AP Calculus AB 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. Its position at time t ≥ 0 is given by the function s(t) = t³ - 6t² + 9t. What is the velocity of the particle at time t = 2?

A)-3B)2C)3D)-21

All Questions (7)

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

A) -3

B) 2

C) 3

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 - 3t² - 12t + 1. What is the acceleration of the particle at time t = 1?

A) -12

B) 0

C) 6

D) 12

Correct Answer: C

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

A particle's position is described by the function s(t) = t³ - 6t² + 5 for t > 0. At what time t is the particle momentarily at rest?

A) t = 2

B) t = 3

C) t = 4

D) The particle is never at rest for t > 0.

Correct Answer: C

A particle is at rest when its velocity is zero. The velocity function is the derivative of the position function: v(t) = s'(t) = 3t² - 12t. Set v(t) = 0 and solve for t: 3t² - 12t = 0 → 3t(t - 4) = 0. The solutions are t = 0 and t = 4. Since the question specifies t > 0, the particle is at rest at t = 4.

The position of a particle moving along a line is given by s(t) = t³ - 9t² + 24t for t ≥ 0. For what time interval is the particle moving to the left?

A) (0, 2)

B) (2, 4)

C) (4, ∞)

D) (0, 2) U (4, ∞)

Correct Answer: B

The particle moves to the left when its velocity is negative. First, find the velocity function: v(t) = s'(t) = 3t² - 18t + 24. To find where v(t) < 0, find the roots of v(t) = 0: 3(t² - 6t + 8) = 0 → 3(t - 2)(t - 4) = 0. The roots are t = 2 and t = 4. Testing the intervals, we find that v(t) is negative between t = 2 and t = 4. Thus, the particle is moving left on the interval (2, 4).

A particle moves along a straight line with velocity given by v(t) = t² - 8t + 12. At t = 3, is the particle speeding up or slowing down?

A) Speeding up because velocity and acceleration have the same sign.

B) Slowing down because velocity and acceleration have different signs.

C) Slowing down because velocity is negative.

D) Speeding up because acceleration is negative.

Correct Answer: A

To determine if the particle is speeding up or slowing down, we compare the signs of velocity and acceleration at t = 3. First, find the velocity: v(3) = (3)² - 8(3) + 12 = 9 - 24 + 12 = -3. Next, find the acceleration function a(t) = v'(t) = 2t - 8. Then, find the acceleration at t = 3: a(3) = 2(3) - 8 = 6 - 8 = -2. Since both v(3) and a(3) are negative (they have the same sign), the particle's speed is increasing.

A diver jumps from a platform. The diver's height above the water, in feet, is given by h(t) = -16t² + 16t + 32, where t is the time in seconds after jumping. What is the diver's velocity upon impact with the water?

A) -32 ft/s

B) -48 ft/s

C) -64 ft/s

D) 2 s

Correct Answer: B

First, find the time of impact by setting h(t) = 0: -16t² + 16t + 32 = 0. Dividing by -16 gives t² - t - 2 = 0, which factors to (t - 2)(t + 1) = 0. The physically relevant time is t = 2 seconds. Next, find the velocity function by taking the derivative of the position function: v(t) = h'(t) = -32t + 16. Finally, evaluate the velocity at the time of impact, t = 2: v(2) = -32(2) + 16 = -64 + 16 = -48 ft/s.

The position of a particle is given by s(t) = (1/3)t³ - 4t² + 12t for t ≥ 0. What is the acceleration of the particle each time the velocity is zero?

A) -4 and 4

B) 2 and 6

C) 0

D) -8

Correct Answer: A

First, find the velocity function: v(t) = s'(t) = t² - 8t + 12. Set the velocity to zero to find the times when the particle is at rest: t² - 8t + 12 = 0 → (t - 2)(t - 6) = 0. The velocity is zero at t = 2 and t = 6. Next, find the acceleration function: a(t) = v'(t) = 2t - 8. Finally, evaluate the acceleration at t = 2 and t = 6. a(2) = 2(2) - 8 = -4. a(6) = 2(6) - 8 = 4. The accelerations are -4 and 4.