Unit 1 · 10–15% of the exam Beta

Unit 1 practice test: Kinematics

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value; data sets share one table, graph or scene diagram.

Covers: Scalars and Vectors on a Line; Displacement, Velocity and Acceleration; Representing Motion; Reference Frames and Relative Motion; Vectors and Motion in Two Dimensions.

Graph

Two runners on a straight path

Runners A and B move along the same straight path. The graph shows each runner's position, measured from a lamppost, against time.

051015202530354045012345678Time (s)Position (m)

Runner ARunner B

Data table
Time (s)Runner ARunner B
0010
1512.5
21015
31517.5
42020
52522.5
63025
73527.5
84030

1. At t = 4.0 s, the two lines cross. What is true at that instant?

  1. The runners have the same velocity
  2. The runners are at the same position and have the same velocity
  3. Runner B is just catching up to runner A
  4. The runners are at the same position, but A is faster
Show the answer

On an x-t graph, the height is the position and the slope is the velocity. Crossing lines share a position, not a velocity.

  • The runners have the same velocity: Equal velocity would mean equal slopes. The lines cross, which means equal positions.
  • The runners are at the same position and have the same velocity: The positions match, but the slopes (5.0 and 2.5 m/s) do not.
  • Runner B is just catching up to runner A: B started ahead at 10 m; it is A who catches up with B at 4.0 s.
  • Correct: The runners are at the same position, but A is faster: Right: the crossing means equal x; A's line is steeper, so A has the larger velocity.

2. Which runner has the greater acceleration?

  1. Runner A, because A's line is steeper
  2. Runner B, because B started farther along
  3. Neither; both accelerations are zero
  4. Runner A, because A covers more distance
Show the answer

Straight x-t lines mean constant velocities. A constant velocity means zero acceleration for both.

  • Runner A, because A's line is steeper: A steeper x-t line means a larger velocity, not an acceleration. A straight line has a constant velocity.
  • Runner B, because B started farther along: A starting position does not affect the acceleration.
  • Correct: Neither; both accelerations are zero: Right: both x-t graphs are straight lines, so each runner's velocity is constant and neither accelerates.
  • Runner A, because A covers more distance: More distance in the same time means a larger velocity, not an acceleration.

3. Which describes the two runners' velocity-time graphs?

  1. Two horizontal lines, A's higher than B's
  2. Two lines crossing at t = 4.0 s
  3. Two rising lines, A's steeper
  4. A horizontal line for A and a rising line for B
Show the answer

A straight x-t line has a constant slope, so its v-t graph is a horizontal line at that slope: 5.0 m/s for A, 2.5 m/s for B.

  • Correct: Two horizontal lines, A's higher than B's: Right: each velocity is constant (5.0 and 2.5 m/s), so each v-t graph is flat, and A's is higher.
  • Two lines crossing at t = 4.0 s: The x-t lines cross, but the velocities never become equal.
  • Two rising lines, A's steeper: Rising v-t lines would mean speeding up; these runners keep a steady pace.
  • A horizontal line for A and a rising line for B: Both x-t graphs are straight, so both velocities are constant.

Data table

A cart rolls up a ramp and back

A student gives a cart a quick push up a long, straight ramp and lets go at t = 0. A motion detector at the bottom of the ramp records the cart's velocity, with up the ramp as positive.

Velocity of the cart after release
Time (s)Velocity (m/s)
0.0+1.20
0.5+0.80
1.0+0.40
1.50.00
2.0−0.40
2.5−0.80

4. The cart's velocity changes by the same amount in each 0.5 s interval. What is its acceleration at t = 1.5 s, the instant its velocity is zero?

  1. Zero, because the cart is momentarily at rest
  2. The same as at the other readings: negative, not zero
  3. Positive, because the cart is about to turn around
  4. It is undefined, because Δt is zero at a single instant
Show the answer

Acceleration is the rate of change of velocity, not the velocity. From 1.0 s to 2.0 s the velocity goes +0.40 → −0.40 m/s, the same steady change as before, including at the instant v = 0.

  • Zero, because the cart is momentarily at rest: A zero velocity is not a zero acceleration. The velocity is still changing at that instant: it was positive and becomes negative.
  • Correct: The same as at the other readings: negative, not zero: Right: the velocity keeps changing by −0.40 m/s every 0.5 s, straight through zero, so the acceleration has not changed.
  • Positive, because the cart is about to turn around: The velocity goes from positive to negative, so Δv is negative; the acceleration does not flip sign at the top.
  • It is undefined, because Δt is zero at a single instant: The table shows the same steady change before and after 1.5 s, so the acceleration there has the same value as at the other readings.

5. From t = 1.5 s to t = 2.5 s, the cart rolls back down the ramp. During this time its acceleration is negative. What is happening to its speed?

  1. It decreases, because a negative acceleration means slowing down
  2. It stays the same, because the acceleration does not change
  3. It decreases, because the velocity values are below zero
  4. It increases, because the velocity and acceleration have the same sign
Show the answer

Speeding up: velocity and acceleration point the same way. Slowing down: they point opposite ways. Here both are negative after 1.5 s, so the cart speeds up down the ramp.

