OxfordAQA International AS and A-level Mathematics: Kinematics in a straight line: constant and variable acceleration (9660) – 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 sections M1.1 and M1.2 (straight-line motion with constant and variable acceleration) of OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, which belong to International AS Unit PSM1. Take g = 9.8 m/s². Give non-exact answers to 3 significant figures.

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Questions

1. A runner goes 120 m east along a straight track, then 50 m west, taking 40 s in total. Find the distance travelled, the displacement, the average speed and the average velocity. [4]

2. A car’s speed increases uniformly from 6 m/s to 21 m/s in 5 s. Find its acceleration and the distance it travels in this time. [4]

3. A motorbike slows uniformly from 30 m/s to 12 m/s while travelling 126 m. Find its acceleration and the time taken. [4]

4. A stone is dropped from rest into a well and hits the water after 2.5 s. Find the depth of the water below the release point and the speed of the stone as it hits the water. [4]

5. A ball is thrown vertically upwards at 12 m/s from a point 1.5 m above the ground. Find (a) the greatest height above the ground, [3] (b) the time before the ball hits the ground. [3]

6. A boat starts from rest and accelerates uniformly to 9 m/s in 6 s. It then travels at 9 m/s for T seconds before slowing uniformly to rest in 3 s. The total distance is 180 m. (a) Sketch the velocity–time graph and find T. [4] (b) Find its average speed for the whole journey. [2]

7. Particle A moves in a straight line at a constant 6 m/s, passing a point O at time zero. Particle B starts from rest at O, 6 s later, and moves along the same line in the same direction with constant acceleration 1.5 m/s². Find how long after B starts it catches A, and the distance from O at which this happens. [6]

8. A particle moves on a straight line so that its displacement from O at time t seconds is s = t³ − 12t² + 36t metres, for t ≥ 0. (a) Find expressions for its velocity and acceleration. [2] (b) Find the times at which it is instantaneously at rest and its acceleration at each. [3] (c) Find the total distance it travels in the first 7 s. [3]

9. A particle moves on a straight line with acceleration a = 12 − 6t m/s². When t = 0 it is at O with velocity 4 m/s. (a) Find expressions for its velocity and its displacement from O. [3] (b) Find its velocity and displacement when t = 3. [2] (c) Find the greatest velocity and the time at which it occurs. [2]

10. A particle moves on a straight line with velocity v = 6√t − 2t m/s, for t ≥ 0, and is at O when t = 0. (a) Find the acceleration in terms of t and the time at which it is zero. [3] (b) Show that the particle is at rest again at t = 9 and find its displacement from O then. [3]

11. A train travels between two stations in 60 s. It accelerates uniformly from rest at 1.5 m/s² to a speed of 12 m/s, travels at 12 m/s, then decelerates uniformly at 2 m/s² to rest. (a) Find the time spent at constant speed. [3] (b) Find the distance between the stations. [3] (c) Find the average speed. [2]

Answers

1. Distance = 120 + 50 = 170 m [1]. Displacement = 120 − 50 = 70 m east [1]. Average speed = 170 ÷ 40 = 4.25 m/s [1]. Average velocity = 70 ÷ 40 = 1.75 m/s east [1]. [4] Examiner insight: Stating the direction of the displacement and velocity is part of a complete answer, and quoting 170 m as the displacement shows the distance and displacement have been confused.

2. a = (21 − 6) ÷ 5 [1] = 3 m/s² [1]. s = ½(6 + 21) × 5 [1] = 67.5 m [1]. [4] Examiner insight: Several equations give the distance; choose one that uses the given values rather than your calculated a, so the two parts are checked independently.

3. v² = u² + 2as: 144 = 900 + 252a [1], a = −3 m/s² (deceleration 3 m/s²) [1]. v = u + at: 12 = 30 − 3t [1], t = 6 s [1]. [4] Examiner insight: A bare “3” with no sign or the word deceleration leaves the meaning unclear, so say which you mean.

4. Down positive: u = 0, a = 9.8, t = 2.5. s = ½ × 9.8 × 2.5² [1] = 30.6 m (30.625) [1]. v = 9.8 × 2.5 [1] = 24.5 m/s [1]. [4] Examiner insight: Show the substitution into the equation, because an answer with no working is difficult to credit if the number is wrong.

5. (a) Up positive: u = 12, a = −9.8, v = 0 at the top. 0 = 144 − 19.6s [1], s = 7.35 m above the launch point [1]. Height above ground = 7.35 + 1.5 = 8.85 m [1]. (b) At the ground s = −1.5: −1.5 = 12t − 4.9t² [1], so 4.9t² − 12t − 1.5 = 0 and t = (12 + √(144 + 29.4)) ÷ 9.8 [1] = 2.57 s (the negative root is rejected) [1]. [6] Examiner insight: Remember to add the 1.5 m in part (a), and in part (b) state clearly that the negative root is rejected because time cannot be negative.

