A Level Physics 2.1 Equations of Motion MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Free practice questions for Cambridge International AS & A Level Physics (9702), Section 2.1 Equations of Motion. Work through SUVAT calculations, motion graphs, free fall and projectile motion, then open the answer toggle on each question to check the full worked solution.

Formula recap: v = u + at  |  s = ut + 0.5at2  |  v2 = u2 + 2as  |  s = 0.5(u + v)t  |  s = vt − 0.5at2. Keep every equation and when to use it on one page with the A Level Physics Formula Sheet.

The Equations of Motion (SUVAT)

Q1. A car starts from rest and accelerates uniformly at 3.0 m/s2 for 6.0 s. What is its final speed?

  • (A) 0.50 m/s
  • (B) 18 m/s
  • (C) 54 m/s
  • (D) 108 m/s
Show Answer

Answer: B – Use v = u + at with u = 0: v = 0 + 3.0 × 6.0 = 18 m/s. (54 m is the distance travelled, a common mix-up.)

Q2. A train travelling at 25 m/s brakes with a uniform deceleration of 1.25 m/s2. Taking the direction of travel as positive, how long does it take to come to rest?

  • (A) 5.0 s
  • (B) 10 s
  • (C) 20 s
  • (D) 31 s
Show Answer

Answer: C – v = u + at gives 0 = 25 + (−1.25)t, so t = 25 / 1.25 = 20 s.

Q3. An object moving at 3.0 m/s accelerates uniformly at 2.0 m/s2 over a distance of 40 m. What is its final speed?

  • (A) 11 m/s
  • (B) 13 m/s
  • (C) 15 m/s
  • (D) 169 m/s
Show Answer

Answer: B – Time is not given, so use v2 = u2 + 2as = 3.02 + 2 × 2.0 × 40 = 9 + 160 = 169, giving v = 13 m/s.

Q4. A trolley starts from rest and accelerates uniformly at 4.0 m/s2 for 3.5 s. How far does it travel?

  • (A) 7.0 m
  • (B) 14 m
  • (C) 49 m
  • (D) 24.5 m
Show Answer

Answer: D – Use s = ut + 0.5at2 with u = 0: s = 0.5 × 4.0 × 3.52 = 0.5 × 4.0 × 12.25 = 24.5 m.

Q5. A car travelling at 20 m/s brakes and decelerates uniformly at 5.0 m/s2 until it stops. What is its braking distance?

  • (A) 4.0 m
  • (B) 20 m
  • (C) 40 m
  • (D) 80 m
Show Answer

Answer: C – v2 = u2 + 2as gives 0 = 202 + 2 × (−5.0)s, so s = 400 / 10 = 40 m.

Q6. A driver travelling at 30 m/s has a reaction time of 0.70 s before applying the brakes. What is the thinking distance (the distance travelled during the reaction time)?

  • (A) 21 m
  • (B) 43 m
  • (C) 30 m
  • (D) 9.0 m
Show Answer

Answer: A – During the reaction time the car continues at constant speed, so distance = speed × time = 30 × 0.70 = 21 m.

Q7. A student knows the initial speed u, the final speed v and the acceleration a of an object, but does not know the time t. Which equation gives the displacement s directly?

  • (A) v = u + at
  • (B) s = ut + 0.5at2
  • (C) s = 0.5(u + v)t
  • (D) v2 = u2 + 2as
Show Answer

Answer: D – v2 = u2 + 2as is the only equation of motion that does not contain t, so it links v, u, a and s directly. A good rule: pick the equation that contains the three knowns plus the one unknown.

Q8. Taking the direction of motion as positive, a car has initial velocity +15 m/s and decelerates at 3.0 m/s2, so its acceleration is −3.0 m/s2. What is its velocity after 4.0 s?

  • (A) +27 m/s
  • (B) +3.0 m/s
  • (C) −3.0 m/s
  • (D) +12 m/s
Show Answer

Answer: B – v = u + at = 15 + (−3.0)(4.0) = 15 − 12 = +3.0 m/s. The car is still moving forwards, just more slowly. Keep the sign of a consistent with your chosen positive direction.

Q9. A cyclist accelerates uniformly from 6.0 m/s to 14 m/s in 8.0 s. What distance does she travel during this time?

  • (A) 48 m
  • (B) 64 m
  • (C) 80 m
  • (D) 112 m
Show Answer

Answer: C – Use s = 0.5(u + v)t = 0.5 × (6.0 + 14) × 8.0 = 0.5 × 20 × 8.0 = 80 m. The acceleration is not needed because the average speed is (u + v) / 2.

Q10. A train accelerates uniformly from rest to 40 m/s over a distance of 800 m. What is its acceleration?

  • (A) 1.0 m/s2
  • (B) 0.50 m/s2
  • (C) 2.0 m/s2
  • (D) 20 m/s2
Show Answer

Answer: A – v2 = u2 + 2as gives 402 = 0 + 2 × a × 800, so a = 1600 / 1600 = 1.0 m/s2.

