AP Physics C Mechanics Quiz: Displacement Velocity And Acceleration

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Displacement Velocity And AccelerationQuestion 1 of 20

A particle starts from rest at the origin and moves along the x-axis with an acceleration $a_x(t) = C t^{1/2}$ , where $C$ is a positive constant. Which of the following represents the particle's position $x(t)$ as a function of time?

\frac{2C}{3}t^{3/2}

\frac{C}{2}t^{-1/2}

\frac{4C}{15}t^{5/2}

C t^{3/2}

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Displacement Velocity And Acceleration

Practice Displacement Velocity And Acceleration in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Displacement Velocity And Acceleration, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Question 1

$\frac{2C}{3}t^{3/2}$
$\frac{C}{2}t^{-1/2}$
$\frac{4C}{15}t^{5/2}$ (correct answer)
$C t^{3/2}$

Explanation: To find position from acceleration, we must integrate twice. First, find velocity:

v_x(t) = \int a_x(t) dt = \int C t^{1/2} dt = C \frac{t^{3/2}}{3/2} + v_0

. Since the particle starts from rest,

v_0=0

, so

v_x(t) = \frac{2C}{3}t^{3/2}

. Next, find position:

x(t) = \int v_x(t) dt = \int \frac{2C}{3}t^{3/2} dt = \frac{2C}{3} \frac{t^{5/2}}{5/2} + x_0

. Since it starts at the origin,

x_0=0

. Thus,

x(t) = \frac{4C}{15}t^{5/2}

Question 2

A particle moves along the x-axis such that its position is given by the function $x(t) = At^3 - Bt^2 + C$ , where $A=2.0 \text{ m/s}^3$ , $B=3.0 \text{ m/s}^2$ , and $C=1.0 \text{ m}$ . What is the instantaneous velocity of the particle at $t = 2.0 \text{ s}$ ?

$12 \text{ m/s}$ (correct answer)
$5.0 \text{ m/s}$
$24 \text{ m/s}$
$36 \text{ m/s}$

Explanation: Instantaneous velocity is the time derivative of the position function,

v(t) = \frac{dx}{dt}

. Differentiating

x(t)

with respect to time gives

v(t) = 3At^2 - 2Bt

. Substituting the given values and

t=2.0 \text{ s}

yields

v(2.0) = 3(2.0)(2.0)^2 - 2(3.0)(2.0) = 24 - 12 = 12 \text{ m/s}

Question 3

An object's acceleration along the x-axis is given by $a(t) = 6t - 4$ , in $m/s^2$ . If the object's initial velocity at $t=0$ is $v(0) = 5 \text{ m/s}$ , what is its velocity at $t=3 \text{ s}$ ?

$14 \text{ m/s}$
$15 \text{ m/s}$
$20 \text{ m/s}$ (correct answer)
$27 \text{ m/s}$

Explanation: Velocity is the integral of acceleration with respect to time.

v(t) = \int a(t) dt = \int (6t - 4) dt = 3t^2 - 4t + C

. We use the initial condition

v(0) = 5 \text{ m/s}

to find the constant of integration

C

v(0) = 3(0)^2 - 4(0) + C = 5

, so

C=5

. The velocity function is

v(t) = 3t^2 - 4t + 5

. At

t=3 \text{ s}

v(3) = 3(3)^2 - 4(3) + 5 = 27 - 12 + 5 = 20 \text{ m/s}

Question 4

An object moves along a straight line with a constant, non-zero acceleration. Which of the following statements correctly describes the object's velocity and position as functions of time, $v(t)$ and $x(t)$ ?

$v(t)$ is a constant function, and $x(t)$ is a linear function of time.
$v(t)$ is a linear function of time, and $x(t)$ is a quadratic function of time. (correct answer)
$v(t)$ is a quadratic function of time, and $x(t)$ is a cubic function of time.
$v(t)$ is a linear function of time, and $x(t)$ is also a linear function of time.

Explanation: If acceleration

a

is constant, integrating with respect to time gives velocity:

v(t) = \int a dt = at + v_0

, which is a linear function of time. Integrating velocity gives position:

x(t) = \int (at + v_0) dt = \frac{1}{2}at^2 + v_0t + x_0

, which is a quadratic function of time.

Question 5

The velocity of a particle is given by $v(t) = \frac{1}{3}t^3 - 2t^2 + 4t$ . At what time $t>0$ is the particle's acceleration equal to zero?

