AP Physics C Mechanics Quiz: Representing Motion

20 questions · exam conditions

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Representing MotionQuestion 1 of 20

The acceleration of a particle is constant and negative. The particle starts from rest at the origin at $t=0$ . Which of the following best describes the particle's position-time graph?

A straight line with a constant negative slope starting from the origin.

A parabola opening downward that starts at the origin.

A horizontal line on the time axis, since it starts from rest.

A parabola opening upward that starts at the origin.

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Representing Motion

Practice Representing Motion 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 Representing Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The acceleration of a particle is constant and negative. The particle starts from rest at the origin at $t=0$ . Which of the following best describes the particle's position-time graph?

A straight line with a constant negative slope starting from the origin.
A parabola opening downward that starts at the origin. (correct answer)
A horizontal line on the time axis, since it starts from rest.
A parabola opening upward that starts at the origin.

Explanation: With constant negative acceleration

a = -A

and starting from rest at the origin, the position is given by

x(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 0 + 0 + \frac{1}{2}(-A)t^2 = -\frac{1}{2}At^2

. This is the equation of a parabola that opens downward and starts at the origin.

Question 2

A velocity-time graph of a moving object is a curve that is always in the first quadrant, and its slope is continuously decreasing, approaching zero as time increases. Which of the following is a correct statement about the object's motion?

The object is slowing down because its acceleration is decreasing.
The object moves with constant velocity since the acceleration eventually becomes zero.
The object is moving in the positive direction with decreasing acceleration, approaching a constant final velocity. (correct answer)
The object is speeding up at a constant rate throughout the motion.

Explanation: The graph in the first quadrant means velocity

v

is always positive. The slope of the v-t graph is acceleration

a

. The slope is decreasing and approaching zero, implying acceleration is positive but decreasing towards zero. Since

v

and

a

are both positive, the object is speeding up, but at a decreasing rate. As

a

approaches zero, the velocity approaches a constant value.

Question 3

An object's acceleration increases linearly with time, according to $a(t) = kt$ , where $k$ is a positive constant. If the object starts from rest at the origin at $t=0$ , what is its position $x$ as a function of time?

$x(t) = \frac{1}{2}kt^2$
$x(t) = \frac{1}{6}kt^3$ (correct answer)
$x(t) = kt$
$x(t) = k$

Explanation: We must integrate twice. First, find velocity:

v(t) = \int a(t) dt = \int kt dt = \frac{1}{2}kt^2 + C_1

. Since

v(0)=0

C_1=0

. Next, find position:

x(t) = \int v(t) dt = \int \frac{1}{2}kt^2 dt = \frac{1}{6}kt^3 + C_2

. Since

x(0)=0

C_2=0

. Thus,

x(t) = \frac{1}{6}kt^3

Question 4

A velocity-time graph for a particle moving in one dimension shows a straight line with a positive slope that does not pass through the origin. Which statement accurately describes the particle's motion?

The particle is moving with constant positive acceleration. (correct answer)
The particle is moving with constant positive velocity.
The particle's position is constant during the motion.
The particle's acceleration is increasing at a constant rate.

Explanation: The slope of a velocity-time graph represents acceleration. A straight line has a constant slope. Since the slope is positive, the particle has a constant positive acceleration. The fact that it doesn't pass through the origin means its initial velocity was not zero.

Question 5

The position of an object as a function of time is represented by a graph of $x$ versus $t$ . If the graph is a parabola opening upward, which of the following statements about the object's motion is correct?

The object has a constant positive velocity because the graph is a smooth curve.
The object has a constant positive acceleration because the rate of change of velocity is constant. (correct answer)
The object's acceleration is increasing with time because its velocity is increasing.
The object's velocity is constant but its acceleration is not, as indicated by the parabolic shape.

Explanation: A parabolic position-time graph, specifically one opening upward, corresponds to a quadratic function

x(t) = At^2 + Bt + C

with

A > 0

. The velocity is the derivative,

v(t) = 2At + B

, which is a linearly increasing function. The acceleration is the second derivative,

a(t) = 2A

, which is a positive constant.

Question 6

A projectile is launched vertically upward with an initial speed $v_0$ from the ground. Neglecting air resistance, which equation represents its position $y$ as a function of time $t$ ? Let the upward direction be positive.

