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AP Physics 1 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

    Object A moves at a constant velocity v. Object B starts from rest and accelerates uniformly to a final velocity of 2v over the same time interval t. Which statement correctly compares the magnitude of displacement Δx for the two objects?

    Δx_A is equal to Δx_B because the average velocity for both objects is v.

    Δx_A is less than Δx_B because Object B reaches a higher final velocity of 2v.

    Δx_A is greater than Δx_B because Object A maintains constant velocity throughout the interval.

    Δx_A is less than Δx_B because Object B accelerates while Object A does not change velocity.

    Show answer

    Answer: (A)

    Calculating the average velocity for uniform acceleration (v_avg = (v_i + v_f)/2) shows that Object B averages v (from 0 to 2v), matching Object A's constant velocity v. Since both travel for time t with the same average velocity, their displacements are identical.

  2. Question 2

    An athlete runs along a circular track of radius R. In Scenario 1, the athlete runs exactly half a lap. In Scenario 2, the athlete runs exactly one full lap. Which statement correctly compares the magnitude of displacement Δx between the two scenarios?

    Δx in Scenario 1 is greater because the final position differs from the initial position.

    Δx in Scenario 1 is less because the total distance traveled is smaller than in Scenario 2.

    Δx in Scenario 1 is equal because the motion occurs on the same circular track in both.

    Δx in Scenario 1 is greater because the average speed is higher than in Scenario 2.

    Show answer

    Answer: (A)

    Displacement is the straight-line distance between start and end points; Scenario 1 has a displacement of 2R (diameter), while Scenario 2 has a displacement of zero because the athlete returns to the start.

  3. Question 3

    A block starts from rest and slides down a frictionless incline. Scenario 1 is the block's motion during the first second (t=0 to t=1). Scenario 2 is the block's motion during the second second (t=1 to t=2). Which statement correctly compares the magnitude of displacement Δx in Scenario 2 to Scenario 1?

    Δx in Scenario 2 is greater because the average velocity is higher.

    Δx in Scenario 2 is equal because the acceleration is constant.

    Δx in Scenario 2 is less because the acceleration is decreasing.

    Δx in Scenario 2 is equal because the time interval is the same.

    Show answer

    Answer: (A)

    Since the block accelerates uniformly from rest, its velocity increases continuously. Therefore, the average velocity during the second interval is higher than during the first, resulting in greater displacement for the same duration.

  4. Question 4

    A cyclist travels 12 km12 \text{ km} East at 12 km/h12 \text{ km/h}, then immediately turns around and travels 6.0 km6.0 \text{ km} West at 12 km/h12 \text{ km/h}. What is the average velocity for the entire trip?

    4.0 km/h4.0 \text{ km/h} directed East

    4.0 km/h4.0 \text{ km/h} directed West

    12.0 km/h12.0 \text{ km/h} directed East

    12.0 km/h12.0 \text{ km/h} directed West

    Show answer

    Answer: (A)

    Calculating average velocity requires dividing total displacement by total time. The displacement is +12 km−6 km=+6 km+12 \text{ km} - 6 \text{ km} = +6 \text{ km}, and the total time is 1.0 h+0.5 h=1.5 h1.0 \text{ h} + 0.5 \text{ h} = 1.5 \text{ h}, yielding 4.0 km/h4.0 \text{ km/h} East.

  5. Question 5

    A tennis ball moving horizontally at 25 m/s25 \text{ m/s} strikes a vertical wall and rebounds in the opposite direction at 20 m/s20 \text{ m/s}. If the contact time is 0.050 s0.050 \text{ s}, what is the magnitude of the average acceleration?

    100 m/s2100 \text{ m/s}^2

    500 m/s2500 \text{ m/s}^2

    900 m/s2900 \text{ m/s}^2

    1100 m/s21100 \text{ m/s}^2

    Show answer

    Answer: (C)

    Calculating acceleration requires the change in velocity vector: vf−viv_f - v_i. Taking the initial direction as positive, Δv=−20 m/s−25 m/s=−45 m/s\Delta v = -20 \text{ m/s} - 25 \text{ m/s} = -45 \text{ m/s}, so magnitude is 45/0.050=900 m/s245 / 0.050 = 900 \text{ m/s}^2.

