OCR A Level Mathematics A: Quantities, units, and kinematics with constant acceleration (H240) – 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 from your board or school.

These questions cover quantities, units and kinematics with constant acceleration in OCR A Level Mathematics A (H240), content references 3.01a to 3.01c and 3.02a to 3.02e of the OCR specification, Version 3.1 (August 2026). Most of this content is Stage 1, common to AS and A Level. Parts marked A Level only use the unit for moment or vectors in two dimensions. The marks shown are our own guide to the size of each step. If a topic is unfamiliar, start with the study guide; the revision notes recap each method. Related links: the course hub, the checklist and the free diagnostics.

Questions

1. State the SI unit, with its symbol, of each quantity.

(a) mass [1] (b) acceleration [1] (c) the moment of a force (A Level only) [1]

2. A skier covers 3.6 km in 12 minutes at a steady speed. Find the speed in m s⁻¹. [3]

3. A cyclist rides along a straight path. Measured from a fixed point O, she starts at position −3 m, reaches +10 m and then stops at +2 m. The whole ride takes 20 s. Find her displacement, the distance she travels, her average speed and her average velocity. [4]

4. A particle moves in a straight line. Its velocity-time graph is made of straight segments joining (0, 0), (4, 10), (10, 10) and (16, −2), with t in seconds and v in m s⁻¹.

(a) Find the acceleration in the first stage and in the last stage. [2] (b) Find the displacement from the start at t = 16. [2] (c) Find the total distance travelled. [2]

5. Starting from v = u + at and s = ½(u + v)t, show that v² = u² + 2as. [3]

6. A car travelling at 24 m s⁻¹ brakes with constant deceleration and stops after 72 m. Find the deceleration and the time the car takes to stop. [3]

7. A particle moves in a straight line with constant acceleration. It passes O with velocity 5 m s⁻¹. Four seconds later it is 36 m from O, further along the line. Find the acceleration and the velocity at that time. [4]

8. A particle moves on a straight line with constant acceleration −1.5 m s⁻². At t = 0 it is at O with velocity 9 m s⁻¹ in the positive direction.

(a) Find the time at which it is instantaneously at rest and its displacement from O then. [2] (b) Find the times at which it is 24 m from O in the positive direction. [3] (c) Find the distance it has travelled by the time it returns to O. [2]

9. Two skaters move along the same straight line. At time t = 0, skater A is 16 m ahead of skater B. A moves at a constant 4 m s⁻¹. B starts from rest and moves with constant acceleration 1.5 m s⁻².

(a) Write expressions for the distance each skater has moved from B’s starting point after t seconds. [2] (b) Find the time at which B catches A. [2] (c) Find B’s speed and the distance B has travelled at that time. [2]

10. (A Level only) A particle P has position vector (−2, 5) m at t = 0 and velocity (4, −1) m s⁻¹. It moves with constant acceleration (−0.5, 0.5) m s⁻².

(a) Find the velocity and position vector of P at t = 6. [4] (b) Find the time at which P moves parallel to the i direction and its position vector then. [3] (c) Find the speed of P at t = 4. [1]

11. A tram starts from rest at stop A, accelerates uniformly for 20 s to speed V m s⁻¹, travels at constant speed, then decelerates uniformly to rest at stop B in 10 s. The whole journey takes 100 s and the distance AB is 1190 m.

(a) Find the time spent at constant speed. [1] (b) Find V. [3] (c) Find the acceleration in the first stage and in the last stage. [2] (d) Find the distance travelled at constant speed. [1]

Answers

1. (a) kilogram, kg [1] (b) metres per second squared, m s⁻² [1] (c) newton metre, N m [1] Examiner insight: Give the correct symbol or name in full; “N/m” or “m/s” for acceleration describes a different quantity.

2. 3.6 km = 3600 m and 12 minutes = 720 s [1] speed = 3600 ÷ 720 [1] 5 m s⁻¹ [1] Examiner insight: Show both conversions separately. A correct final answer with no visible conversion is a risk if one conversion slips.

3. Displacement = 2 − (−3) = 5 m [1] Distance = 13 + 8 = 21 m [1] Average speed = 21 ÷ 20 = 1.05 m s⁻¹ [1] Average velocity = 5 ÷ 20 = 0.25 m s⁻¹ in the positive direction [1] Examiner insight: Be sure of which quantity each word asks for. Using displacement in the average speed is an easy slip to make.

