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Free Response · Position, velocity, and acceleration along a line

[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.

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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.

Past AP Exams

Real AP exam questions on this topic

Official AP Calculus AB free-response questions, each with a full step-by-step solution.

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