$t=1 \text{ s}$
$t=2 \text{ s}$ (correct answer)
$t=3 \text{ s}$
$t=4 \text{ s}$

Explanation: Acceleration is the derivative of velocity with respect to time.

a(t) = \frac{dv}{dt} = t^2 - 4t + 4 = (t-2)^2

. To find when acceleration is zero, we set

a(t)=0

. This gives

(t-2)^2 = 0

, so

t=2 \text{ s}

Question 6

A car starts from rest and its speed is given by $v(t) = kt^2$ , where $k$ is a positive constant. How does its average speed $v_{avg}$ over the interval from $t=0$ to $t=T$ compare to its instantaneous speed $v(T/2)$ at the midpoint of the interval?

$v_{avg} < v(T/2)$
$v_{avg} = v(T/2)$
$v_{avg} > v(T/2)$ (correct answer)
The relationship depends on the value of $k$ .

Explanation: First, find the instantaneous speed at

t=T/2

v(T/2) = k(T/2)^2 = \frac{1}{4}kT^2

. Next, find the average speed. Average speed is total distance divided by time. Distance is the integral of speed:

x(T) = \int_0^T kt^2 dt = [\frac{1}{3}kt^3]_0^T = \frac{1}{3}kT^3

. The average speed is

v_{avg} = \frac{x(T)}{T} = \frac{\frac{1}{3}kT^3}{T} = \frac{1}{3}kT^2

. Comparing the two,

\frac{1}{3}kT^2 > \frac{1}{4}kT^2

, so

v_{avg} > v(T/2)

Question 7

An object starts from rest and experiences an acceleration given by $a(t) = A\cos(\frac{\pi}{2}t)$ , where $A$ is a constant. What is the object's change in velocity during the time interval from $t=0$ to $t=1$ s?

$0$
$A$
$\frac{\pi A}{2}$
$\frac{2A}{\pi}$ (correct answer)

Explanation: The change in velocity is the definite integral of acceleration over the time interval.

\Delta v = \int_0^1 a(t) dt = \int_0^1 A\cos(\frac{\pi}{2}t) dt

. The integral of

\cos(kt)

\frac{1}{k}\sin(kt)

. So,

\Delta v = A [\frac{2}{\pi}\sin(\frac{\pi}{2}t)]_0^1 = \frac{2A}{\pi}[\sin(\frac{\pi}{2}) - \sin(0)] = \frac{2A}{\pi}[1 - 0] = \frac{2A}{\pi}

Question 8

A particle moves along the x-axis. At a particular instant, its velocity is negative and its acceleration is positive. Which of the following statements correctly describes the particle's motion at that instant?

The particle is moving in the positive direction and speeding up.
The particle is moving in the positive direction and slowing down.
The particle is moving in the negative direction and speeding up.
The particle is moving in the negative direction and slowing down. (correct answer)

Explanation: The sign of velocity indicates the direction of motion. Since velocity is negative, the particle is moving in the negative direction. When velocity and acceleration have opposite signs, the object's speed is decreasing. Therefore, the particle is moving in the negative direction and slowing down.

Question 9

A particle moves along the x-axis with a velocity given by $v_x(t) = 4t^3 - 2t$ . If the particle is at $x=1$ at time $t=1$ , what is its position at time $t=2$ ?

$12$
$13$ (correct answer)
$11$
$14$

Explanation: Position is the integral of velocity.

x(t) = \int v_x(t) dt = \int (4t^3 - 2t) dt = t^4 - t^2 + C

. Use the condition

x(1)=1

to find

C

1 = (1)^4 - (1)^2 + C

, which gives

1 = 0 + C

, so

C=1

. The position function is

x(t) = t^4 - t^2 + 1

. At

t=2

, the position is

x(2) = (2)^4 - (2)^2 + 1 = 16 - 4 + 1 = 13

Question 10

An object's position is given by the function $x(t) = 2t^3 - 15t^2 + 36t - 8$ . At which of the following times is the object momentarily at rest?

$t=1.5 \text{ s}$ and $t=6 \text{ s}$
$t=2.5 \text{ s}$ only
$t=2 \text{ s}$ and $t=3 \text{ s}$ (correct answer)
$t=6 \text{ s}$ only

Explanation: The object is momentarily at rest when its velocity is zero. We find the velocity function by differentiating the position function:

v(t) = \frac{dx}{dt} = 6t^2 - 30t + 36

. Set

v(t) = 0

and solve for

t

6(t^2 - 5t + 6) = 0

, which factors to

6(t-2)(t-3) = 0

. The solutions are

t=2 \text{ s}

and

t=3 \text{ s}

Question 11

A particle moves along a straight line such that its position is given by $x(t) = 2t^3 - 9t^2 + 12t + 5$ meters, where $t$ is in seconds. At what time(s) does the particle momentarily come to rest?