$y(t) = v_0 - gt$
$y(t) = v_0 t - \frac{1}{2}gt^2$ (correct answer)
$y(t) = v_0 t + gt$
$y(t) = v_0 t + \frac{1}{2}gt^2$

Explanation: Using the standard kinematic equation for position under constant acceleration:

y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2

. Here, the initial position

y_0 = 0

, the initial velocity

v_{0y} = v_0

, and the acceleration

a_y = -g

because gravity acts downward. Substituting these values gives

y(t) = v_0 t - \frac{1}{2}gt^2

Question 7

A model rocket launches vertically from $y_0=0\ \text{m}$ with initial speed $v_0=18\ \text{m/s}$ upward. After launch, the engine cuts off immediately, so the only force is gravity with $a=-9.8\ \text{m/s}^2$ . Air resistance is neglected, and the motion is one-dimensional along the vertical axis. Analyze the motion from $t=0$ until the rocket reaches its peak. Based on the scenario described, determine the maximum height reached by the projectile.

$8.3\ \text{m}$ above launch point
$16.5\ \text{m}$ above launch point (correct answer)
$33.1\ \text{m}$ above launch point
$16.5\ \text{m/s}$ above launch point

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through vertical projectile analysis. Vertical motion with constant acceleration follows kinematic equations where maximum height occurs when velocity reaches zero. In this scenario, a rocket launches upward at 18 m/s against gravity (-9.8 m/s²), reaching maximum height when v = 0, using v² = v0² - 2gh gives h = v0²/(2g) = (18)²/(2×9.8) ≈ 16.5 m. Choice B is correct because it properly calculates the maximum height using energy conservation or kinematic equations. Choice D is incorrect because it confuses height (measured in meters) with velocity (measured in m/s), a common dimensional error when students rush through problems. To help students: Emphasize checking units in final answers. Practice recognizing when velocity equals zero at turning points in motion.

Question 8

Two carts move on a frictionless track along the $x$ axis toward each other. Cart 1 starts at $x_1=0\ \text{m}$ with constant velocity $+2.0\ \text{m/s}$ , and cart 2 starts at $x_2=10\ \text{m}$ with constant velocity $-1.0\ \text{m/s}$ . External forces are negligible before the collision, so each cart's acceleration is zero until impact. Analyze the motion from $t=0$ until the carts collide. Based on the scenario described, calculate the displacement of the object after 5 seconds.

Cart 1: $+10\ \text{m}$ , Cart 2: $-5\ \text{m}$ (correct answer)
Cart 1: $+5\ \text{m}$ , Cart 2: $-10\ \text{m}$
Cart 1: $+10\ \text{m}$ , Cart 2: $+5\ \text{m}$
Cart 1: $+10\ \text{m/s}$ , Cart 2: $-5\ \text{m/s}$

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through relative motion analysis. When objects move with constant velocities, displacement equals velocity times time for each object independently. In this scenario, Cart 1 moves at +2.0 m/s for 5 seconds, giving displacement Δx1 = 2.0 × 5 = +10 m, while Cart 2 moves at -1.0 m/s for 5 seconds, giving displacement Δx2 = -1.0 × 5 = -5 m. Choice A is correct because it properly calculates displacement as velocity times time for each cart with correct signs. Choice D is incorrect because it confuses displacement (measured in meters) with velocity (measured in m/s), showing the velocity values instead of calculating displacement. To help students: Emphasize the difference between instantaneous quantities (velocity) and accumulated quantities (displacement). Practice problems with multiple objects to build comfort with relative motion.

Question 9

A tennis ball is thrown straight upward from $y_0=1.5\ \text{m}$ with initial velocity $v_0=14\ \text{m/s}$ . Gravity provides constant acceleration $a=-9.8\ \text{m/s}^2$ , and air resistance is neglected. The motion is one-dimensional, and upward is defined as the positive direction. Analyze the motion for the first $t=3.0\ \text{s}$ after release. Based on the scenario described, describe the motion of the object using kinematic equations.

