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AP Physics C: Mechanics Unit 1 practice questions

Try 10 multiple-choice questions from Unit 1 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. 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?

    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.

    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.

    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.

    The claim is incorrect because displacement represents the magnitude of the path traveled and must be a positive value for any time interval.

    Show answer

    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.

  2. 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?

    Correct, because displacement is the magnitude of the vector sum of the individual component vectors.

    Incorrect, because displacement is the scalar sum of the path lengths, which equals 7.0 m in this case.

    Incorrect, because displacement is the average of the component vectors, which yields 3.5 m in this case.

    Correct, because displacement is the algebraic sum of the magnitudes of the perpendicular vectors.

    Show answer

    Answer: (A)

    Displacement is the straight-line vector connecting the initial and final positions. The magnitude is found using the Pythagorean theorem (42+32=5\sqrt{4^2 + 3^2} = 5), representing the vector sum of the north and east components.

  3. 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?

    Displacement is the difference between the final position y(2) and the initial position y(0), regardless of the path.

    Displacement is the sum of the absolute distances traveled between the turning points at t = 1 s and t = 2 s.

    Displacement is the integral of the speed function |v(t)| calculated over the specific time interval from start to end.

    Displacement is the average position of the drone calculated over the time interval using the mean value theorem.

    Show answer

    Answer: (A)

    Displacement is defined strictly as Δy=yf−yi\Delta y = y_f - y_i. Calculating y(2)=4y(2) = 4 and y(0)=0y(0) = 0 yields a displacement of 4 m, independent of any turning points or distance traveled.

  4. 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?

    Correct, because average velocity is the net displacement of 12 m divided by the total time of 6 s.

    Incorrect, because average velocity is the total distance of 24 m divided by the total time of 6 s.

    Incorrect, because average velocity is the arithmetic mean of the interval velocities, which is 4.0 m/s.

    Correct, because average velocity is the average of the initial and final instantaneous velocities.

    Show answer

    Answer: (A)

    Average velocity is defined as total displacement divided by total time (Δx/Δt\Delta x / \Delta t). Using the endpoints (12−0=1212 - 0 = 12 m) and total time (6 s) yields 2.0 m/s.

  5. Question 5

    A projectile is launched with initial speed v0v_0 at an angle of 60∘60^\circ above the horizontal (Scenario 1) and then with the same speed at 30∘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?

    The speed in Scenario 2 is 3\sqrt{3} times the speed in Scenario 1

    The speed in Scenario 2 is 33 times the speed in Scenario 1

    The speed in Scenario 2 is equal to the speed in Scenario 1

    The speed in Scenario 2 is 1/31/\sqrt{3} times the speed in Scenario 1

    Show answer

    Answer: (A)

    At the peak, vertical velocity is zero, so speed equals the horizontal component vx=v0cos⁡θv_x = v_0 \cos \theta. Comparing cos⁡(30∘)=3/2\cos(30^\circ) = \sqrt{3}/2 and cos⁡(60∘)=1/2\cos(60^\circ) = 1/2 yields a ratio of 3\sqrt{3}.

  6. Question 6

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

    The magnitude in Scenario 2 is 2\sqrt{2} times that in Scenario 1

    The magnitude in Scenario 2 is 22 times that in Scenario 1

    The magnitude in Scenario 2 is 44 times that in Scenario 1

    The magnitude in Scenario 2 is equal to that in Scenario 1

    Show answer

    Answer: (A)

    In Scenario 1, ∣Δv⃗∣=v2+v2=v2|\Delta \vec{v}| = \sqrt{v^2+v^2} = v\sqrt{2}. In Scenario 2, ∣Δv⃗∣=∣v−(−v)∣=2v|\Delta \vec{v}| = |v - (-v)| = 2v. The ratio 2v/(v2)2v / (v\sqrt{2}) simplifies to 2\sqrt{2}.

  7. Question 7

    A constant force F⃗=5i^\vec{F} = 5\hat{i} N acts on an object. In Scenario 1, the object undergoes displacement d⃗1=4i^\vec{d}_1 = 4\hat{i} m. In Scenario 2, it undergoes displacement d⃗2=(4i^+3j^)\vec{d}_2 = (4\hat{i} + 3\hat{j}) m. How does the work done by the force compare in these scenarios?

    Work in Scenario 2 is equal because the parallel component matches

    Work in Scenario 2 is greater due to larger displacement magnitude

    Work in Scenario 2 is less due to the perpendicular force component

    Work in Scenario 2 is undefined without knowing the path duration

    Show answer

    Answer: (A)

    Work is the dot product F⃗⋅d⃗\vec{F} \cdot \vec{d}. Since F⃗\vec{F} has only an x-component, only the x-component of displacement contributes. Both scenarios have dx=4d_x = 4 m.

  8. Question 8

    Object A has position r⃗A(t)=(2t3)i^\vec{r}_A(t) = (2t^3)\hat{i}. Object B has position r⃗B(t)=(3t2)i^\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=1t = 1 s?

    Magnitude for A is twice the magnitude for B

    Magnitude for A is four times the magnitude for B

    Magnitude for A is half the magnitude for B

    Magnitude for A is equal to the magnitude for B

    Show answer

    Answer: (A)

    Differentiating twice: aA(t)=12ta_A(t) = 12t and aB(t)=6a_B(t) = 6. At t=1t=1, aA=12a_A = 12 and aB=6a_B = 6, so A is twice B.

  9. Question 9

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

    Acceleration in Scenario 2 is 3\sqrt{3} times that in Scenario 1

    Acceleration in Scenario 2 is 22 times that in Scenario 1

    Acceleration in Scenario 2 is 1/31/\sqrt{3} times that in Scenario 1

    Acceleration in Scenario 2 is equal to that in Scenario 1

    Show answer

    Answer: (A)

    Acceleration is gsin⁡θg \sin \theta. Comparing sin⁡(60∘)=3/2\sin(60^\circ) = \sqrt{3}/2 and sin⁡(30∘)=1/2\sin(30^\circ) = 1/2 gives a ratio of 3\sqrt{3}.

  10. Question 10

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

    Magnitudes are equal because displacement magnitudes are identical

    Magnitudes are equal because path lengths traveled are identical

    Magnitude in Scenario 1 is greater due to the y-component

    Magnitude in Scenario 2 is greater due to the x-component

    Show answer

    Answer: (A)

    Displacement magnitude in S1 is 32+42=5\sqrt{3^2+4^2}=5. In S2 it is 5. Since time is equal, average velocity magnitudes (d/td/t) are equal.

What's on the AP Physics C: Mechanics Unit 1 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

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

Questions are drawn from 1122 multiple-choice questions and 29 FRQs written for Unit 1, Kinematics.

Practice AP Physics C: Mechanics Unit 1 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 1 before you start

Other AP Physics C: Mechanics unit tests