OCR A Level Mathematics A: Variable acceleration, motion under gravity and projectiles (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 content references 3.02f to 3.02i of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Questions 1 to 3 are Stage 1 (shared with AS). Questions 4 to 12 are A Level only. Take g = 9.8 m s⁻² and give decimals to 3 significant figures. Learn the methods first from the study guide. The free diagnostics show where to focus.

Questions

1. A particle moves on a straight line with acceleration a = 12t − 8 m s⁻². When t = 1 its velocity is 3 m s⁻¹ and its displacement from O is 2 m. Find an expression for its displacement s at time t. [4]

2. A particle moves on a straight line with displacement s = 2t³ − 21t² + 60t metres from O at time t seconds.

(a) Find the values of t at which the particle is instantaneously at rest. [3] (b) Find the acceleration at the later of these times, and the total distance travelled in the first 5 seconds. [4]

3. A particle moves on a straight line with velocity v = 12 − 3√t m s⁻¹ for t ≥ 0 and starts at O.

(a) Find the acceleration when t = 4. [2] (b) Show that the displacement at time t is 12t − 2t^(3/2). [3] (c) Find the time at which the particle is first at rest, and its displacement then. [2]

4. (A Level only) A particle has position r = (t² − 6t)i + (t³ − 12t)j metres at time t seconds.

(a) Show that at t = 2 the particle moves parallel to i. [3] (b) Find the magnitude of the acceleration at t = 2. [3] (c) Find the time at which the particle moves parallel to j, and its position then. [2]

5. (A Level only) A particle has acceleration a = (12t − 6)i + 6tj m s⁻². At t = 0 its velocity is (4i − 12j) m s⁻¹ and its position is (i + 10j) m.

(a) Find the velocity at time t. [3] (b) Find the position vector at time t. [3] (c) Show that when t = 2 the particle moves parallel to i, and find its speed then. [3]

6. (A Level only) A ball moves with acceleration −9.8j m s⁻² and is projected from the origin with velocity (4i + 14j) m s⁻¹.

(a) Find its position vector at t = 1.5. [2] (b) Find the times at which it is 6 m above the level of projection. [3] (c) Find the greatest height reached, and the ball’s speed at that point. [2]

7. (A Level only) A ball is kicked from level ground with speed 32 m s⁻¹ at 25° above the horizontal.

(a) Find the time of flight. [3] (b) Find the horizontal range. [1] (c) Find the greatest height reached. [2]

8. (A Level only) A stone is thrown horizontally at 14 m s⁻¹ from the top of a cliff 60 m above the sea.

(a) Find the time taken to reach the sea. [2] (b) Find the horizontal distance from the foot of the cliff at which it lands. [1] (c) Find the speed of the stone as it lands and the angle its path makes with the horizontal. [3]

9. (A Level only) A projectile is launched from the origin with speed u at angle α above the horizontal. Show that its path has equation y = x tan α − g x² / (2u² cos²α). [4]

10. (A Level only) A projectile is launched from the origin at 22 m s⁻¹ and passes through the point (30, 10), where x is horizontal and y vertical, both in metres.

(a) Find the two possible angles of projection, to 1 decimal place. [5] (b) Explain why there are two angles. [1]

11. (A Level only) A ball is kicked from level ground with speed 21 m s⁻¹ at 50° above the horizontal towards a vertical wall 12 m high, whose base is 18 m away on the same ground.

(a) Find the height of the ball when it is level with the wall. [3] (b) State whether it clears the wall, and by how much. [1] (c) Show that the ball is still rising as it passes the wall. [2] (d) Give one reason why a real ball might not clear the wall. [1]

12. (A Level only) A particle is projected from a point A, 20 m above level ground, with velocity (8i + 6j) m s⁻¹, where i is horizontal and j is upward.

(a) Find the time taken to reach the ground. [3] (b) Find the horizontal distance travelled in that time. [1] (c) Find the speed on landing and the angle of the velocity below the horizontal. [3] (d) Find the time at which the particle moves at 45° below the horizontal. [2] (e) State one assumption of the model that makes part (a) unrealistic. [1]

Answers

1. v = ∫(12t − 8) dt = 6t² − 8t + C [1]. At t = 1, 6 − 8 + C = 3 so C = 5 [1]. s = ∫(6t² − 8t + 5) dt = 2t³ − 4t² + 5t + D [1]. At t = 1, 2 − 4 + 5 + D = 2, so D = −1 and s = 2t³ − 4t² + 5t − 1 [1] [4] Examiner insight: Each integration needs its own constant.

2. (a) v = ds/dt = 6t² − 42t + 60 [1]. Set v = 0: 6(t − 2)(t − 5) = 0 [1]. t = 2 and t = 5 [1] [3] (b) a = 12t − 42, so at t = 5, a = 18 m s⁻² [1]. s(2) = 52 [1] and s(5) = 25 [1]. It moves 52 m, then returns 27 m: 52 + 27 = 79 m [1] [4] Examiner insight: For distance travelled, show where the direction changes. Here s(5) − s(0) = 25 is only the displacement.

3. (a) a = dv/dt = −(3/2)t^(−1/2) [1], so at t = 4, a = −3/4 m s⁻² [1] [2] (b) s = ∫(12 − 3t^(1/2)) dt = 12t − 2t^(3/2) + C [1]. s = 0 at t = 0 gives C = 0 [1]. So s = 12t − 2t^(3/2) [1] [3] (c) v = 0 gives √t = 4, so t = 16 s [1]. s = 12(16) − 2(64) = 64 m [1] [2] Examiner insight: Apply the starting condition to find the constant before quoting the given result.

