AQA A-Level Mathematics: Q: Kinematics (7357) – 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 Section Q: Kinematics (Q1 to Q5) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification assesses Section Q on Paper 2. A calculator is required in every 7357 paper. Take g = 9.8 m s⁻² and give answers to 3 significant figures unless a question says otherwise. In projectiles, i is horizontal and j vertically upwards.

Learn the content first with the Kinematics study guide and the Kinematics revision notes. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/. Not sure where to start? Try the free 10-minute diagnostics.

Questions

1. A particle starts at O and moves 9 m along a straight line in the positive direction to A. It then moves 15 m in the negative direction to B. The whole journey takes 12 s.

(a) State the displacement of B from O and the distance travelled. [2] (b) Calculate the average speed and the average velocity for the journey. [2]

2. A displacement–time graph for a particle moving in a straight line is made of straight segments joining (0, 0), (4, 12), (7, 12) and (13, 0), with t in seconds and s in metres.

(a) Find the velocity of the particle in each of the three stages. [3] (b) Find the average speed over the 13 seconds. [1]

3. A train starts from rest at station A and accelerates uniformly for 40 s until its speed is V m s⁻¹. It travels at this speed for 180 s, then decelerates uniformly at 0.5 m s⁻² until it stops at station B. The distance from A to B is 4.4 km.

(a) Use a velocity–time graph to show that V² + 200V − 4400 = 0. [3] (b) Find V and the acceleration in the first stage. [2] (c) Find the total time for the journey. [1]

4. Starting from v = u + at and s = ut + ½at², show that s = vt − ½at² for motion with constant acceleration. [3]

5. A cyclist rides along a straight road at a constant 8 m s⁻¹. She passes a parked car at a point P. Two seconds later the car sets off from rest after her, with constant acceleration 2 m s⁻².

(a) Find the time, measured from when the cyclist passes P, at which the car draws level with her. [4] (b) Find how far from P this happens. [1]

6. At time t = 0 a particle is at the point with position vector (2i + 3j) m and has velocity (4i − 6j) m s⁻¹. Its acceleration is constant at (−i + 2j) m s⁻².

(a) Find the velocity of the particle at time t seconds. [2] (b) Find the time at which the particle is moving parallel to i. [2] (c) Find the distance of the particle from the origin at that time. [3]

7. A particle moves along a straight line. It starts at O, and its velocity at time t seconds is v = (t² − 7t + 10) m s⁻¹ for 0 ≤ t ≤ 6.

(a) Find the times at which the particle is instantaneously at rest. [2] (b) Find the acceleration of the particle when t = 4. [2] (c) Find the total distance travelled in the 6 seconds. [4]

8. A particle starts from rest at O and moves along a straight line. Its acceleration at time t seconds is a = (6√t − 2) m s⁻².

(a) Find an expression for the velocity at time t. [2] (b) Find the value of t > 0 at which the particle is instantaneously at rest. [2] (c) Find the displacement of the particle from O when t = 4. [3]

9. A particle P moves in a horizontal plane. At time t seconds its acceleration is (4i − 6t j) m s⁻². At t = 0, P is at the origin O with velocity (−2i + 9j) m s⁻¹.

(a) Find the velocity of P at time t. [2] (b) Find the time at which P is moving parallel to j, and state its velocity then. [2] (c) Find the position vector of P when t = 2. [3] (d) Find the speed of P when t = 2. [2]

10. A ball is struck from a point A, 1.5 m above horizontal ground, with velocity (24i + 10j) m s⁻¹. A vertical wall 4 m high stands 36 m horizontally from A, across the ball’s path. Model the ball as a particle.

(a) Show that the ball reaches the wall 1.5 s after being struck. [1] (b) Show that the ball passes over the wall, and find by how much it clears it. [3] (c) Find the greatest height of the ball above the ground. [2] (d) Find the speed of the ball and its direction of motion as it passes over the wall. [3] (e) Suggest one refinement to the model. [1]

11. A particle is projected from a point on horizontal ground with speed U m s⁻¹ at angle α above the horizontal. It lands on the ground.

(a) Show that its time of flight is 2U sin α / g, and hence that its range is U² sin 2α / g. [4] (b) U = 20. Find the two possible values of α for which the range is 30 m. [2]

Answers

1. (a) Displacement = 9 − 15 = −6 m [1] Distance = 9 + 15 = 24 m [1] (b) Average speed = 24 ÷ 12 = 2 m s⁻¹ [1] Average velocity = −6 ÷ 12 = −0.5 m s⁻¹ [1] Examiner insight: the sign is part of a displacement or velocity; “6 m” with no direction loses the accuracy mark.

2. (a) Gradient of each segment [1] 0 ≤ t ≤ 4: 12 ÷ 4 = 3 m s⁻¹; 4 ≤ t ≤ 7: 0 (at rest) [1] 7 ≤ t ≤ 13: (0 − 12) ÷ 6 = −2 m s⁻¹ [1] (b) Distance = 12 + 12 = 24 m, average speed = 24 ÷ 13 = 1.85 m s⁻¹ [1] Examiner insight: the last stage has a negative gradient; writing +2 m s⁻¹ loses the accuracy mark.

3. (a) Deceleration time = V ÷ 0.5 = 2V seconds [1] Area = ½(40)V + 180V + ½(2V)V = 200V + V² [1] 200V + V² = 4400, so V² + 200V − 4400 = 0 [1] (b) (V − 20)(V + 220) = 0, V > 0, so V = 20 [1] Acceleration = 20 ÷ 40 = 0.5 m s⁻² [1] (c) 40 + 180 + 40 = 260 s [1] Examiner insight: convert 4.4 km to 4400 m first; a “show that” earns nothing for a fudged last line.