$y(t)=1.5+14t-4.9t^2$ with $v(t)=14-9.8t$ (correct answer)
$y(t)=1.5+14t+4.9t^2$ with $v(t)=14+9.8t$
$y(t)=1.5+9.8t-4.9t^2$ with $v(t)=9.8-14t$
$y(t)=14+1.5t-9.8t^2$ with $v(t)=1.5-9.8t$

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through kinematic equation application. One-dimensional motion with constant acceleration follows position and velocity equations derived from calculus or kinematic relationships. In this scenario, with y0 = 1.5 m, v0 = 14 m/s, and a = -9.8 m/s², the position equation is y(t) = y0 + v0t + ½at² = 1.5 + 14t - 4.9t², and velocity is v(t) = v0 + at = 14 - 9.8t. Choice A is correct because it properly applies both kinematic equations with correct signs for upward initial velocity and downward acceleration. Choice B is incorrect because it uses positive acceleration (+4.9t²), failing to recognize that gravity acts downward (negative in the upward-positive convention). To help students: Consistently define coordinate systems before solving. Practice deriving velocity from position equations to verify consistency.

Question 10

A ball is thrown from ground level at $v_0=18\,\text{m/s}$ and $\theta=50^\circ$ above the horizontal, with $g=9.8\,\text{m/s}^2$ downward and no air resistance. Take $t=0$ at launch and analyze motion during the first $2.0\,\text{s}$ . The only force during flight is gravity, so acceleration is constant and vertical. Initial position is $\vec r_0=\langle 0,0\rangle\,\text{m}$ , and $v_x$ remains constant. Use component kinematics to model the motion. Based on the scenario described, describe the motion of the object using kinematic equations.

$x(t)=(v_0\cos\theta)t,\; y(t)=(v_0\sin\theta)t-\tfrac12gt^2$ (correct answer)
$x(t)=\tfrac12(v_0\cos\theta)t^2,\; y(t)=(v_0\sin\theta)t-gt^2$
$x(t)=(v_0\cos\theta)t-\tfrac12gt^2,\; y(t)=(v_0\sin\theta)t$
$x(t)=(v_0\cos\theta)t,\; y(t)=(v_0\sin\theta)t-\tfrac12g^2t^2$

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through parametric equations for projectile motion. Projectile motion requires separate kinematic equations for horizontal and vertical components, with constant horizontal velocity and constant vertical acceleration. In this scenario, a ball is thrown at an angle, requiring component analysis with x(t) for horizontal motion and y(t) for vertical motion. Choice A is correct because horizontal motion has constant velocity: x(t)=(v₀cosθ)t with no acceleration term, while vertical motion includes initial velocity and gravitational acceleration: y(t)=(v₀sinθ)t-½gt². Choice B is incorrect because it incorrectly includes ½ in the horizontal equation, suggesting acceleration where none exists, and omits the ½ factor in the vertical equation. To help students: Emphasize that horizontal motion has zero acceleration in projectile problems. Create tables showing initial conditions and accelerations for each component, and practice deriving position equations from first principles.

Question 11

A soccer ball is kicked from ground level at $v_0=20\ \text{m/s}$ and $\theta=35^\circ$ , with $g=9.8\ \text{m/s}^2$ downward and no air resistance. The ball starts at $x_0=0\ \text{m}$ and $y_0=0\ \text{m}$ and moves as a projectile. Only gravitational force acts after launch, so acceleration is constant. Analyze the motion from $t=0$ to $t=2.0\ \text{s}$ . Based on the scenario described, determine the maximum height reached by the projectile.

$2.7\ \text{m}$ above launch level
$6.6\ \text{m}$ above launch level (correct answer)
$13.3\ \text{m}$ above launch level
$6.6\ \text{m/s}$ above launch level

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through projectile motion analysis. Projectile motion involves analyzing the independent horizontal and vertical components of motion, where only gravity acts in the vertical direction. In this scenario, a soccer ball is kicked at 20 m/s at 35° above horizontal, creating initial velocity components vx = 20cos(35°) ≈ 16.4 m/s and vy = 20sin(35°) ≈ 11.5 m/s. Choice B is correct because the maximum height occurs when vy = 0, using vy² = v0y² - 2gh gives h = v0y²/(2g) = (11.5)²/(2×9.8) ≈ 6.6 m. Choice C is incorrect because it doubles the correct answer, a common error when students confuse the maximum height formula with total vertical displacement. To help students: Use component analysis diagrams to visualize velocity vectors. Practice identifying when vertical velocity equals zero at the peak of trajectory.

Question 12

A soccer ball is kicked from ground level at $v_0=20\,\text{m/s}$ and $\theta=35^\circ$ , with $g=9.8\,\text{m/s}^2$ downward and no air resistance; take $t=0$ at launch and analyze motion over $0\le t\le 3\,\text{s}$ . The only force during flight is gravity, so acceleration is constant and vertical. Initial position is $\vec r_0=\langle 0,0\rangle\,\text{m}$ and initial velocity components follow from the launch angle. Students may represent motion using kinematic equations for each component. Based on the scenario described, determine the maximum height reached by the projectile.