4. (a) v = (2t − 6)i + (3t² − 12)j [1]. At t = 2 the j-component is 3(4) − 12 = 0 [1]. So v = −2i, which is parallel to i [1] [3] (b) a = 2i + 6tj [1]. At t = 2, a = 2i + 12j [1]. |a| = √(4 + 144) = √148 = 12.2 m s⁻² [1] [3] (c) The i-component is zero when 2t − 6 = 0, so t = 3 [1]. Then r = (9 − 18)i + (27 − 36)j = −9i − 9j [1] [2] Examiner insight: For “parallel to i”, say which component you set to zero.

5. (a) Integrate each component: v = (6t² − 6t)i + 3t²j + c [1]. At t = 0, c = 4i − 12j [1]. So v = (6t² − 6t + 4)i + (3t² − 12)j [1] [3] (b) r = (2t³ − 3t² + 4t)i + (t³ − 12t)j + d [1]. At t = 0, d = i + 10j [1]. So r = (2t³ − 3t² + 4t + 1)i + (t³ − 12t + 10)j [1] [3] (c) At t = 2 the j-component is 3(4) − 12 = 0 [1], so the particle moves parallel to i. The i-component is 24 − 12 + 4 = 16 [1]. Speed = 16 m s⁻¹ [1] [3] Examiner insight: The constant of integration is a vector, so write it as one.

6. (a) r = (4i + 14j)(1.5) + ½(−9.8j)(1.5²) [1] = 6i + 9.98j (m) [1] [2] (b) 14t − 4.9t² = 6, so 4.9t² − 14t + 6 = 0 [1]. t = (14 ± √78.4)/9.8 [1]. t = 0.525 s and t = 2.33 s [1] [3] (c) At the top 14 − 9.8t = 0, so t = 1.43 s, and the height is 14²/19.6 = 10.0 m [1]. The vertical velocity is 0 there, so the speed is the horizontal component, 4 m s⁻¹ [1] [2] Examiner insight: Both roots are valid here: the ball passes 6 m going up and coming down.

7. (a) Vertical component 32 sin 25° = 13.52 [1]. Solve 13.52t − 4.9t² = 0 [1]: T = 2.76 s [1] [3] (b) Horizontal component 32 cos 25° = 29.0, so range = 29.0 × 2.760 = 80.0 m [1] [1] (c) H = 13.52²/(2 × 9.8) [1] = 9.33 m [1] [2] Examiner insight: Carry unrounded values forward.

8. (a) The initial vertical velocity is 0, so 60 = ½(9.8)t² [1], giving t = 3.50 s [1] [2] (b) Distance = 14 × 3.499 = 49.0 m [1] [1] (c) v_y = 9.8 × 3.499 = 34.3 [1]. Speed = √(14² + 34.3²) = 37.0 m s⁻¹ [1]. Angle = tan⁻¹(34.3/14) = 67.8° below the horizontal [1] [3] Examiner insight: State what the angle is measured from.

9. Horizontally x = (u cos α)t [1]. Vertically y = (u sin α)t − ½gt² [1]. Eliminating t = x/(u cos α) gives y = (u sin α)x/(u cos α) − ½g x²/(u² cos²α) [1]. Since sin α / cos α = tan α, y = x tan α − g x² / (2u² cos²α) [1] [4] Examiner insight: Show the substitution for t and the use of tan α = sin α / cos α.

10. (a) Substitute: 10 = 30 tan α − 9.8(900) / (2 × 22² × cos²α) [1]. Use 1/cos²α = 1 + tan²α, T = tan α: 10 = 30T − 9.1116(1 + T²) [1]. Rearranged: 9.11T² − 30T + 19.11 = 0 [1]. T = 0.864 or T = 2.43 [1]. α = 40.8° or α = 67.6° [1] [5] (b) A flatter path and a steeper path can pass through the same point at one launch speed [1] [1] Examiner insight: Replacing 1/cos²α by 1 + tan²α makes this a quadratic. Name the identity.

11. (a) y = x tan α − g x² / (2u² cos²α) [1]. With x = 18: 18 tan 50° = 21.45 and the second term is 9.8(324)/(2 × 441 × cos²50°) = 8.71 [1]. y = 12.7 m [1] [3] (b) 12.74 − 12 = 0.74, so the ball clears the wall by about 0.74 m [1] [1] (c) Time to the wall = 18/(21 cos 50°) = 1.33 s [1]. v_y = 21 sin 50° − 9.8(1.333) = 3.02 > 0, so it is still rising [1] [2] (d) Air resistance would slow the ball, so it may be lower at the wall [1] [1] Examiner insight: A margin of 0.74 m is small, so keep extra digits until the end.

12. (a) 20 + 6t − 4.9t² = 0 [1]. t = (6 + √(36 + 392))/9.8 [1]. t = 2.72 s (negative root rejected) [1] [3] (b) Distance = 8 × 2.723 = 21.8 m [1] [1] (c) v_y = 6 − 9.8(2.723) = −20.7 [1]. Speed = √(8² + 20.7²) = 22.2 m s⁻¹ [1]. Angle = tan⁻¹(20.7/8) = 68.9° below the horizontal [1] [3] (d) 45° below the horizontal means v_y = −8 [1]. 6 − 9.8t = −8, so t = 1.43 s [1] [2] (e) Air resistance is ignored, so a real flight would be shorter and slower on landing [1] [1] Examiner insight: Reject the negative root and interpret the sign of v_y.

Where marks are usually lost

• Dropping the constant after integrating, or using the wrong condition.
• Giving displacement when distance travelled is asked.
• Setting the wrong component to zero for “parallel to i”.
• Using u instead of its components in the equations.
• Using the range formula for different launch and landing heights.

Next steps

• Revisit the revision notes and the study guide.
• Use the course hub and checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

OCR, A Level in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, content references 3.02f to 3.02i.