AP Physics C: Mechanics Practice Quiz: Representing Motion

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

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

On a graph of an object's velocity as a function of time, what does a horizontal, non-zero line represent?

A)The object is at rest.B)The object is moving with constant, non-zero velocity.C)The object is moving with constant, non-zero acceleration.D)The object is moving with changing acceleration.

All Questions (10)

On a graph of an object's velocity as a function of time, what does a horizontal, non-zero line represent?

A) The object is at rest.

B) The object is moving with constant, non-zero velocity.

C) The object is moving with constant, non-zero acceleration.

D) The object is moving with changing acceleration.

Correct Answer: B

A horizontal line on a velocity-time graph means the velocity value is constant over time. Since the line is non-zero, the object is moving with a constant, non-zero velocity. This is a description of the object's motion based on a graphical representation.

A car starts from rest and accelerates uniformly at 3.0 m/s² in a straight line. What is the car's velocity after 5.0 s?

A) 0.6 m/s

B) 1.67 m/s

C) 15 m/s

D) 37.5 m/s

Correct Answer: C

This is a constant acceleration problem. We are given initial velocity (v_x0 = 0 m/s), acceleration (a_x = 3.0 m/s²), and time (t = 5.0 s). We can use the kinematic equation v_x = v_x0 + a_x*t. Plugging in the values: v_x = 0 + (3.0 m/s²)(5.0 s) = 15 m/s.

An object initially moving at 4.0 m/s accelerates uniformly at 2.0 m/s² for 3.0 s. What is the object's displacement during this time?

A) 10 m

B) 12 m

C) 18 m

D) 21 m

Correct Answer: D

This problem requires finding the displacement under constant acceleration. We are given initial velocity (v_x0 = 4.0 m/s), acceleration (a_x = 2.0 m/s²), and time (t = 3.0 s). We use the kinematic equation x = x_0 + v_x0*t + (1/2)a_x*t². The displacement is Δx = x - x_0. So, Δx = v_x0*t + (1/2)a_x*t² = (4.0 m/s)(3.0 s) + (1/2)(2.0 m/s²)(3.0 s)² = 12 m + 9.0 m = 21 m.

A particle accelerates from 5.0 m/s to 15 m/s with a constant acceleration of 2.5 m/s². What is the distance it travels during this acceleration?

A) 2.0 m

B) 4.0 m

C) 40 m

D) 200 m

Correct Answer: C

This problem involves constant acceleration but does not provide time. We are given initial velocity (v_x0 = 5.0 m/s), final velocity (v_x = 15 m/s), and acceleration (a_x = 2.5 m/s²). The appropriate kinematic equation is v_x² = v_x0² + 2a_x(x - x_0). We solve for the displacement, Δx = x - x_0. Rearranging the equation: Δx = (v_x² - v_x0²) / (2a_x) = ((15 m/s)² - (5.0 m/s)²) / (2 * 2.5 m/s²) = (225 - 25) / 5.0 = 200 / 5.0 = 40 m.

The velocity of an object as a function of time is shown on a graph. The graph is a horizontal line at v = 5 m/s from t = 2 s to t = 6 s. What is the displacement of the object during this time interval?

A) 4 m

B) 5 m

C) 20 m

D) 30 m

Correct Answer: C

The displacement is the area under the velocity-time graph. The described graph forms a rectangle with a height of 5 m/s and a width (time interval) of 6 s - 2 s = 4 s. The area is height × width = (5 m/s) × (4 s) = 20 m. This corresponds to the principle that displacement is the area under the v-t curve.

An object's acceleration as a function of time is represented by a graph showing a constant acceleration of 4.0 m/s² from t = 0 s to t = 3.0 s. If the object's initial velocity is 2.0 m/s, what is its final velocity at t = 3.0 s?

A) 6.0 m/s

B) 12 m/s

C) 14 m/s

D) 20 m/s

Correct Answer: C

The change in velocity is the area under the acceleration-time graph. The area is a rectangle with height 4.0 m/s² and width 3.0 s, so Δv_x = (4.0 m/s²)(3.0 s) = 12 m/s. The change in velocity is Δv_x = v_x - v_x0. Therefore, the final velocity is v_x = v_x0 + Δv_x = 2.0 m/s + 12 m/s = 14 m/s. This also directly corresponds to the equation v_x = v_x0 + a_x*t.

A race car, starting from rest, reaches a final velocity of 60 m/s after traveling 180 m with constant acceleration. What was the car's acceleration?

A) 3.0 m/s²

B) 6.0 m/s²

C) 10 m/s²

D) 12 m/s²

Correct Answer: C

This problem requires finding acceleration given initial velocity (v_x0 = 0), final velocity (v_x = 60 m/s), and displacement (Δx = 180 m). The time is not given, so the most direct equation to use is v_x² = v_x0² + 2a_x(x - x_0). We rearrange to solve for a_x: a_x = (v_x² - v_x0²) / (2Δx) = ((60 m/s)² - (0 m/s)²) / (2 * 180 m) = 3600 / 360 = 10 m/s².

Which of the following quantities is represented by the slope of a velocity-versus-time graph?

A) Position

B) Displacement

C) Instantaneous velocity

D) Instantaneous acceleration

Correct Answer: D

Acceleration is defined as the rate of change of velocity, a = Δv/Δt. On a graph of velocity versus time, the slope is calculated as the change in the vertical axis (velocity) divided by the change in the horizontal axis (time), which is the definition of acceleration. This question tests the description of acceleration using a representation of motion.

An object starts from rest and accelerates uniformly. Its velocity-time graph is a straight line passing through the origin (t=0, v=0) and the point (t=4s, v=8 m/s). What is the object's displacement from t=0 to t=4 s?

A) 2 m

B) 8 m

C) 16 m

D) 32 m

Correct Answer: C

The displacement is the area under the velocity-time graph. The described graph forms a triangle with a base of 4 s and a height of 8 m/s. The area of a triangle is (1/2) × base × height. Therefore, the displacement is Δx = (1/2)(4 s)(8 m/s) = 16 m.

A ball is thrown directly upward with an initial velocity of 20 m/s. It experiences a constant downward acceleration of 10 m/s². How much time does it take for the ball to reach its maximum height?

A) 1.0 s

B) 2.0 s

C) 4.0 s

D) 10 s

Correct Answer: B

At its maximum height, the instantaneous velocity of the ball is 0 m/s. We are given the initial velocity (v_x0 = 20 m/s), the final velocity (v_x = 0 m/s), and the constant acceleration (a_x = -10 m/s², it's negative because it's downward, opposing the initial upward velocity). We can use the kinematic equation v_x = v_x0 + a_x*t to find the time. Rearranging for t: t = (v_x - v_x0) / a_x = (0 m/s - 20 m/s) / (-10 m/s²) = 2.0 s.