4. (a) First stage a = 10 ÷ 4 = 2.5 m s⁻² [1] Last stage a = (−2 − 10) ÷ 6 = −2 m s⁻² [1] (b) The velocity is zero when 10 − 2(t − 10) = 0, so t = 15 [1] Areas: 20, 60, 25 and −1 give displacement 104 m [1] (c) The particle changes direction at t = 15, so the last area counts as distance [1] Distance = 20 + 60 + 25 + 1 = 106 m [1] Examiner insight: Find where the line crosses the axis before taking areas. Adding a negative area into the distance is a common slip.

5. From v = u + at, t = (v − u)/a [1] Substitute: s = ½(u + v)(v − u)/a [1] 2as = v² − u², so v² = u² + 2as [1] Examiner insight: A “show that” needs every line of the algebra. Stating the final result alone does not show the reasoning.

6. Take the positive direction as the direction of motion: u = 24, v = 0, s = 72 [1] 0 = 576 + 2a × 72, so a = −4 [1] Deceleration 4 m s⁻², and t = 24 ÷ 4 = 6 s [1] Examiner insight: Say whether the answer is a deceleration or an acceleration. A bare “−4” or “4” invites doubt about the direction.

7. u = 5, t = 4, s = 36 [1] 36 = 20 + ½ × a × 16 [1] a = 2 m s⁻² [1] v = 5 + 2 × 4 = 13 m s⁻¹ [1] Examiner insight: Write down the list of known values first. It shows the formula choice, and a later arithmetic slip is easier to follow.

8. (a) v = 0: 0 = 9 − 1.5t, so t = 6 [1] s = 9 × 6 − 0.75 × 36 = 27 m [1] (b) 9t − 0.75t² = 24 [1] t² − 12t + 32 = 0 [1] t = 4 and t = 8 [1] (c) It returns to O when 9t − 0.75t² = 0, so t = 12 [1] Distance = 27 + 27 = 54 m [1] Examiner insight: Quote both roots of the quadratic in part (b) and say what each means. For part (c), the distance is not the displacement, which is zero.

9. (a) Take B’s starting point as origin. A: 16 + 4t. B: ½ × 1.5 × t² = 0.75t² [1] [1] (b) 0.75t² = 16 + 4t, so 3t² − 16t − 64 = 0 [1] (3t + 8)(t − 8) = 0, so t = 8 s (the negative root is rejected) [1] (c) v = 1.5 × 8 = 12 m s⁻¹ [1] s = 0.75 × 64 = 48 m [1] Examiner insight: Reject the negative root with a reason. Check the position of A as well, which is 16 + 32 = 48 m, to confirm the answer.

10. (a) v = (4, −1) + 6(−0.5, 0.5) = (1, 2) m s⁻¹ [1] [1] r = (−2, 5) + 6(4, −1) + 18(−0.5, 0.5) [1] = (13, 8) m [1] (b) Parallel to i means the j-component of velocity is zero: −1 + 0.5t = 0, so t = 2 [1] r = (−2, 5) + 2(4, −1) + 2(−0.5, 0.5) [1] = (5, 4) m [1] (c) At t = 4, v = (2, 1), so speed = √5 ≈ 2.24 m s⁻¹ [1] Examiner insight: The position vector includes the starting position. Giving only the displacement from the starting point answers a different question.

11. (a) 100 − 20 − 10 = 70 s [1] (b) Area of the trapezium: ½ × V × (100 + 70) = 1190 [1] [1] V = 14 [1] (c) First stage 14 ÷ 20 = 0.7 m s⁻² [1] Last stage −14 ÷ 10 = −1.4 m s⁻² [1] (d) 14 × 70 = 980 m [1] Examiner insight: Sketch the velocity-time graph and label the three durations. The parallel sides of the trapezium are 100 and 70, which is easy to mix up.

Where marks are usually lost

• Not converting km, minutes or grams to SI before substituting.
• Using displacement where a question wants distance, and the reverse.
• Taking the area below the time axis as positive when finding displacement.
• Using the constant acceleration formulae for motion that has stages with different accelerations, instead of treating each stage separately.
• Losing the sign of a deceleration when moving in the positive direction.
• Giving only one root of a quadratic in t and ignoring the second valid time.
• Leaving out units, or giving a speed as a vector.
• Forgetting the starting position when asked for a position vector.
• Skipping lines in a “show that” derivation.

Next steps

• Read the revision notes for the formulae and methods.
• Reread the study guide for the worked examples.
• Use the course hub and the checklist to plan what to cover next.
• Try the free diagnostics for a quick check.
• Book a free trial class at /trial/.

Official syllabus

These questions follow the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.