$2.0\,\text{m}$ above launch point
$6.6\,\text{m}$ above launch point (correct answer)
$3.3\,\text{m}$ above launch point
$66\,\text{m}$ above launch point

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through kinematic analysis of projectile motion. Projectile motion involves analyzing the independent horizontal and vertical components of motion, where the vertical component experiences constant acceleration due to gravity while the horizontal component maintains constant velocity. In this scenario, a soccer ball is kicked at 20 m/s at a 35° angle, requiring us to find the maximum height using kinematic equations. Choice B is correct because at maximum height, the vertical velocity becomes zero, so using v²=v₀²+2aΔy with v_y=0, v₀y=20sin(35°)≈11.5 m/s, and a=-9.8 m/s², we get Δy=v₀y²/(2g)≈6.6 m. Choice C is incorrect because it appears to use half the correct value, possibly from misapplying the equation or using the wrong initial velocity component. To help students: Emphasize decomposing initial velocity into components using trigonometry. Use energy methods as an alternative check, and practice identifying when vertical velocity equals zero at the peak.

Question 13

A car moves around a flat circular track of radius $r=80\,\text{m}$ at constant speed $v=16\,\text{m/s}$ for a time interval of $0\le t\le 25\,\text{s}$ . The car's velocity changes direction continuously while its speed stays constant. The net force is horizontal and points toward the center, producing centripetal acceleration. Ignore banking and assume tires provide sufficient static friction. Based on the scenario described, what is the velocity of the object at time $t=10\,\text{s}$ ?

$16\,\text{m/s}$ , tangent to the circle (correct answer)
$3.2\,\text{m/s}$ , radially inward
$16\,\text{m/s}^2$ , tangent to the circle
$0\,\text{m/s}$ because speed is constant

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through understanding the distinction between velocity and acceleration in circular motion. Uniform circular motion involves constant speed but changing velocity direction, requiring students to understand velocity as a vector quantity with both magnitude and direction. In this scenario, a car moves at constant speed around a circular track, and we need to identify the velocity at a specific time. Choice A is correct because velocity in circular motion has magnitude equal to the speed (16 m/s) and direction tangent to the circle at any instant, regardless of the specific time. Choice B is incorrect because it confuses velocity with centripetal acceleration, which points radially inward with magnitude v²/r=3.2 m/s². To help students: Use vector diagrams to show velocity tangent to the path and acceleration pointing toward the center. Emphasize that constant speed does not mean constant velocity when direction changes.

Question 14

A drone flies in the horizontal plane with initial position $\vec r_0=\langle 0,0\rangle\,\text{m}$ and initial velocity $\vec v_0=\langle 6,2\rangle\,\text{m/s}$ . A constant acceleration from its motors acts only in the $x$ -direction: $\vec a=\langle 1.5,0\rangle\,\text{m/s}^2$ for $0\le t\le 6\,\text{s}$ . Neglect wind and assume the acceleration remains constant throughout the interval. Students model each component using kinematic equations. Based on the scenario described, calculate the displacement of the object after $5\,\text{s}$ .

$\langle 30,10\rangle\,\text{m}$
$\langle 48.75,10\rangle\,\text{m}$ (correct answer)
$\langle 48.75,0\rangle\,\text{m}$
$\langle 48.75,10\rangle\,\text{m/s}$

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through vector kinematics with constant acceleration in two dimensions. Two-dimensional motion with constant acceleration requires applying kinematic equations independently to each component, then combining results as vectors. In this scenario, a drone has initial velocity components and constant acceleration only in the x-direction, requiring separate analysis of x and y motions. Choice B is correct because using Δr=v₀t+½at² for each component: Δx=(6)(5)+½(1.5)(25)=30+18.75=48.75 m and Δy=(2)(5)+½(0)(25)=10 m, giving displacement ⟨48.75,10⟩ m. Choice D is incorrect because it has units of velocity rather than displacement, showing confusion between kinematic quantities. To help students: Set up separate kinematic equations for x and y components before solving. Use tables to organize initial conditions, accelerations, and results for each component, and always verify units match the quantity requested.

Question 15

The position of a particle is given by the function $x(t)$ . Which of the following mathematical expressions represents the particle's instantaneous acceleration at time $t$ ?