6. (a) Graph: a straight line from (0, 0) to (6, 9), horizontal to (6 + T, 9), then a straight line down to (9 + T, 0) [1]. Area: ½ × 6 × 9 + 9T + ½ × 3 × 9 [1] = 27 + 9T + 13.5 = 180 [1], so T = 15.5 [1]. (b) Total time = 6 + 15.5 + 3 = 24.5 s [1]. Average speed = 180 ÷ 24.5 = 7.35 m/s [1]. [6] Examiner insight: Label the axes and the key values on the sketch, and remember the total time includes all three stages, not just the constant-speed stage.

7. Let B have moved for u seconds, so A has moved for u + 6 seconds. A: distance = 6(u + 6) [1]. B: distance = ½ × 1.5 × u² = 0.75u² [1]. Equate: 0.75u² = 6u + 36 [1], u² − 8u − 48 = 0 [1], (u − 12)(u + 4) = 0, so u = 12 s (reject −4) [1]. Distance = 0.75 × 144 = 108 m (check: 6 × 18 = 108) [1]. [6] Examiner insight: Define your time variable clearly for each particle; using the same t for both, with the 6 s delay left out, gives a different and incorrect equation.

8. (a) v = ds/dt = 3t² − 24t + 36 [1]; a = dv/dt = 6t − 24 [1]. (b) v = 0: 3(t² − 8t + 12) = 0 [1], 3(t − 2)(t − 6) = 0, so t = 2 s and t = 6 s [1]. a(2) = −12 m/s², a(6) = 12 m/s² [1]. (c) s(0) = 0, s(2) = 32, s(6) = 0, s(7) = 7 [1]. Legs: 32, then 32, then 7 [1]. Distance = 32 + 32 + 7 = 71 m [1]. [8] Examiner insight: Finding s(7) − s(0) = 7 gives the displacement only; the distance needs each leg between the times at rest, summed as positive values.

9. (a) v = ∫(12 − 6t) dt = 12t − 3t² + c, and v = 4 at t = 0 gives c = 4, so v = 4 + 12t − 3t² [1]. s = ∫v dt = 4t + 6t² − t³ + k [1], and s = 0 at t = 0 gives s = 4t + 6t² − t³ [1]. (b) v(3) = 4 + 36 − 27 = 13 m/s [1]; s(3) = 12 + 54 − 27 = 39 m [1]. (c) v is greatest when a = 0 [1]: 12 − 6t = 0 gives t = 2, and v(2) = 4 + 24 − 12 = 16 m/s [1]. (Since da/dt = −6 < 0, this is a maximum.) [7] Examiner insight: Each integration needs its own constant, found from its own condition, and the maximum should be justified rather than just assumed.

10. (a) v = 6t^(1/2) − 2t, so a = dv/dt = 3t^(−1/2) − 2 = 3/√t − 2 [1]. a = 0: 3/√t = 2 [1], √t = 1.5, so t = 2.25 s [1]. (b) v = 0: 6√t = 2t, so t = 0 or √t = 3 [1], giving t = 9 (at rest again). s = ∫v dt = 4t^(3/2) − t² [1] (the constant is zero as s = 0 at t = 0). s(9) = 4 × 27 − 81 = 27 m [1]. [6] Examiner insight: In a “show that” question, state the equation you solve and the value it gives, rather than only substituting t = 9 without a conclusion; writing the powers as indices keeps the calculus accurate.

11. (a) Speed-up time = 12 ÷ 1.5 = 8 s [1]; slow-down time = 12 ÷ 2 = 6 s [1]; constant-speed time = 60 − 8 − 6 = 46 s [1]. (b) Distance = ½ × 12 × 8 + 12 × 46 + ½ × 12 × 6 [1] = 48 + 552 + 36 [1] = 636 m [1]. (c) Average speed = 636 ÷ 60 [1] = 10.6 m/s [1]. [8] Examiner insight: A velocity–time sketch with the three times marked makes the area calculation a quick sum and helps you spot an omitted stage.

Where marks are usually lost

• Not choosing and stating a positive direction, so signs of a, v and s clash.
• Using a = +9.8 for upward motion when up is positive.
• Quoting a distance when the displacement is required, or the reverse.
• Not splitting the motion at the times when v = 0 for a distance question.
• Leaving out the constant of integration, or not using the condition given.
• Using the constant acceleration equations when a is a function of t.
• Rounding intermediate values and then losing accuracy in the final answer.
• Giving a negative or zero root of a quadratic in time as a valid answer.
• Writing no units or the wrong units, such as m for a velocity.

Next steps

• Revise with the revision notes.
• Read the study guide.
• Go to the course hub or the printable checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Sections M1.1 and M1.2 of Unit PSM1.