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AP Calculus AB Practice Quiz: Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

A particle moves along the x-axis with a velocity given by v(t) = 3t^2 - 6t for t ≥ 0. What is the displacement of the particle over the time interval 0 ≤ t ≤ 3?

All Questions (7)

A particle moves along the x-axis with a velocity given by v(t) = 3t^2 - 6t for t ≥ 0. What is the displacement of the particle over the time interval 0 ≤ t ≤ 3?

A) 0

B) 4

C) 8

D) 9

Correct Answer: A

Displacement is the definite integral of the velocity function over the given interval. Displacement = ∫[0, 3] (3t^2 - 6t) dt = [t^3 - 3t^2] from 0 to 3 = (3^3 - 3(3^2)) - (0^3 - 3(0^2)) = (27 - 27) - 0 = 0.

A particle moves along the x-axis with a velocity given by v(t) = 3t^2 - 6t for t ≥ 0. What is the total distance traveled by the particle over the time interval 0 ≤ t ≤ 3?

A) 0

B) 4

C) 8

D) 9

Correct Answer: C

Total distance traveled is the integral of the speed, |v(t)|. First, find where v(t) is negative: 3t(t-2) = 0 gives t=0, t=2. The velocity is negative on (0, 2). Total distance = ∫[0, 3] |3t^2 - 6t| dt = -∫[0, 2] (3t^2 - 6t) dt + ∫[2, 3] (3t^2 - 6t) dt. This evaluates to -[t^3 - 3t^2] from 0 to 2 + [t^3 - 3t^2] from 2 to 3 = -((8-12)-0) + ((27-27)-(8-12)) = 4 + 4 = 8.

At time t=0, a particle moving along the x-axis is at position x=5. The velocity of the particle is given by v(t) = 2t + 1. What is the position of the particle at time t=2?

A) 6

B) 9

C) 10

D) 11

Correct Answer: D

The final position is the initial position plus the displacement. The displacement is the integral of velocity from t=0 to t=2. Position x(2) = x(0) + ∫[0, 2] (2t + 1) dt. The integral evaluates to [t^2 + t] from 0 to 2, which is (2^2 + 2) - 0 = 6. Therefore, x(2) = 5 + 6 = 11.

The velocity of a particle moving in a straight line is given by v(t). Which of the following expressions represents the total distance the particle traveled from time t=a to t=b?

A) ∫[a, b] v(t) dt

B) v(b) - v(a)

C) ∫[a, b] |v(t)| dt

D) (v(b) - v(a)) / (b - a)

Correct Answer: C

The definite integral of velocity, ∫[a, b] v(t) dt, represents the particle's displacement (net change in position). The total distance traveled is the integral of the speed, which is the absolute value of velocity, |v(t)|. Therefore, the total distance is ∫[a, b] |v(t)| dt.

A particle moves along a straight line with velocity v(t) = cos(πt). If the particle's position is x=4 at t=1, what is the position of the particle at t=2?

A) 4

B) 4 - 2/π

C) 4 + 1/π

D) 5

Correct Answer: A

The position at t=2 is the position at t=1 plus the displacement from t=1 to t=2. Position x(2) = x(1) + ∫[1, 2] cos(πt) dt. The integral is [ (1/π)sin(πt) ] from 1 to 2 = (1/π)sin(2π) - (1/π)sin(π) = 0 - 0 = 0. So, x(2) = 4 + 0 = 4.

The velocity of a particle moving along the x-axis is given by v(t). For the interval 0 ≤ t ≤ 4, it is known that ∫[0, 4] v(t) dt = 5 and ∫[0, 4] |v(t)| dt = 9. Which of the following must be true about the particle's motion on this interval?

A) The particle only moved in the positive direction.

B) The particle ended at position 9.

C) The particle changed direction at least once on the interval 0 < t < 4.

D) The particle's final speed was greater than its initial speed.

Correct Answer: C

The displacement, ∫[0, 4] v(t) dt, is 5. The total distance traveled, ∫[0, 4] |v(t)| dt, is 9. Since the displacement (5) is not equal to the total distance traveled (9), the particle must have changed direction at some point. If it had not changed direction, the displacement and total distance would be equal.

A particle is at position x=2 at time t=0 and moves along the x-axis with velocity v(t) = t^2 - 4 for t ≥ 0. What is the position of the particle when its velocity is first equal to 0?

A) -14/3

B) -10/3

C) 2

D) 22/3

Correct Answer: B

First, find the time when the velocity is 0: v(t) = t^2 - 4 = 0 gives t=2 (since t ≥ 0). Next, find the position at t=2. The position is the initial position plus the displacement: x(2) = x(0) + ∫[0, 2] (t^2 - 4) dt. The integral is [t^3/3 - 4t] from 0 to 2 = (8/3 - 8) - 0 = -16/3. The position is x(2) = 2 + (-16/3) = 6/3 - 16/3 = -10/3.