---
title: "AP Physics C: Mechanics Unit 1 Practice Test"
description: "A timed AP Physics C: Mechanics Unit 1 (Kinematics) practice test in the AP format, drawn from 1,122 multiple-choice questions and 29 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-physics-c-mechanics/unit-test/1"
type: "unit-test"
subject: "AP Physics C: Mechanics"
unit: "Unit 1"
---

# AP Physics C: Mechanics Unit 1 Practice Test

## Overview

A timed AP Physics C: Mechanics Unit 1 (Kinematics) practice test in the AP format, drawn from 1,122 multiple-choice questions and 29 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

## Unit FRQs by Type

- [FRQ 1 – Mathematical Routines](/ap-physics-c-mechanics/frqs/mathematical-routines?unit=1): 9 questions
- [FRQ 2 – Translation Between Representations](/ap-physics-c-mechanics/frqs/translation-between-representations?unit=1): 10 questions
- [FRQ 3 – Experimental Design](/ap-physics-c-mechanics/frqs/experimental-design?unit=1): 3 questions
- [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-c-mechanics/frqs/qualitative-quantitative-translation?unit=1): 7 questions

## Practice Questions

### Question 1

A particle moves along the x-axis with a velocity given by v(t) = 3t^2 - 12 m/s. A student claims that the displacement of the particle from t = 0 s to t = 3 s is -9 m. Which of the following statements correctly evaluates this claim with appropriate physical justification?

- (A) The claim is correct because the definite integral of velocity from t = 0 to t = 3 accounts for direction and yields the net change in position.
- (B) The claim is correct because the definite integral of speed from t = 0 to t = 3 accounts for magnitude and yields the total path length traveled.
- (C) The claim is incorrect because the particle changes direction at t = 2 s, requiring the integral to be split to find the total distance traveled.
- (D) The claim is incorrect because displacement represents the magnitude of the path traveled and must be a positive value for any time interval.

**Answer: (A)** Calculating displacement requires integrating the velocity function v(t) over the time interval, which sums the signed areas under the curve to account for direction. The integral of 3t^2 - 12 from 0 to 3 yields (27 - 36) - 0 = -9 m.

### Question 2

A robot travels 4.0 m due north and then 3.0 m due east. A student claims the magnitude of the robot's total displacement is 5.0 m. Which of the following correctly evaluates this claim with appropriate physical justification?

- (A) Correct, because displacement is the magnitude of the vector sum of the individual component vectors.
- (B) Incorrect, because displacement is the scalar sum of the path lengths, which equals 7.0 m in this case.
- (C) Incorrect, because displacement is the average of the component vectors, which yields 3.5 m in this case.
- (D) Correct, because displacement is the algebraic sum of the magnitudes of the perpendicular vectors.

**Answer: (A)** Displacement is the straight-line vector connecting the initial and final positions. The magnitude is found using the Pythagorean theorem ($\sqrt{4^2 + 3^2} = 5$), representing the vector sum of the north and east components.

### Question 3

A drone moves vertically with a position function y(t) = 2t^3 - 9t^2 + 12t, where y is in meters and t is in seconds. A student claims the displacement from t = 0 s to t = 2 s is 4.0 m. Which reasoning correctly supports this claim?

- (A) Displacement is the difference between the final position y(2) and the initial position y(0), regardless of the path.
- (B) Displacement is the sum of the absolute distances traveled between the turning points at t = 1 s and t = 2 s.
- (C) Displacement is the integral of the speed function |v(t)| calculated over the specific time interval from start to end.
- (D) Displacement is the average position of the drone calculated over the time interval using the mean value theorem.

**Answer: (A)** Displacement is defined strictly as $\Delta y = y_f - y_i$. Calculating $y(2) = 4$ and $y(0) = 0$ yields a displacement of 4 m, independent of any turning points or distance traveled.

### Question 4

An object moves along a straight track with positions recorded as x = 0 m at t = 0 s, x = 10 m at t = 1 s, x = 4 m at t = 4 s, and x = 12 m at t = 6 s. A student claims the average velocity is 2.0 m/s. Which statement correctly evaluates this claim?

- (A) Correct, because average velocity is the net displacement of 12 m divided by the total time of 6 s.
- (B) Incorrect, because average velocity is the total distance of 24 m divided by the total time of 6 s.
- (C) Incorrect, because average velocity is the arithmetic mean of the interval velocities, which is 4.0 m/s.
- (D) Correct, because average velocity is the average of the initial and final instantaneous velocities.

**Answer: (A)** Average velocity is defined as total displacement divided by total time ($\Delta x / \Delta t$). Using the endpoints ($12 - 0 = 12$ m) and total time (6 s) yields 2.0 m/s.

### Question 5

A projectile is launched with initial speed $v_0$ at an angle of $60^\circ$ above the horizontal (Scenario 1) and then with the same speed at $30^\circ$ above the horizontal (Scenario 2). How does the speed of the projectile at the highest point of its trajectory compare between the two scenarios?

- (A) The speed in Scenario 2 is $\sqrt{3}$ times the speed in Scenario 1
- (B) The speed in Scenario 2 is $3$ times the speed in Scenario 1
- (C) The speed in Scenario 2 is equal to the speed in Scenario 1
- (D) The speed in Scenario 2 is $1/\sqrt{3}$ times the speed in Scenario 1

**Answer: (A)** At the peak, vertical velocity is zero, so speed equals the horizontal component $v_x = v_0 \cos \theta$. Comparing $\cos(30^\circ) = \sqrt{3}/2$ and $\cos(60^\circ) = 1/2$ yields a ratio of $\sqrt{3}$.

