51 < t < 5 (moving right).

Compare signs and conclude

From 1<t<21 < t < 2, vHv_H is negative while vJv_J is positive. Therefore, the particles move in opposite directions on the interval (1,2)(1, 2).

AP Scoring Note — P3 + P4 + P5

P3: Considers the sign by setting vH(t)=0v_H(t)=0 or vJ(t)=0v_J(t)=0.

P4: Correct analysis of motion for at least one particle.

P5: Requires correct analysis for both particles and the final interval (1,2)(1, 2).
CPart C
Medium
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.
Need a Hint?
Speed increases if velocity and acceleration (v′v') have the same sign. Speed decreases if they have opposite signs.
Show Solution

Determine the sign of velocity

vJ(2)=2(2)(22−1)3=4(3)3=108v_J(2) = 2(2)(2^2 - 1)^3 = 4(3)^3 = 108, which is >0> 0.

Compare with acceleration

We are given vJ′(2)>0v_J'(2) > 0. Since both vJ(2)v_J(2) and vJ′(2)v_J'(2) are positive (same sign), the speed is increasing.

AP Scoring Note — P6

P6: Must state that velocity and acceleration have the same sign at t=2t=2. Numerical values for vJ(2)v_J(2) or vJ′(2)v_J'(2) are not required but must be correct if provided.
DPart D
Hard
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.
Need a Hint?
Use the Fundamental Theorem of Calculus: x(2)=x(0)+∫02vJ(t) dtx(2) = x(0) + \int_0^2 v_J(t) \, dt. Use uu-substitution for the integral.
Show Solution

Set up the integral

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

Integrate using substitution

Let u=t2−1,du=2t dtu = t^2 - 1, du = 2t \, dt. Limits: t=0→u=−1t=0 \to u=-1, t=2→u=3t=2 \to u=3.
∫−13u3 du=[14u4]−13=14(34−(−1)4)=14(80)=20\int_{-1}^3 u^3 \, du = [\frac{1}{4}u^4]_{-1}^3 = \frac{1}{4}(3^4 - (-1)^4) = \frac{1}{4}(80) = 20.

Final Position

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

AP Scoring Note — P7 + P8 + P9

P7: Earned for the correct integrand vJ(t)v_J(t) in a definite or indefinite integral.

P8: Earned for a correct antiderivative of the form k(t2−1)4k(t^2-1)^4.

P9: Earned for the final answer 27. Arithmetic simplification is not required to bank the point.

Question 2

2024 AP Calculus AB Exam — FRQ 2

Medium
A particle moves along the x-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.
aPart a
Easy
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.
Need a Hint?
The particle is at rest when v(t)=0v(t) = 0. Use a graphing calculator to find the zero in the specified interval. Check the sign of v(t)v(t) to determine direction.
Show Solution

Find the rest time $t_R$

v(t)=ln⁡(t2−4t+5)−0.2t=0⇒t=1.425610v(t) = \ln(t^2 - 4t + 5) - 0.2t = 0 \Rightarrow t = 1.425610. Therefore, tR≈1.426t_R \approx 1.426 (or 1.4251.425).

Determine direction

For 0<t<1.4260 < t < 1.426, v(t)>0v(t) > 0. Because the velocity is positive on this interval, the particle is moving to the right.

AP Scoring Note — P1 + P2

P1: Earned for considering v(t)=0v(t) = 0 and reporting the correct value of tRt_R.

P2: Earned for the correct direction with a valid reason based on the sign of v(t)v(t).
bPart b
Medium
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.
Need a Hint?
Acceleration is v′(t)v'(t). Use your calculator to find v′(1.5)v'(1.5). Speed increases if velocity and acceleration have the same sign.
Show Solution

Calculate acceleration

a(1.5)=v′(1.5)=−1a(1.5) = v'(1.5) = -1 (or −0.999-0.999).

Determine speed behavior

v(1.5)≈−0.076856<0v(1.5) \approx -0.076856 < 0. Since both a(1.5)<0a(1.5) < 0 and v(1.5)<0v(1.5) < 0 have the same sign, the speed of the particle is increasing at t=1.5t = 1.5.

AP Scoring Note — P3 + P4

P3: Requires showing the setup (like v′(1.5)v'(1.5)) and the numerical acceleration value.

P4: Requires a conclusion consistent with the signs of both velocity and acceleration.
cPart c
Medium
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.
Need a Hint?
Use the Fundamental Theorem of Calculus: x(b)=x(a)+∫abv(t)dtx(b) = x(a) + \int_a^b v(t) dt.
Show Solution

Set up the position integral

x(4)=x(1)+∫14v(t)dt=−3+∫14(ln⁡(t2−4t+5)−0.2t)dtx(4) = x(1) + \int_{1}^{4} v(t) dt = -3 + \int_{1}^{4} (\ln(t^2 - 4t + 5) - 0.2t) dt

Calculate final position

x(4)=−3+0.197117=−2.802883x(4) = -3 + 0.197117 = -2.802883. The position is approximately −2.803-2.803 (or −2.802-2.802).

AP Scoring Note — P5 + P6 + P7

P5: Earned for the definite integral of v(t)v(t).

P6: Earned for correctly using the initial condition x(1)=−3x(1) = -3.

P7: Earned for the correct final numerical answer.
dPart d
Medium
Find the total distance traveled by the particle over the interval 1≤t≤41 \leq t \leq 4. Show the setup for your calculations.
Need a Hint?
Total distance is the integral of speed: ∫ab∣v(t)∣dt\int_a^b |v(t)| dt.
Show Solution

Set up total distance integral

Total Distance =∫14∣v(t)∣dt= \int_{1}^{4} |v(t)| dt

Calculate the value

Using a calculator, ∫14∣v(t)∣dt≈0.958\int_{1}^{4} |v(t)| dt \approx 0.958.