$\int x(t) dt$
$\frac{dx}{dt}$
$\frac{d^2x}{dt^2}$ (correct answer)
$\frac{d}{dt} \left( \frac{1}{2} \left( \frac{dx}{dt} \right)^2 \right)$

Explanation: Instantaneous velocity is the first derivative of position with respect to time,

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

. Instantaneous acceleration is the first derivative of velocity with respect to time,

a(t) = \frac{dv}{dt}

. Substituting the expression for velocity gives

a(t) = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2x}{dt^2}

Question 16

The velocity of an object is given by $v(t) = 2t^3 - 9t^2 + 12t$ . The average acceleration of the object from $t=0$ s to $t=2$ s is $\bar{a}$ , and the instantaneous acceleration at $t=1$ s is $a(1)$ . Which statement is correct?

$\bar{a} > a(1)$ (correct answer)
$a(1) > \bar{a}$
$\bar{a} = a(1)$
The relationship cannot be determined without the initial position.

Explanation: Average acceleration is

\bar{a} = \frac{v(2)-v(0)}{2-0}

v(2) = 2(8) - 9(4) + 12(2) = 4

m/s.

v(0)=0

. So

\bar{a} = \frac{4-0}{2} = 2

m/s². Instantaneous acceleration is

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

. At

t=1

a(1) = 6 - 18 + 12 = 0

m/s². Therefore,

\bar{a} > a(1)

Question 17

The position-time graph of an object is a cubic function of time, $x(t) = At^3$ , where $A$ is a positive constant. Which of the following statements correctly describes the acceleration of the object?

The acceleration is zero because the motion is smooth.
The acceleration is constant and non-zero.
The acceleration increases linearly with time. (correct answer)
The acceleration is proportional to the square of time.

Explanation: To find acceleration, we take the second derivative of the position function. First derivative (velocity):

v(t) = \frac{dx}{dt} = 3At^2

. Second derivative (acceleration):

a(t) = \frac{dv}{dt} = 6At

. This shows that acceleration

a

is directly proportional to time

t

, meaning it increases linearly.

Question 18

The acceleration of a car is shown on a graph of $a$ versus $t$ . The car starts with an initial velocity $v_0$ . What does the area under the acceleration-time graph from $t=0$ to a time $t_f$ represent?

The average velocity of the car during the interval.
The final velocity of the car, $v_f$ , at time $t_f$ .
The displacement of the car during the interval.
The change in the car's velocity, $v_f - v_0$ , during the interval. (correct answer)

Explanation: From the definition of acceleration

a = dv/dt

, we can write

dv = a dt

. Integrating both sides from the initial state to the final state gives

\int_{v_0}^{v_f} dv = \int_{0}^{t_f} a(t) dt

. This yields

v_f - v_0 = \Delta v

, which is the area under the a-t graph.

Question 19

An object is dropped from rest near the surface of a planet where the acceleration due to gravity is $g$ . Assuming negligible air resistance, which of the following correctly represents its position $y$ as a function of time $t$ , with $y=0$ at the initial position and downward as the positive direction?

$y(t) = gt$
$y(t) = \frac{1}{2}gt^2$ (correct answer)
$y(t) = mg - gt$
$y(t) = g$

Explanation: For an object in free fall from rest, the motion is described by the constant acceleration kinematic equation

y = y_0 + v_{0y}t + \frac{1}{2}a_y t^2

. With

y_0=0

v_{0y}=0

, and

a_y=g

(since downward is positive), the equation simplifies to

y(t) = \frac{1}{2}gt^2

Question 20

A car starts from rest and accelerates uniformly. It then travels at a constant velocity, and finally decelerates uniformly to a stop. Which description best matches the position-time graph for this motion?

A straight line with positive slope, a horizontal line, then a straight line with negative slope.
A parabolic curve opening upward, a straight line with positive slope, then a parabolic curve opening downward. (correct answer)
A horizontal line, a straight line with positive slope, then a horizontal line.
A parabolic curve opening downward, a straight line with negative slope, then a parabolic curve opening upward.

Explanation: Uniform acceleration from rest means velocity increases linearly, so position

x

changes quadratically (a parabola opening upward). Constant velocity means position changes linearly (a straight line with positive slope). Uniform deceleration means velocity decreases linearly, so position changes quadratically, but the curve flattens (a parabola opening downward).