4. From v = u + at, u = v − at [1] s = (v − at)t + ½at² [1] = vt − at² + ½at² = vt − ½at² [1] Examiner insight: each substitution must be shown; quoting the result earns no marks.

5. (a) Cyclist: s = 8t [1] Car (moving for t − 2 seconds): s = ½(2)(t − 2)² = (t − 2)² [1] (t − 2)² = 8t gives t² − 12t + 4 = 0 [1] t = 6 ± 4√2; t = 6 − 4√2 ≈ 0.343 is before the car starts, so t = 6 + 4√2 = 11.7 s [1] (b) s = 8 × 11.657 = 93.3 m [1] Examiner insight: reject the smaller root with a reason; choosing without comment usually costs the final mark.

6. (a) v = u + at [1] = (4 − t)i + (2t − 6)j m s⁻¹ [1] (b) Parallel to i when the j-component is 0: 2t − 6 = 0 [1] t = 3 s; the i-component is then 1 ≠ 0 [1] (c) r = (2i + 3j) + (4i − 6j)t + ½(−i + 2j)t² [1] At t = 3: r = (2 + 12 − 4.5)i + (3 − 18 + 9)j = 9.5i − 6j [1] Distance = √(9.5² + 6²) = √126.25 = 11.2 m [1] Examiner insight: leaving out the starting position (2i + 3j) in (c) puts the particle at the wrong point.

7. (a) (t − 2)(t − 5) = 0 [1] t = 2 s and t = 5 s [1] (b) a = dv/dt = 2t − 7 [1] At t = 4, a = 1 m s⁻² [1] (c) x = t³/3 − 7t²/2 + 10t (x = 0 at t = 0) [1] x(2) = 26/3, x(5) = 25/6, x(6) = 6 [1] Distance = 26/3 + (26/3 − 25/6) + (6 − 25/6) [1] = 15 m [1] Examiner insight: integrating v from 0 to 6 in one go gives the displacement, 6 m, not the distance.

8. (a) v = ∫(6t^½ − 2) dt = 4t^(3/2) − 2t + c [1] v = 0 at t = 0, so c = 0: v = 4t^(3/2) − 2t [1] (b) 2t(2√t − 1) = 0 [1] √t = ½, t = 0.25 s [1] (c) x = ∫(4t^(3/2) − 2t) dt = (8/5)t^(5/2) − t² + k [1] x = 0 at t = 0, so k = 0 [1] x(4) = (8/5)(32) − 16 = 35.2 m [1] Examiner insight: write √t as t^½ before integrating; a wrong power carries through every later part.

9. (a) Integrate each component: v = 4t i − 3t² j + c [1] c = −2i + 9j, so v = (4t − 2)i + (9 − 3t²)j m s⁻¹ [1] (b) 4t − 2 = 0, t = 0.5 s [1] v = 8.25j m s⁻¹, which is non-zero [1] (c) r = (2t² − 2t)i + (9t − t³)j + d [1] r = 0 at t = 0, so d = 0 [1] r = 4i + 10j m when t = 2 [1] (d) v = 6i − 3j [1] Speed = √45 = 6.71 m s⁻¹ [1] Examiner insight: the constant of integration is a vector; adding it to one component only loses the mark in (a).

10. (a) Horizontal velocity is constant: 36 ÷ 24 = 1.5 s [1] (b) Vertical displacement: 10(1.5) − 4.9(1.5)² = 3.975 m above A [1] Height above ground = 3.975 + 1.5 = 5.475 m, which is more than 4 m, so it clears the wall [1] Clearance = 5.475 − 4 = 1.475 m [1] (c) 0 = 10² − 2(9.8)h, h = 5.10 m above A [1] Greatest height = 6.60 m above the ground [1] (d) v = 24i + (10 − 9.8 × 1.5)j = 24i − 4.7j [1] Speed = √(24² + 4.7²) = 24.5 m s⁻¹ [1] Angle = tan⁻¹(4.7/24) = 11.1° below the horizontal [1] (e) Any one: include air resistance; allow for the ball’s size or spin [1] Examiner insight: in (b), compare with 4 m and state the conclusion; a height alone is incomplete.

11. (a) Vertically: 0 = (U sin α)T − ½gT² [1] T ≠ 0, so T = 2U sin α / g [1] R = (U cos α)T = 2U² sin α cos α / g [1] Using sin 2α = 2 sin α cos α, R = U² sin 2α / g [1] (b) 30 = 400 sin 2α / 9.8, so sin 2α = 0.735 [1] 2α = 47.3° or 132.7°, so α = 23.7° or 66.3° [1] Examiner insight: “two possible values” means you need the second solution of sin 2α = 0.735.

Where marks are usually lost

• Giving a distance when a displacement is asked for, or dropping its sign.
• Using suvat in questions 7, 8 or 9, where the acceleration depends on t.
• Taking upwards as positive and then using a = +9.8 m s⁻².
• Dropping or mishandling a constant of integration, especially a vector one.
• Giving only the speed when the velocity or direction of motion is asked for.

Next steps

• Recap with the Kinematics revision notes.
• Re-read any weak section in the Kinematics study guide.
• Then try the Forces and Newton’s laws practice and Moments practice.
• Course hub: /boards/aqa/a-level/mathematics/.
• Printable checklist: /checklists/aqa/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.18 Q: Kinematics (Q1 to Q5), with Appendix A: mathematical notation (mechanics) and Appendix B: mathematical formulae and identities (kinematics).