AS Physics: Kinematics and Equations of Motion — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Kinematics revision notes

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Questions

1. Distinguish between distance and displacement, and between speed and velocity. [2]

2. State the condition under which the suvat equations may be used. [1]

3. State what the gradient and the area represent on (a) a displacement–time graph, (b) a velocity–time graph. [3]

4. A car accelerates uniformly from 8.0 m s⁻¹ to 26 m s⁻¹ over a distance of 170 m.

(a) Calculate the acceleration. [3] (b) Calculate the time taken. [2]

5. A ball is thrown vertically upwards at 18 m s⁻¹ from ground level. (g = 9.81 m s⁻²)

(a) Calculate the maximum height reached. [3] (b) Calculate the total time before it returns to the thrower’s hand. [2] (c) State the velocity and the acceleration at the highest point. [2]

6. A stone is thrown horizontally at 12 m s⁻¹ from a cliff 45 m high.

(a) Explain why the horizontal and vertical motions can be treated separately. [2] (b) Calculate the time to reach the ground. [3] (c) Calculate the horizontal distance travelled. [2] (d) Calculate the speed on impact. [3]

7. Derive the equation v = u + at from the definition of acceleration. [2]

8. An object is dropped from rest and falls freely for 1.5 s. Using g = 9.81 m s⁻², calculate the distance fallen and its velocity at that point. [3]

9. Describe an experimental method to determine g in the laboratory, including how random error is reduced. [3]

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10. A car accelerates uniformly from rest to a speed of 18 m s⁻¹ in 12 s. Calculate the distance travelled in this time. [2]

Answers

1. Distance is a scalar — the total path length; displacement is a vector — the straight-line distance in a stated direction [1]. Speed is a scalar; velocity is speed in a stated direction [1].

2. Only when the acceleration is uniform (constant) [1].

3. (a) Gradient = velocity; area has no physical meaning [1]. (b) Gradient = acceleration [1]; area = displacement [1].

4. (a) v² = u² + 2as [1] 26² = 8.0² + 2a(170) → 676 = 64 + 340a [1] a = 612 ÷ 340 = 1.80 m s⁻² [1]. (b) v = u + at → 26 = 8.0 + 1.80t [1] t = 18 ÷ 1.80 = 10.0 s [1].

5. (a) At maximum height v = 0. v² = u² + 2as [1] 0 = 18² − 2(9.81)s [1] s = 324 ÷ 19.62 = 16.5 m [1]. (b) Time up: v = u + at → 0 = 18 − 9.81t, t = 1.83 s [1]. Total = 2 × 1.83 = 3.67 s [1]. (c) Velocity = zero [1]; acceleration = 9.81 m s⁻² downwards — it is unchanged [1]. The acceleration is not zero at the top. This is examined often.

6. (a) Assuming negligible air resistance, there is no horizontal force, so horizontal velocity stays constant, and gravity acts only vertically, giving a constant vertical acceleration g [1] independent of the horizontal motion; the two perpendicular components can therefore be treated separately, with time as the only quantity linking them [1]. (Being perpendicular makes the components mathematically resolvable, but it is the absence of a horizontal force and gravity’s purely vertical direction — not perpendicularity alone — that makes the two motions independent.) (b) Vertical: s = ut + ½at², with u = 0 [1] 45 = ½(9.81)t² → t² = 9.174 [1] t = 3.03 s [1]. (c) Horizontal: s = vt = 12 × 3.03 [1] = 36.4 m [1]. (d) Vertical velocity: v = at = 9.81 × 3.03 = 29.7 m s⁻¹ [1] Resultant = √(12² + 29.7²) [1] = √(144 + 882) = 32.0 m s⁻¹ [1].

7. Acceleration is defined as a = (v − u) ÷ t, the rate of change of velocity [1]. Rearranging directly: at = v − u, so v = u + at [1]. The remaining suvat equations can similarly be derived by combining this result with the definition of velocity as displacement over time.

8. s = ut + ½at², with u = 0: s = ½ × 9.81 × 1.5² [1] = 11.0 m [1]. v = u + at = 0 + 9.81 × 1.5 = 14.7 m s⁻¹ [1]. Air resistance is assumed negligible throughout this calculation.

9. Drop an object through a known, measured height and time the fall electronically, using a light gate or a timer released by an electromagnet [1]. Using s = ½gt² (since u = 0), rearranged to g = 2s/t² [1]. Repeating the drop and averaging t reduces the effect of random timing error [1]. Electronic timing is preferred over a hand-operated stopwatch: a person’s reaction time is a much larger source of timing uncertainty over such a short fall, and it has both a consistent (roughly systematic) average delay and trial-to-trial (random) variation, whereas electronic timing removes both.

10. s = ½(u + v)t [1] = ½ × (0 + 18) × 12 = 108 m [1].

Examiner insight (June 2025): weaker answers used s = vt, which only applies at constant speed, instead of s = ½vt for uniform acceleration from rest. Try the real question next: Cambridge International AS & A Level Physics 9702, June 2025, Paper 22, Question 1.

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Where marks are usually lost

• Using suvat where acceleration is not uniform.
• Saying acceleration is zero at the top of a vertical throw.
• Mixing horizontal and vertical components.
• Forgetting the sign convention for upward and downward motion.
• Defining acceleration loosely, e.g. “rate of change of velocity per unit time” — that’s a rate of a rate. It is simply the rate of change of velocity; don’t add “per unit time” on top of “rate of”.
• When a projectile is launched at an angle (not purely horizontally), using its total initial speed directly in a vertical suvat equation instead of resolving it into a vertical component first.
• Timing only a single drop when measuring g experimentally, rather than repeating and averaging to reduce random error.
• Forgetting that the suvat equations themselves can be derived from the definitions of velocity and acceleration, not simply recalled from memory.