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AP Calculus BC Practice Quiz: Straight-Line Motion: Connecting Position, Velocity, and Acceleration

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 a straight line with its position at time t ≥ 0 given by the function s(t) = t³ - 6t² + 9t. What is the velocity of the particle at time t = 2?

All Questions (7)

A particle moves along a straight line with its position at time t ≥ 0 given by the function s(t) = t³ - 6t² + 9t. What is the velocity of the particle at time t = 2?

A) -3

B) 3

C) 2

D) -21

Correct Answer: A

Velocity is the derivative of the position function. First, find the velocity function v(t) by differentiating s(t): v(t) = s'(t) = 3t² - 12t + 9. Then, substitute t = 2 into the velocity function: v(2) = 3(2)² - 12(2) + 9 = 12 - 24 + 9 = -3.

The position of a particle moving on the x-axis is given by x(t) = 2t³ - 9t² + 12t - 4 for t ≥ 0. What is the acceleration of the particle at time t = 1?

A) 0

B) 6

C) -6

D) 5

Correct Answer: C

Acceleration is the second derivative of the position function. First, find the velocity function: v(t) = x'(t) = 6t² - 18t + 12. Then, find the acceleration function by differentiating the velocity function: a(t) = v'(t) = 12t - 18. Finally, evaluate the acceleration at t = 1: a(1) = 12(1) - 18 = -6.

At a certain time t, a particle moving along a line has a velocity v(t) = -5 and an acceleration a(t) = -2. Which of the following statements is true about the particle's motion at this time?

A) The particle is speeding up.

B) The particle is slowing down.

C) The particle is momentarily at rest.

D) The particle is changing direction.

Correct Answer: A

The speed of a particle is increasing when its velocity and acceleration have the same sign. In this case, both velocity (-5) and acceleration (-2) are negative. Since they have the same sign, the particle's speed is increasing.

A particle's velocity is described by the function v(t) = t² - 5t + 4 for t ≥ 0. At what time(s) t is the particle momentarily at rest?

A) t = 0 only

B) t = 1 only

C) t = 4 only

D) t = 1 and t = 4

Correct Answer: D

A particle is at rest when its velocity is zero. Set the velocity function equal to zero and solve for t: v(t) = t² - 5t + 4 = 0. Factoring the quadratic gives (t - 1)(t - 4) = 0. The solutions are t = 1 and t = 4.

The position of a particle moving along a line is given by s(t) = (1/3)t³ - 3t² + 8t + 1 for t ≥ 0. What is the acceleration of the particle each time its velocity is zero?

A) a = 0

B) a = 2 only

C) a = -2 only

D) a = -2 and a = 2

Correct Answer: D

First, find the velocity function: v(t) = s'(t) = t² - 6t + 8. Set v(t) = 0 to find when the particle is at rest: t² - 6t + 8 = (t - 2)(t - 4) = 0, which gives t = 2 and t = 4. Next, find the acceleration function: a(t) = v'(t) = 2t - 6. Finally, evaluate the acceleration at t = 2 and t = 4. a(2) = 2(2) - 6 = -2. a(4) = 2(4) - 6 = 2.

The position of a particle moving along a line is given by a function s(t), where s is in meters and t is in seconds. Which of the following represents the rate of change of the particle's position with respect to time?

A) The acceleration of the particle, s''(t).

B) The velocity of the particle, s'(t).

C) The average position of the particle.

D) The total distance traveled by the particle.

Correct Answer: B

The derivative of a function represents its instantaneous rate of change. In the context of rectilinear motion, the rate of change of the position function s(t) is the velocity function, which is denoted by s'(t) or v(t).

The position of a particle is given by x(t) = 2t³ - 15t² + 36t for t ≥ 0. During which of the following time intervals is the particle slowing down?

A) (2, 3)

B) (0, 2) and (2.5, 3)

C) (2, 2.5) and (3, ∞)

D) (0, 2.5)

Correct Answer: B

A particle is slowing down when its velocity and acceleration have opposite signs. First, find v(t) and a(t). v(t) = x'(t) = 6t² - 30t + 36 = 6(t-2)(t-3). a(t) = v'(t) = 12t - 30 = 6(2t-5). The critical points are t=2, t=3 for velocity and t=2.5 for acceleration. We analyze the signs on the intervals: 1) (0, 2): v(t) > 0, a(t) < 0. Opposite signs, so slowing down. 2) (2, 2.5): v(t) < 0, a(t) < 0. Same signs, so speeding up. 3) (2.5, 3): v(t) < 0, a(t) > 0. Opposite signs, so slowing down. 4) (3, ∞): v(t) > 0, a(t) > 0. Same signs, so speeding up. Therefore, the particle is slowing down on the intervals (0, 2) and (2.5, 3).