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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.

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    Answer: (A)

    Displacement is the net change in position, found by integrating the signed velocity v(t) over the interval, so direction is accounted for. The integral of 3t2−123t^2 - 12 from 0 to 3 is (27−36)−0=−9(27 - 36) - 0 = -9 m, so the claim is correct.

  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.

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    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.

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    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.

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    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

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    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

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    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 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

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    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.

  8. Question 8

    Vector A⃗=2i^−2j^\vec{A} = 2\hat{i} - 2\hat{j} and vector B⃗=−3i^+3j^\vec{B} = -3\hat{i} + 3\hat{j}. Which of the following correctly compares the directions of these two vectors?

    The vectors point in exactly opposite directions

    The vectors point in exactly the same direction

    The vectors are perpendicular to each other

    The vectors are separated by an angle of 45 degrees

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    Answer: (A)

    Vector B⃗\vec{B} is a scalar multiple of A⃗\vec{A} by −1.5-1.5. The negative scalar indicates they are antiparallel (180∘180^\circ apart).

  9. Question 9

    Train A moves East at speed vv. Train B moves West at speed vv. How does the velocity of A relative to B (v⃗AB\vec{v}_{AB}) compare to the velocity of B relative to A (v⃗BA\vec{v}_{BA})?

    Magnitudes are equal; directions are opposite

    Magnitudes are equal; directions are identical

    Magnitudes are different; directions are opposite

    Magnitudes are different; directions are identical

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    Answer: (A)

    v⃗AB=v⃗A−v⃗B=vi^−(−vi^)=2vi^\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = v\hat{i} - (-v\hat{i}) = 2v\hat{i}. v⃗BA=v⃗B−v⃗A=−2vi^\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = -2v\hat{i}. Magnitudes are 2v2v, directions are opposite.

  10. Question 10

    Vector A⃗\vec{A} has magnitude 4 and Vector B⃗\vec{B} has magnitude 3. In Scenario 1, the angle between them is 0∘0^\circ. In Scenario 2, the angle is 180∘180^\circ. How does the scalar product A⃗⋅B⃗\vec{A} \cdot \vec{B} compare?

    Product in S1 is positive; S2 is negative with same magnitude

    Product in S1 is positive; S2 is negative with different magnitude

    Product in S1 is zero; S2 is zero due to symmetry

    Product in S1 is equal to the product in Scenario 2

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    Answer: (A)

    Dot product is ABcos⁡θAB \cos \theta. S1: 12cos⁡(0)=1212 \cos(0) = 12. S2: 12cos⁡(180)=−1212 \cos(180) = -12. Magnitudes are equal, signs opposite.

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 884 multiple-choice questions and 26 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