### Question 6

A particle moves in a circle with constant speed $v$. In Scenario 1, it moves through a quarter circle ($90^\circ$). In Scenario 2, it moves through a half circle ($180^\circ$). How does the magnitude of the change in velocity vector $|\Delta \vec{v}|$ compare between the scenarios?

- (A) The magnitude in Scenario 2 is $\sqrt{2}$ times that in Scenario 1
- (B) The magnitude in Scenario 2 is $2$ times that in Scenario 1
- (C) The magnitude in Scenario 2 is $4$ times that in Scenario 1
- (D) The magnitude in Scenario 2 is equal to that in Scenario 1

**Answer: (A)** In Scenario 1, $|\Delta \vec{v}| = \sqrt{v^2+v^2} = v\sqrt{2}$. In Scenario 2, $|\Delta \vec{v}| = |v - (-v)| = 2v$. The ratio $2v / (v\sqrt{2})$ simplifies to $\sqrt{2}$.

### Question 7

A constant force $\vec{F} = 5\hat{i}$ N acts on an object. In Scenario 1, the object undergoes displacement $\vec{d}_1 = 4\hat{i}$ m. In Scenario 2, it undergoes displacement $\vec{d}_2 = (4\hat{i} + 3\hat{j})$ m. How does the work done by the force compare in these scenarios?

- (A) Work in Scenario 2 is equal because the parallel component matches
- (B) Work in Scenario 2 is greater due to larger displacement magnitude
- (C) Work in Scenario 2 is less due to the perpendicular force component
- (D) Work in Scenario 2 is undefined without knowing the path duration

**Answer: (A)** Work is the dot product $\vec{F} \cdot \vec{d}$. Since $\vec{F}$ has only an x-component, only the x-component of displacement contributes. Both scenarios have $d_x = 4$ m.

### Question 8

Object A has position $\vec{r}_A(t) = (2t^3)\hat{i}$. Object B has position $\vec{r}_B(t) = (3t^2)\hat{i}$. How does the magnitude of acceleration for Object A compare to that of Object B at time $t = 1$ s?

- (A) Magnitude for A is twice the magnitude for B
- (B) Magnitude for A is four times the magnitude for B
- (C) Magnitude for A is half the magnitude for B
- (D) Magnitude for A is equal to the magnitude for B

**Answer: (A)** Differentiating twice: $a_A(t) = 12t$ and $a_B(t) = 6$. At $t=1$, $a_A = 12$ and $a_B = 6$, so A is twice B.

### Question 9

A block sits on a frictionless incline. In Scenario 1, the incline angle is $30^\circ$. In Scenario 2, the incline angle is $60^\circ$. How does the magnitude of the acceleration vector pointing down the incline compare?

- (A) Acceleration in Scenario 2 is $\sqrt{3}$ times that in Scenario 1
- (B) Acceleration in Scenario 2 is $2$ times that in Scenario 1
- (C) Acceleration in Scenario 2 is $1/\sqrt{3}$ times that in Scenario 1
- (D) Acceleration in Scenario 2 is equal to that in Scenario 1

**Answer: (A)** Acceleration is $g \sin \theta$. Comparing $\sin(60^\circ) = \sqrt{3}/2$ and $\sin(30^\circ) = 1/2$ gives a ratio of $\sqrt{3}$.

### Question 10

A particle travels from the origin to $(3, 4)$ in 5 seconds (Scenario 1). Another particle travels from the origin to $(5, 0)$ in 5 seconds (Scenario 2). How do the magnitudes of the average velocity vectors compare?

- (A) Magnitudes are equal because displacement magnitudes are identical
- (B) Magnitudes are equal because path lengths traveled are identical
- (C) Magnitude in Scenario 1 is greater due to the y-component
- (D) Magnitude in Scenario 2 is greater due to the x-component

**Answer: (A)** Displacement magnitude in S1 is $\sqrt{3^2+4^2}=5$. In S2 it is 5. Since time is equal, average velocity magnitudes ($d/t$) are equal.

## Review the Unit

- [1.1 Scalars and Vectors](/ap-physics-c-mechanics/unit-1/1-scalars-and-vectors/study-guide/rVQeOgdT8itcgCoV)
- [1.2 Displacement, Velocity, and Acceleration](/ap-physics-c-mechanics/unit-1/2-displacement-velocity-and-acceleration/study-guide/robnlCwaanT6NImP)
- [1.3 Representing Motion](/ap-physics-c-mechanics/unit-1/3-representing-motion/study-guide/ZIECLULiWCrBlX16)
- [1.4 Reference Frames and Relative Motion](/ap-physics-c-mechanics/unit-1/4-reference-frames-and-relative-motion/study-guide/MhWvdpnoJuVbZ0WW)
- [1.5 Motion in Two or Three Dimensions](/ap-physics-c-mechanics/unit-1/5-motion-in-two-or-three-dimensions/study-guide/FnxHaY283LuHyd54)

## Other Unit Tests

- [Unit 2: Force and Translational Dynamics](/ap-physics-c-mechanics/unit-test/2)
- [Unit 3: Work, Energy, and Power](/ap-physics-c-mechanics/unit-test/3)
- [Unit 4: Linear Momentum](/ap-physics-c-mechanics/unit-test/4)
- [Unit 5: Torque and Rotational Dynamics](/ap-physics-c-mechanics/unit-test/5)
- [Unit 6: Energy and Momentum of Rotating Systems](/ap-physics-c-mechanics/unit-test/6)
- [Unit 7: Oscillations](/ap-physics-c-mechanics/unit-test/7)

## About This Document

Canonical Fiveable pages are available as Markdown at the same path plus `.md`.

- [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns
- [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing
- [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`)
- [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)