Motion Graphs & Free Fall

Q11. On a velocity-time graph, the velocity of an object increases uniformly in a straight line from 0 to 24 m/s over 8.0 s. What is the acceleration of the object?

  • (A) 3.0 m/s2
  • (B) 0.33 m/s2
  • (C) 24 m/s2
  • (D) 192 m/s2
Show Answer

Answer: A – Acceleration is the gradient of a velocity-time graph: a = 24 / 8.0 = 3.0 m/s2. (192 would come from wrongly using the area, which gives displacement.)

Q12. The velocity-time graph for a car forms a triangle with a base of 10 s and a maximum height of 30 m/s. What distance does the car travel?

  • (A) 3.0 m
  • (B) 30 m
  • (C) 300 m
  • (D) 150 m
Show Answer

Answer: D – Displacement is the area under a velocity-time graph. Area of triangle = 0.5 × 10 × 30 = 150 m.

Q13. A displacement-time graph for an object is a curve whose gradient becomes steeper and steeper with time. Which statement correctly describes the motion?

  • (A) The velocity is constant
  • (B) The velocity is decreasing
  • (C) The velocity is increasing
  • (D) The object is at rest
Show Answer

Answer: C – The gradient of a displacement-time graph is the velocity. A gradient that grows steeper means the velocity is increasing, so the object is accelerating.

Q14. An acceleration-time graph shows a constant acceleration of 4.0 m/s2 lasting for 5.0 s. What is the change in velocity of the object?

  • (A) 0.80 m/s
  • (B) 20 m/s
  • (C) 25 m/s
  • (D) 45 m/s
Show Answer

Answer: B – The area under an acceleration-time graph is the change in velocity: Δv = 4.0 × 5.0 = 20 m/s.

Q15. The velocity-time graph of an object is a curve. At one instant, the tangent to the curve passes through velocity values of 4.0 m/s and 16 m/s separated by a time interval of 4.0 s. What is the instantaneous acceleration at that instant?

  • (A) 1.0 m/s2
  • (B) 4.0 m/s2
  • (C) 12 m/s2
  • (D) 3.0 m/s2
Show Answer

Answer: D – For a curved velocity-time graph, draw the tangent at the point and find its gradient: a = (16 − 4.0) / 4.0 = 12 / 4.0 = 3.0 m/s2.

Q16. A stone is dropped from rest and falls freely with g = 9.81 m/s2 (air resistance neglected). What is its speed after 2.0 s?

  • (A) 9.8 m/s
  • (B) 19.6 m/s
  • (C) 29.4 m/s
  • (D) 39.2 m/s
Show Answer

Answer: B – v = u + gt = 0 + 9.81 × 2.0 = 19.6 m/s.

Q17. A ball is dropped from rest from a cliff and falls freely for 3.0 s with g = 9.81 m/s2. How far has it fallen in this time?

  • (A) 44 m
  • (B) 29 m
  • (C) 88 m
  • (D) 15 m
Show Answer

Answer: A – s = ut + 0.5gt2 = 0 + 0.5 × 9.81 × 3.02 = 0.5 × 9.81 × 9.0 = 44 m. (29 m/s is its speed at that moment, not the distance.)

Q18. A ball is thrown vertically upwards at 19.6 m/s. Taking g = 9.81 m/s2, how long does it take to reach its highest point?

  • (A) 1.0 s
  • (B) 4.0 s
  • (C) 2.0 s
  • (D) 9.8 s
Show Answer

Answer: C – At the highest point v = 0. Taking upwards as positive, v = u + at gives 0 = 19.6 − 9.81t, so t = 19.6 / 9.81 = 2.0 s.

Q19. The same ball is thrown vertically upwards at 19.6 m/s with g = 9.81 m/s2. What is the maximum height it reaches?

  • (A) 19.6 m
  • (B) 9.8 m
  • (C) 29.4 m
  • (D) 39.2 m
Show Answer

Answer: A – v2 = u2 + 2as with v = 0 and a = −9.81: 0 = 19.62 − 2 × 9.81 × s, so s = 384.16 / 19.62 = 19.6 m.

Q20. A velocity-time graph for a bouncing ball shows the ball hitting the ground and rebounding, with the positive velocity peak just after the bounce smaller than the speed just before the bounce. Why is the rebound speed smaller?

  • (A) The acceleration of free fall decreases at the bounce
  • (B) The mass of the ball decreases at the bounce
  • (C) Air resistance removes all of the kinetic energy
  • (D) Some kinetic energy is transferred to internal energy and sound at the impact, so less remains for the rebound
Show Answer

Answer: D – The bounce is not perfectly elastic: part of the kinetic energy becomes internal energy of the ball and ground plus sound. With less kinetic energy after impact, the rebound speed (and the next bounce height) is lower. Between bounces the graph sections are straight lines of gradient −9.81 m/s2, because the acceleration is always g downwards.