AP Scoring Note — P8 + P9

P8: Earned for the integral of the absolute value of velocity.

P9: Earned for the correct numerical answer.

Question 3

2023 AP Calculus AB Exam — FRQ 2

Medium
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(\frac{\pi}{56}t\right), where tt is measured in seconds and v(t)v(t) is measured in meters per second.
APart A
Easy
Find all times tt in the interval 0<t<900 < t < 90 at which Stephen changes direction. Give a reason for your answer.
Need a Hint?
A change in direction occurs when the velocity v(t)v(t) changes its sign. Set v(t)=0v(t) = 0 and solve for tt.
Show Solution

Find when velocity is zero

Set v(t)=2.38e−0.02tsin⁡(π56t)=0v(t) = 2.38e^{-0.02t}\sin\left(\frac{\pi}{56}t\right) = 0. Since 2.38e−0.02t2.38e^{-0.02t} is never zero, we solve sin⁡(π56t)=0\sin\left(\frac{\pi}{56}t\right) = 0 for 0<t<900 < t < 90.

Identify the time and verify sign change

The solution is π56t=π\frac{\pi}{56}t = \pi, which gives t=56t = 56 seconds. At t=56t = 56, v(t)v(t) changes from positive to negative.

Conclusion

Stephen changes direction at t=56t = 56 seconds because his velocity changes sign at this time.

AP Scoring Note — P1 + P2

P1: Earned for considering the sign of v(t)v(t) or setting v(t)=0v(t) = 0 .

P2: Earned for the correct answer t=56t = 56 with a reason involving the velocity changing sign.
BPart B
Medium
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.
Need a Hint?
Acceleration is the derivative of velocity: a(t)=v′(t)a(t) = v'(t). Speeding up requires velocity and acceleration to have the same sign.
Show Solution

Calculate acceleration

Using a calculator, a(60)=v′(60)≈−0.036a(60) = v'(60) \approx -0.036 meters per second per second.

Determine velocity sign at t = 60

Evaluate v(60)=2.38e−0.02(60)sin⁡(π56⋅60)≈−0.1595v(60) = 2.38e^{-0.02(60)}\sin\left(\frac{\pi}{56}\cdot 60\right) \approx -0.1595, which is negative.

Compare signs for speeding/slowing

Since both v(60)<0v(60) < 0 and a(60)<0a(60) < 0 have the same sign, Stephen is speeding up.

AP Scoring Note — P3 + P4 + P5

P3: Earned for a(60)a(60) with setup (showing v′(60)v'(60)) .

P4: Earned for correct units (m/s² or meters per second per second) .

P5: Earned for correctly concluding 'speeding up' with a reason based on velocity and acceleration signs.
CPart C
Medium
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.
Need a Hint?
The change in position (displacement) is found by integrating the velocity function: ∫t1t2v(t) dt\int_{t_1}^{t_2} v(t) \, dt.
Show Solution

Set up the displacement integral

The distance between the two positions is ∣∫2080v(t) dt∣\left| \int_{20}^{80} v(t) \, dt \right|.

Calculate the result

∫2080v(t) dt≈23.384\int_{20}^{80} v(t) \, dt \approx 23.384 meters (or 23.38323.383 meters).

AP Scoring Note — P6 + P7

P6: Earned for the correct definite integral setup ∫2080v(t) dt\int_{20}^{80} v(t) \, dt .

P7: Earned for the correct numerical answer 23.38423.384.
DPart D
Medium
Find the total distance Stephen swims over the time interval 0≤t≤900 \leq t \leq 90 seconds. Show the setup for your calculations.
Need a Hint?
Total distance is the integral of speed (the absolute value of velocity): ∫090∣v(t)∣ dt\int_{0}^{90} |v(t)| \, dt.
Show Solution

Set up the total distance integral

Total distance =∫090∣v(t)∣ dt= \int_{0}^{90} |v(t)| \, dt.

Calculate the total distance

Using a calculator, ∫090∣v(t)∣ dt≈62.164\int_{0}^{90} |v(t)| \, dt \approx 62.164 meters.

AP Scoring Note — P8 + P9

P8: Earned for the integral of the absolute value of velocity or splitting the integral at the turn point: ∫056v(t) dt−∫5690v(t) dt\int_{0}^{56} v(t) \, dt - \int_{56}^{90} v(t) \, dt .

P9: Earned for the correct final answer 62.16462.164.

Timed Practice

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Common Pitfalls
  • ⚠
    The Initial Condition Trap (+C)When finding position s(t), students often forget to add the initial value s(k). Remember: Final = Initial + Net Change. Missing s(k) is a guaranteed point loss.
  • ⚠
    Vague Interpretations of Definite IntegralsWhen explaining the meaning of an integral (e.g., total distance), you MUST include: (1) The Value, (2) The Units, and (3) The Time Interval (from t=a to t=b). Example: 'The total distance in meters from t=0 to t=10 seconds.'
  • ⚠
    Missing or Incorrect UnitsAlways check the last part of the FRQ for unit requirements. Position (meters), Velocity (m/sec), Acceleration (m/sec²). A missing unit can cost you an entire point.
  • ⚠
    Confusing Displacement vs. Total DistanceDisplacement can be negative; Total Distance is always non-negative. Never use |\int v\,dt| to find total distance—you must integrate the absolute value: \int |v|\,dt.
  • ⚠
    Assuming v = 0 means a change in directionThe particle only changes direction if v(t) changes sign. If v(t) touches zero but stays positive, the particle just momentarily paused and continued forward.

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