  6. Question 6

    A particle travels +15 m+15 \text{ m} in 3.0 s3.0 \text{ s}, stops for 2.0 s2.0 \text{ s}, and then travels −25 m-25 \text{ m} in 5.0 s5.0 \text{ s}. What is the average velocity over the entire 10.0 s10.0 \text{ s} interval?

    1.0 m/s1.0 \text{ m/s} in the negative x-direction

    1.0 m/s1.0 \text{ m/s} in the positive x-direction

    4.0 m/s4.0 \text{ m/s} in the negative x-direction

    4.0 m/s4.0 \text{ m/s} in the positive x-direction

    Show answer

    Answer: (A)

    Calculating average velocity requires total displacement divided by total time. The sum is +15 m+0 m−25 m=−10 m+15 \text{ m} + 0 \text{ m} - 25 \text{ m} = -10 \text{ m}, divided by 10 s10 \text{ s}, yielding −1.0 m/s-1.0 \text{ m/s}.

  7. Question 7

    Two carts on a straight track move toward each other. Cart A moves right at 4.0 m/s4.0 \text{ m/s} and Cart B moves left at 6.0 m/s6.0 \text{ m/s}. If they are 50 m50 \text{ m} apart at t=0t=0, when do they collide?

    2.5 s2.5 \text{ s}

    5.0 s5.0 \text{ s}

    12.5 s12.5 \text{ s}

    25.0 s25.0 \text{ s}

    Show answer

    Answer: (B)

    Calculating the collision time requires the relative speed of approach. Since they move in opposite directions, relative speed is 4.0−(−6.0)=10.0 m/s4.0 - (-6.0) = 10.0 \text{ m/s}. Time is distance divided by rate: 50/10=5.0 s50 / 10 = 5.0 \text{ s}.

  8. Question 8

    A stone is thrown vertically upward with an initial speed of 20 m/s20 \text{ m/s}. Taking g=10 m/s2g = 10 \text{ m/s}^2 and the upward direction as positive, what is the displacement of the stone after 3.0 s3.0 \text{ s}?

    15 m15 \text{ m} upward from release point

    15 m15 \text{ m} downward from release point

    25 m25 \text{ m} upward from release point

    25 m25 \text{ m} downward from release point

    Show answer

    Answer: (A)

    Calculating displacement uses the kinematic equation Δy=v0t+12at2\Delta y = v_0t + \frac{1}{2}at^2. Substituting values: 20(3)+0.5(−10)(32)=60−45=+15 m20(3) + 0.5(-10)(3^2) = 60 - 45 = +15 \text{ m}. The positive sign indicates upward.

  9. Question 9

    A car traveling at +30 m/s+30 \text{ m/s} applies brakes, creating an acceleration of −6.0 m/s2-6.0 \text{ m/s}^2. What is the car's displacement from the moment braking begins until it stops?

    38 m38 \text{ m} in the direction of motion

    75 m75 \text{ m} in the direction of motion

    150 m150 \text{ m} in the direction of motion

    300 m300 \text{ m} in the direction of motion

    Show answer

    Answer: (B)

    Calculating displacement uses v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x. With v=0v=0, 0=302+2(−6)Δx0 = 30^2 + 2(-6)\Delta x, leading to 900=12Δx900 = 12\Delta x, so Δx=75 m\Delta x = 75 \text{ m}. Direction remains positive.

  10. Question 10

    An athlete runs along a semi-circular path of radius R=50 mR = 50 \text{ m} from point A to point B in 20 s20 \text{ s}. What is the magnitude of the average velocity?

    2.5 m/s2.5 \text{ m/s}

    5.0 m/s5.0 \text{ m/s}

    7.9 m/s7.9 \text{ m/s}

    10.0 m/s10.0 \text{ m/s}

    Show answer

    Answer: (B)

    Calculating average velocity uses the straight-line displacement, which is the diameter (2R=100 m2R = 100 \text{ m}). Velocity magnitude is 100 m/20 s=5.0 m/s100 \text{ m} / 20 \text{ s} = 5.0 \text{ m/s}.

What's on the AP Physics 1 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 1047 multiple-choice questions and 52 FRQs written for Unit 1, Kinematics.

Practice AP Physics 1 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 1 unit tests