$t = 1$ s and $t = 2$ s (correct answer)
$t = 0$ s and $t = 3$ s
$t = 2$ s and $t = 3$ s
$t = 1$ s and $t = 3$ s

Explanation: The particle comes to rest when its velocity equals zero. Taking the derivative:

v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12

. Setting

v(t) = 0

6t^2 - 18t + 12 = 0

, which simplifies to

t^2 - 3t + 2 = 0

. Factoring:

(t-1)(t-2) = 0

, giving

t = 1

s and

t = 2

s. Choice B incorrectly uses the acceleration equation. Choice C uses

t = 2

correctly but incorrectly includes

t = 3

. Choice D correctly identifies

t = 1

but incorrectly includes

t = 3

Question 12

A particle moves along the x-axis with acceleration $a(x) = -kx$ , where $k > 0$ is a constant and $x$ is the position. If the particle starts from rest at position $x_0$ , what is its speed when it reaches position $x = 0$ ?

$\sqrt{\frac{kx_0^2}{2}}$
$x_0\sqrt{\frac{k}{2}}$
$\sqrt{kx_0^2}$
$x_0\sqrt{k}$ (correct answer)

Explanation: Since acceleration depends on position, use

v\frac{dv}{dx} = a(x) = -kx

. Integrating:

\int v dv = \int -kx dx

, giving

\frac{v^2}{2} = -\frac{kx^2}{2} + C

. Using initial condition

v = 0

x = x_0

0 = -\frac{kx_0^2}{2} + C

, so

C = \frac{kx_0^2}{2}

. Therefore

\frac{v^2}{2} = \frac{k}{2}(x_0^2 - x^2)

. At

x = 0

\frac{v^2}{2} = \frac{kx_0^2}{2}

, so

v^2 = kx_0^2

and

v = x_0\sqrt{k}

. Choice A includes an unnecessary factor of 1/2. Choice B incorrectly includes the factor 1/2 under the square root. Choice C has incorrect algebraic form.

Question 13

Two objects A and B move along parallel tracks. Object A has constant velocity $v_A = 10$ m/s. Object B starts from rest with constant acceleration $a_B = 2.0$ m/s². At $t = 0$ , object A is 50 m behind object B. At what time do they have the same velocity, and what is their relative position at that moment?

$t = 5.0$ s; A is 25 m behind B (correct answer)
$t = 5.0$ s; A is 25 m ahead of B
$t = 5.0$ s; A is 50 m ahead of B
$t = 10$ s; A is 50 m ahead of B

Explanation: Object B's velocity is

v_B(t) = 2.0t

. They have equal velocities when

10 = 2.0t

, so

t = 5.0

s. At

t = 5.0

s: Object A's position is

x_A = x_{A0} + 10(5) = (-50) + 50 = 0

m (taking B's initial position as origin). Object B's position is

x_B = 0 + \frac{1}{2}(2.0)(5)^2 = 25

m. Therefore A is at 0 m and B is at 25 m, meaning A is 25 m behind B. Choice B incorrectly claims A is ahead. Choice C uses wrong distance calculation. Choice D uses incorrect time calculation.

Question 14

A ball is thrown vertically upward with initial velocity $v_0$ . At the instant when the ball returns to its starting height, which statement correctly compares the ball's displacement, velocity, and acceleration to their values at the moment of release?

Displacement is zero, velocity has the same magnitude but opposite direction, acceleration is unchanged (correct answer)
Displacement is zero, velocity is zero, acceleration has reversed direction
Displacement is zero, velocity is zero, acceleration is unchanged
Displacement is maximum, velocity has the same magnitude and direction, acceleration is unchanged

Explanation: When the ball returns to its starting height, the displacement is zero by definition. Due to energy conservation and symmetry, the velocity magnitude equals the initial velocity but points downward (opposite direction). Acceleration due to gravity remains constant at

-g

throughout the motion. Choice B incorrectly states velocity is zero (that occurs only at maximum height). Choice C also incorrectly states velocity is zero. Choice D incorrectly states displacement is maximum and velocity direction is the same.

Question 15

Which of the following describes a situation where an object has an instantaneous velocity of zero but a non-zero instantaneous acceleration?

A car traveling at a constant speed of $20 \text{ m/s}$ on a straight road.
A ball thrown vertically upward at the highest point of its trajectory. (correct answer)
A satellite in a stable circular orbit around the Earth.
A block at rest on a horizontal table for an extended period.

