AP Calculus BC 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 - 2t for t ≥ 0. If the particle is at position x = 4 at time t = 1, what is the position of the particle at time t = 3?

A)18B)20C)22D)26

All Questions (7)

A particle moves along the x-axis with a velocity given by v(t) = 3t^2 - 2t for t ≥ 0. If the particle is at position x = 4 at time t = 1, what is the position of the particle at time t = 3?

A) 18

B) 20

C) 22

D) 26

Correct Answer: C

The position of the particle at time t=3 can be found by adding the initial position at t=1 to the displacement from t=1 to t=3. The displacement is the definite integral of the velocity function. Position x(3) = x(1) + ∫(from 1 to 3) (3t^2 - 2t) dt. x(3) = 4 + [t^3 - t^2] from 1 to 3. x(3) = 4 + [(3^3 - 3^2) - (1^3 - 1^2)] = 4 + [(27 - 9) - (1 - 1)] = 4 + [18 - 0] = 22.

A particle moves in a straight line with velocity v(t) = cos(t). What is the displacement of the particle over the time interval [0, π]?

A) 0

B) 1

C) 2

D) π

Correct Answer: A

Displacement is the definite integral of the velocity function over the given interval. Displacement = ∫(from 0 to π) cos(t) dt. The antiderivative of cos(t) is sin(t). Evaluating the definite integral: [sin(t)] from 0 to π = sin(π) - sin(0) = 0 - 0 = 0. The particle ends at the same position it started.

A particle moves along a line with velocity v(t) = 2t - 4 for t ≥ 0. What is the total distance traveled by the particle from t = 0 to t = 3?

A) -3

B) 3

C) 5

D) 9

Correct Answer: C

Total distance traveled is the integral of the speed, which is the absolute value of velocity: ∫(from 0 to 3) |2t - 4| dt. First, find where the velocity is zero: 2t - 4 = 0, so t = 2. The velocity is negative on [0, 2) and positive on (2, 3]. We must split the integral: ∫(from 0 to 2) -(2t - 4) dt + ∫(from 2 to 3) (2t - 4) dt = [-t^2 + 4t] from 0 to 2 + [t^2 - 4t] from 2 to 3 = [(-4+8) - 0] + [(9-12) - (4-8)] = [4] + [(-3) - (-4)] = 4 + 1 = 5.

The velocity v(t) of a particle moving in a straight line is positive for 0 ≤ t < a and negative for a < t ≤ b. Which of the following expressions gives the total distance the particle traveled from t = 0 to t = b?

A) ∫(from 0 to b) v(t) dt

B) ∫(from 0 to a) v(t) dt + ∫(from a to b) v(t) dt

C) ∫(from 0 to a) v(t) dt - ∫(from a to b) v(t) dt

D) |∫(from 0 to b) v(t) dt|

Correct Answer: C

Total distance traveled is the integral of speed, ∫|v(t)|dt. Since v(t) > 0 on [0, a), |v(t)| = v(t). Since v(t) < 0 on (a, b], |v(t)| = -v(t). Therefore, the total distance is ∫(from 0 to a) v(t) dt + ∫(from a to b) -v(t) dt, which simplifies to ∫(from 0 to a) v(t) dt - ∫(from a to b) v(t) dt. Option A represents displacement.

A particle's velocity is given by v(t) = t^2 - 3t + 2. What is the total distance traveled by the particle on the interval [0, 3]?

A) 3/2

B) 11/6

C) 2

D) 5/2

Correct Answer: B

To find the total distance, we integrate the speed, ∫(from 0 to 3) |t^2 - 3t + 2| dt. First, find when v(t) = 0: t^2 - 3t + 2 = (t-1)(t-2) = 0, so t=1 and t=2. We must split the integral based on the sign of v(t): v(t) is positive on [0, 1), negative on (1, 2), and positive on (2, 3]. The integral is ∫(from 0 to 1) (t^2 - 3t + 2) dt + ∫(from 1 to 2) -(t^2 - 3t + 2) dt + ∫(from 2 to 3) (t^2 - 3t + 2) dt. Let F(t) = t^3/3 - 3t^2/2 + 2t. The calculation is (F(1)-F(0)) - (F(2)-F(1)) + (F(3)-F(2)) = (5/6) - (4/6 - 5/6) + (9/6 - 4/6) = 5/6 + 1/6 + 5/6 = 11/6.

The acceleration of a particle moving along the x-axis is given by a(t) = 6t. If the velocity of the particle is 5 at time t = 1, what is the velocity of the particle at time t = 2?

A) 9

B) 12

C) 14

D) 17

Correct Answer: C

The change in velocity from t=1 to t=2 is the definite integral of the acceleration function. The velocity at t=2 is the initial velocity at t=1 plus this change. v(2) = v(1) + ∫(from 1 to 2) a(t) dt. v(2) = 5 + ∫(from 1 to 2) 6t dt = 5 + [3t^2] from 1 to 2 = 5 + [3(2^2) - 3(1^2)] = 5 + [12 - 3] = 5 + 9 = 14.

A particle moves along a straight line. The definite integral of the particle's velocity function, ∫(from a to b) v(t) dt, represents which of the following?

A) The particle's average velocity over the interval [a, b].

B) The particle's total distance traveled over the interval [a, b].

C) The particle's displacement over the interval [a, b].

D) The particle's final position at time t = b.

Correct Answer: C

By definition, for a particle in rectilinear motion, the definite integral of velocity over an interval of time represents the particle’s displacement (net change in position) over that interval. The total distance traveled would be the integral of the speed, ∫(from a to b) |v(t)| dt. The final position would be the initial position plus the displacement.