  • It decreases, because a negative acceleration means slowing down: Negative only gives a direction. Slowing down happens when acceleration and velocity have opposite signs.
  • It stays the same, because the acceleration does not change: A steady acceleration means the velocity changes steadily; the speed goes from 0 to 0.80 m/s.
  • It decreases, because the velocity values are below zero: The speed is the size of the velocity: 0.40 then 0.80 m/s, which is growing.
  • Correct: It increases, because the velocity and acceleration have the same sign: Right: both are negative, so the velocity grows more negative and the speed (its size) grows from 0 to 0.80 m/s.

6. The ramp keeps going and the pattern continues. What velocity will the detector read at t = 3.5 s?

  1. −1.20 m/s
  2. −1.60 m/s
  3. +1.20 m/s
  4. −0.80 m/s
Show the answer

With a = −0.80 m/s², Δv over 1.0 s is −0.80 m/s: −0.80 m/s + (−0.80 m/s) = −1.60 m/s.

  • −1.20 m/s: This adds only one more step of −0.40 m/s; 3.5 s is two half-second steps after 2.5 s.
  • Correct: −1.60 m/s: Right: the velocity changes by −0.40 m/s each 0.5 s, so two more steps after −0.80 m/s give −1.60 m/s.
  • +1.20 m/s: This assumes the cart turns back up the ramp; nothing in the data reverses the steady change.
  • −0.80 m/s: This assumes the velocity stops changing, but the acceleration is the same at every reading.

7. Which statement about a trip along a straight road is correct?

  1. The size of the displacement is at most equal to the distance
  2. The distance is at most equal to the size of the displacement
  3. The distance and the size of the displacement are equal for a trip that ends where it started
  4. The size of the displacement grows each time the traveler turns around
Show the answer

Distance counts every meter of the path; displacement compares the ends. They are equal only for a trip in one direction.

  • Correct: The size of the displacement is at most equal to the distance: Right: the straight-line change in position is the shortest way between the ends, so it cannot exceed the path length.
  • The distance is at most equal to the size of the displacement: This reverses the rule; turning around adds distance without adding displacement.
  • The distance and the size of the displacement are equal for a trip that ends where it started: A round trip has zero displacement but a nonzero distance.
  • The size of the displacement grows each time the traveler turns around: Turning around brings the traveler back toward the start, which can shrink the displacement.

8. A cart on a flatcar speeds up from rest to 0.60 m/s relative to the flatcar in 2.0 s. The flatcar moves at a constant 5.0 m/s relative to the ground. What is the cart's acceleration relative to the ground?

  1. 0.30 m/s², the same as relative to the flatcar
  2. 2.8 m/s², because the flatcar's velocity adds 5.0 m/s
  3. Zero, because the cart rides on the flatcar
  4. 2.5 m/s², the flatcar's velocity divided by 2.0 s
Show the answer

Relative to the ground: v goes from 5.0 to 5.6 m/s in 2.0 s, a = 0.30 m/s². Frames moving at constant velocity relative to each other measure the same acceleration.

  • Correct: 0.30 m/s², the same as relative to the flatcar: Right: the flatcar moves at a constant velocity, so adding it changes every velocity by the same amount and leaves Δv unchanged.
  • 2.8 m/s², because the flatcar's velocity adds 5.0 m/s: A constant added velocity cancels in Δv; it changes velocities, not accelerations.
  • Zero, because the cart rides on the flatcar: The cart speeds up relative to the flatcar, so it also speeds up relative to the ground.
  • 2.5 m/s², the flatcar's velocity divided by 2.0 s: The flatcar's velocity does not change, so it contributes no acceleration.

9. Two balls roll off the edges of two tables at the same horizontal speed. Table B is 4.0 times as high as table A. Ball A lands 0.50 m from its table. How far from its table does ball B land?

Type a number in m.

Show the answer

Fall time t = √(2h/g) ∝ √h, so B falls √4.0 = 2.0 times as long. With the same vₓ, d = vₓt doubles: 1.0 m.

  • Answer: 1.0 m

10. A ball sits at x = −3.0 m on a number line where east is positive. A student redraws the axis with the same origin but west as positive. What is the ball's position now?

  1. −3.0 m
  2. 0 m
  3. +6.0 m
  4. +3.0 m
Show the answer

The ball is 3.0 m west of the origin either way. With west positive, that is written +3.0 m. Positions depend on the coordinate system you choose.

  • −3.0 m: The sign of a position depends on which direction is chosen as positive; flipping the axis flips the sign.
  • 0 m: Only the direction was changed, not the origin, so the ball is still 3.0 m from it.
  • +6.0 m: Changing the positive direction flips the sign; it does not move the ball 3.0 m farther away.
  • Correct: +3.0 m: Right: the ball has not moved, but with west positive its location 3.0 m west of the origin is written +3.0 m.

11. A sprinter starts from rest and reaches 9.0 m/s in 1.8 s. What is her average acceleration?

Type a number and its unit.

Show the answer

a(avg) = Δv/Δt = (9.0 m/s − 0) ÷ 1.8 s = 5.0 m/s².

  • Answer: 5.0 m/s^2

12. A stone dropped from rest takes 0.60 s to fall to the bottom of a dry well. How long would it take to fall from a height 4 times as great? Air resistance is negligible.

Type a number in s.

Show the answer

From rest, Δy = ½gt², so t = √(2Δy/g) ∝ √Δy. Four times the height gives √4 = 2 times the time: 1.2 s.

  • Answer: 1.2 s

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Practice has every question in the unit, with feedback after each one; this unit's lessons are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.