Explanation: At the peak of its trajectory, a vertically thrown ball momentarily stops moving upwards before it begins to fall back down, so its instantaneous velocity is zero. However, the acceleration due to gravity is still acting on it, so its acceleration is approximately

-9.8 \text{ m/s}^2

. Therefore, it has zero velocity and non-zero acceleration.

Question 16

The position of a particle is given by $x(t) = t^3 - 6t^2 + 5$ . For $t>0$ , during which of the following time intervals is the particle slowing down?

$(0, 2)$
$(2, 4)$ (correct answer)
$(4, \infty)$
$(0, 4)$

Explanation: A particle is slowing down when its velocity and acceleration have opposite signs. First, find

v(t) = \frac{dx}{dt} = 3t^2 - 12t = 3t(t-4)

. Then find

a(t) = \frac{dv}{dt} = 6t - 12 = 6(t-2)

. We analyze the signs: For

t \in (2, 4)

v(t)

is negative (e.g., at

t=3

v=27-36=-9

) and

a(t)

is positive (e.g., at

t=3

a=18-12=6

). Since they have opposite signs, the particle is slowing down in this interval.

Question 17

The position of an object is given by $x(t) = 2t^3 - 3t^2 + 4$ . What is the average velocity of the object over the time interval from $t=1 \text{ s}$ to $t=3 \text{ s}$ ?

$10 \text{ m/s}$
$14 \text{ m/s}$ (correct answer)
$26 \text{ m/s}$
$31 \text{ m/s}$

Explanation: Average velocity is defined as displacement divided by the time interval:

v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x(3) - x(1)}{3-1}

. First, calculate the positions:

x(3) = 2(3)^3 - 3(3)^2 + 4 = 54 - 27 + 4 = 31 \text{ m}

x(1) = 2(1)^3 - 3(1)^2 + 4 = 2 - 3 + 4 = 3 \text{ m}

. Then,

v_{avg} = \frac{31 - 3}{2} = \frac{28}{2} = 14 \text{ m/s}

Question 18

The velocity of a particle as a function of time is given by $v(t) = 5t - t^2$ . What is the average acceleration of the particle over the time interval from $t=1 \text{ s}$ to $t=4 \text{ s}$ ?

$0 \text{ m/s}^2$ (correct answer)
$-1 \text{ m/s}^2$
$-2.5 \text{ m/s}^2$
$5 \text{ m/s}^2$

Explanation: Average acceleration is the change in velocity divided by the time interval:

a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v(4) - v(1)}{4-1}

. First, calculate the velocities:

v(4) = 5(4) - (4)^2 = 20 - 16 = 4 \text{ m/s}

v(1) = 5(1) - (1)^2 = 5 - 1 = 4 \text{ m/s}

. Then,

a_{avg} = \frac{4 - 4}{3} = 0 \text{ m/s}^2

Question 19

A particle moves in the xy-plane with a position vector given by $\vec{r}(t) = (2t^3 - 3t)\hat{i} + (t^2 + 1)\hat{j}$ . What is the magnitude of the particle's velocity at $t=2 \text{ s}$ ?

$\sqrt{21}$
$\sqrt{457}$ (correct answer)
$25$
$\sqrt{461}$

Explanation: The velocity vector is the time derivative of the position vector:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t^2 - 3)\hat{i} + (2t)\hat{j}

. At

t=2 \text{ s}

, the velocity vector is

\vec{v}(2) = (6(2)^2 - 3)\hat{i} + (2(2))\hat{j} = (24 - 3)\hat{i} + 4\hat{j} = 21\hat{i} + 4\hat{j}

. The magnitude of the velocity is

|\vec{v}(2)| = \sqrt{21^2 + 4^2} = \sqrt{441 + 16} = \sqrt{457}

Question 20

The velocity of a particle in the xy-plane is given by $\vec{v}(t) = (4t)\hat{i} + (3t^2 - 1)\hat{j}$ . What is the magnitude of the particle's acceleration at $t=1 \text{ s}$ ?

$\sqrt{40}$
$10$
$\sqrt{52}$ (correct answer)
$5$

Explanation: The acceleration vector is the time derivative of the velocity vector:

\vec{a}(t) = \frac{d\vec{v}}{dt} = 4\hat{i} + (6t)\hat{j}

. At

t=1 \text{ s}

, the acceleration vector is

\vec{a}(1) = 4\hat{i} + 6(1)\hat{j} = 4\hat{i} + 6\hat{j}

. The magnitude of the acceleration is

|\vec{a}(1)| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}