Projectile Motion

Q21. A projectile is launched at 40 m/s at an angle of 30° above the horizontal. What is the horizontal component of its initial velocity?

  • (A) 20 m/s
  • (B) 28 m/s
  • (C) 34.6 m/s
  • (D) 40 m/s
Show Answer

Answer: C – Horizontal component = v cosθ = 40 × cos 30° = 40 × 0.866 = 34.6 m/s. (20 m/s is the vertical component, 40 sin 30°.)

Q22. Two balls are released from the same height at the same instant. One is dropped straight down; the other is projected horizontally. Air resistance is negligible. Which statement is correct?

  • (A) The projected ball lands first
  • (B) Both balls land at the same time
  • (C) The dropped ball lands first
  • (D) The landing order depends on the mass of each ball
Show Answer

Answer: B – Horizontal and vertical motion are independent. Both balls have the same vertical motion: zero initial vertical velocity and acceleration g downwards, so they fall through the same height in the same time.

Q23. A projectile is launched at 20 m/s at 30° above the horizontal. Taking g = 9.81 m/s2, what is the maximum height it reaches?

  • (A) 5.1 m
  • (B) 2.6 m
  • (C) 10 m
  • (D) 20 m
Show Answer

Answer: A – Only the vertical component matters: uy = 20 sin 30° = 10 m/s. At the top vy = 0, so h = uy2 / 2g = 102 / (2 × 9.81) = 100 / 19.62 = 5.1 m.

Q24. A projectile is launched at 30 m/s at 60° above the horizontal and lands at the same height. Taking g = 9.81 m/s2, what is its time of flight?

  • (A) 2.6 s
  • (B) 3.1 s
  • (C) 4.6 s
  • (D) 5.3 s
Show Answer

Answer: D – Vertical component uy = 30 sin 60° = 26.0 m/s. Time to the top = 26.0 / 9.81 = 2.65 s, and the flight is symmetrical, so total time of flight = 2 × 2.65 = 5.3 s.

Q25. A projectile is launched at 25 m/s at 45° above the horizontal on level ground. Taking g = 9.81 m/s2, what is its range?

  • (A) 32 m
  • (B) 64 m
  • (C) 88 m
  • (D) 128 m
Show Answer

Answer: B – Range = v2 sin 2θ / g = 252 × sin 90° / 9.81 = 625 / 9.81 = 63.7 m, about 64 m. Equivalently, ux = uy = 17.7 m/s, time of flight = 3.6 s, range = 17.7 × 3.6 = 64 m.

Q26. A projectile is launched at 50 m/s at 60° above the horizontal. Air resistance is negligible. What is the horizontal component of its velocity 3.0 s after launch?

  • (A) 0 m/s
  • (B) 43 m/s
  • (C) 25 m/s
  • (D) 50 m/s
Show Answer

Answer: C – There is no horizontal acceleration, so the horizontal component never changes: ux = 50 cos 60° = 25 m/s at every point of the flight. (43 m/s is the initial vertical component.)

Q27. At one point in its flight, a projectile has a horizontal velocity component of 30 m/s and a vertical velocity component of 40 m/s. What is the magnitude of its velocity at that point?

  • (A) 10 m/s
  • (B) 35 m/s
  • (C) 70 m/s
  • (D) 50 m/s
Show Answer

Answer: D – The components are at right angles, so combine them with Pythagoras: v = √(302 + 402) = √(900 + 1600) = √2500 = 50 m/s.

Q28. A projectile is launched with an initial vertical velocity component of 24 m/s. Taking g = 9.81 m/s2, what maximum height does it reach?

  • (A) 29 m
  • (B) 59 m
  • (C) 15 m
  • (D) 2.4 m
Show Answer

Answer: A – h = uy2 / 2g = 242 / (2 × 9.81) = 576 / 19.62 = 29.4 m, about 29 m.

Q29. A stone is projected horizontally at 12 m/s from the top of a cliff 45 m high. Taking g = 10 m/s2, how far from the base of the cliff does the stone land?

  • (A) 15 m
  • (B) 36 m
  • (C) 45 m
  • (D) 540 m
Show Answer

Answer: B – Vertically: 45 = 0.5 × 10 × t2, so t2 = 9.0 and t = 3.0 s. Horizontally the speed stays 12 m/s, so range = 12 × 3.0 = 36 m.

Q30. A projectile launched at a certain speed at 30° above the horizontal lands at a certain range on level ground. Launched at the same speed, at which of these angles would it land at exactly the same range?

  • (A) 15°
  • (B) 45°
  • (C) 60°
  • (D) 90°
Show Answer

Answer: C – Range depends on sin 2θ. Since sin 60° = sin 120°, launch angles of 30° and 60° (complementary angles, 2θ = 60° and 120°) give the same range at the same launch speed. The 60° shot flies higher and stays in the air longer, but lands at the same point.

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