Free MCAT-style practice, 4A
MCAT Kinematics Practice Questions
Ten free MCAT-style kinematics practice questions on displacement, velocity, acceleration, the kinematic equations, free fall and projectile motion, with full explanations.
Quick concept primer
Kinematics describes motion without asking what causes it. Displacement and velocity are vectors, so direction matters; distance and speed are scalars. With constant acceleration, four relations cover almost every problem: v = v0 + at, Δx = v0t + ½at2, v2 = v02 + 2aΔx, and Δx = ½(v0 + v)t. Pick the one that leaves out the quantity you neither know nor need. On a velocity-time graph, the slope is acceleration and the area under the line is displacement.
A projectile, once released and with air resistance ignored, has only one acceleration: g, straight down. Split the motion into components. The horizontal velocity never changes, the vertical velocity changes by g every second, and the two parts share the same flight time. Launch angles that add to 90 degrees give the same range on level ground, but the steeper one stays up longer.
Traps: velocity is zero at the top of the arc only in the vertical direction, acceleration is never zero in flight, and distance can exceed the size of the displacement. These questions use g = 10 m/s2. The 4A Motion, Forces and Energy pack adds flashcards and more practice on forces, energy and momentum, or try the free sampler first.
10 practice questions
Question 1
A delivery scooter starts from rest at a stoplight and speeds up at a constant 3.0 m/s2 for 4.0 s. How far does it travel in that time?
- 12 m
- 24 m
- 36 m
- 48 m
Show answer and explanation
Answer: B. 24 m
Starting from rest, Δx = ½at2 = 0.5 x 3.0 x 16 = 24 m. The 48 m choice forgets the factor of one half, and 12 m multiplies acceleration by time (giving the final speed, 12 m/s) and mistakes that speed for a distance.
Question 2
A hiker kicks a pebble horizontally off a ledge at 8.0 m/s. The ledge is 20 m above flat ground. Ignoring air resistance, how far from the base of the ledge does the pebble land?
- 8.0 m
- 10 m
- 16 m
- 32 m
Show answer and explanation
Answer: C. 16 m
The vertical motion starts from zero vertical speed, so 20 = ½(10)t2 gives t = 2.0 s. The horizontal speed stays 8.0 m/s the whole time, so the pebble lands 8.0 x 2.0 = 16 m out. Choice A uses 1 s of flight and D uses 4 s.
Question 3
Two identical balls are launched from level ground at the same speed, one at 30 degrees and one at 60 degrees above the horizontal. Ignoring air resistance, which statement is correct?
- The 60 degree ball lands farther away because it rises higher.
- The 30 degree ball lands farther away and also stays in the air longer.
- Both balls land at the same spot and spend the same time in the air.
- Both balls land at the same spot, but the 60 degree ball spends more time in the air.
Show answer and explanation
Answer: D. Both balls land at the same spot, but the 60 degree ball spends more time in the air.
Range on level ground is proportional to sin 2θ, and sin 60 degrees equals sin 120 degrees, so complementary angles give equal ranges. Flight time depends on the vertical launch component, v sin θ, which is larger at 60 degrees, so that ball stays up longer. Height does not by itself add range, which rules out A.
Question 4
A soccer ball is kicked at an angle above level ground and air resistance is negligible. At the highest point of its path, each of the following is true EXCEPT:
- its speed is zero.
- its vertical velocity is zero.
- its acceleration is 10 m/s2 downward.
- its horizontal velocity equals the horizontal component it had at launch.
Show answer and explanation
Answer: A. its speed is zero.
At the peak only the vertical velocity is zero; the horizontal component is unchanged from launch, so the speed equals that horizontal component and is not zero. Gravity still acts at the top, so the acceleration stays 10 m/s2 downward. Choices B, C and D are all true statements.
Question 5
A lab cart moves along a straight track. Its velocity is recorded once per second.
| Time (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Velocity (m/s) | 2 | 5 | 8 | 11 | 14 |
What is the cart's displacement from t = 0 to t = 4 s?
- 8 m
- 28 m
- 32 m
- 56 m
Show answer and explanation
Answer: C. 32 m
Velocity rises by 3 m/s every second, so acceleration is constant and average velocity is (2 + 14)/2 = 8 m/s. Displacement is 8 x 4 = 32 m, the area under the velocity-time line. D multiplies the final velocity by the time, and A uses only the initial velocity.
Question 6
A bolt falls from rest off a high scaffold. Ignoring air resistance, how does the distance it falls during the third second compare with the distance it falls during the first second?
- 3 to 1
- 5 to 1
- 9 to 1
- 3 to 2
Show answer and explanation
Answer: B. 5 to 1
Total distance fallen from rest grows as t2: 5 m after 1 s, 20 m after 2 s, 45 m after 3 s. The third second alone covers 45 − 20 = 25 m against 5 m in the first, a 5 to 1 ratio. The 9 to 1 choice compares total distances after 3 s and 1 s, not the distance within each one-second interval.
Question 7
A juggler tosses a ball straight up at 25 m/s. Ignoring air resistance, about how high above the release point does it rise?
- 31 m
- 25 m
- 63 m
- 125 m
Show answer and explanation
Answer: A. 31 m
At the top the velocity is zero, so v02 = 2gh gives h = 625 / 20 = 31.25 m, about 31 m. Choice C drops the factor of 2 in the denominator (625 / 10), and 125 m comes from multiplying 25 m/s by the 5 s total flight time.
Question 8
A runner completes exactly one lap of a 400 m oval track in 80 s and stops at the starting line. What are the runner's average speed and average velocity for the lap?
- 0 m/s and 0 m/s
- 0 m/s and 5 m/s
- 5 m/s and 5 m/s
- 5 m/s and 0 m/s
Show answer and explanation
Answer: D. 5 m/s and 0 m/s
Average speed is total distance over time, 400 / 80 = 5 m/s. Average velocity is displacement over time, and ending at the start point makes the displacement zero, so average velocity is 0 m/s. C treats the scalar and vector quantities as the same thing.
Question 9
A cart on a straight track has a velocity of +6 m/s. A constant acceleration changes its velocity to −4 m/s over 5.0 s. What total distance does the cart travel during those 5.0 s?
- 5 m
- 10 m
- 13 m
- 25 m
Show answer and explanation
Answer: C. 13 m
The acceleration is (−4 − 6)/5 = −2 m/s2. The cart moves forward until it stops after 3 s, covering 62/(2 x 2) = 9 m, then moves backward for 2 s, covering 42/(2 x 2) = 4 m. Distance is 9 + 4 = 13 m. The 5 m choice is the net displacement, (6 + (−4))/2 x 5, which is not the distance.
Question 10
A ball leaves level ground at 20 m/s, 53 degrees above the horizontal (take sin 53 degrees = 0.80 and cos 53 degrees = 0.60). Ignoring air resistance, about how far away does it land?
- 38 m
- 26 m
- 19 m
- 40 m
Show answer and explanation
Answer: A. 38 m
The vertical component is 16 m/s, so the flight time is 2 x 16 / 10 = 3.2 s. The horizontal component is 12 m/s, giving a range of 12 x 3.2 = 38.4 m, about 38 m. Choice C uses only the 1.6 s rise time, and D is v2/g, the maximum range that only a 45 degree launch reaches.
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