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Kinematics with constant acceleration questions
When acceleration holds steady, five quantities lock together and any three determine the other two. The suvat equations are the bookkeeping, the velocity-time graph is the reason they work, and gravity is the standard example.
8 original questions · 36 marks · the kinematics with constant acceleration notes · Mechanics
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List the five suvat quantities, and name the equation that leaves out s.
Worked answer
Displacement s, initial velocity u, final velocity v, acceleration a and time t. The equation without s is v = u + at. B1 for the list, B1 for the equation. Each of the four standard equations omits exactly one quantity, so choosing by the letter that never appears in the question is the whole routine.A car accelerates uniformly from rest at 2 m s⁻² for 6 s. Find its final speed and the distance it covers.
Worked answer
Taking u = 0, v = u + at = 0 + 2 × 6 = 12 m s⁻¹, and s = ut + ½at2 = 0 + ½ × 2 × 36 = 36 m. M1 A1 for the speed, M1 A1 for the distance. As a check, the average speed over the six seconds is (0 + 12)/2 = 6 m s⁻¹ and 6 × 6 = 36 m, which agrees. 'From rest' is what supplies u = 0, and it is stated in words rather than in symbols on purpose.A car brakes uniformly from 30 m s⁻¹ to rest over a distance of 150 m. Find the deceleration and the time taken.
Worked answer
Time is not mentioned in the first part, so use v2 = u2 + 2as: 0 = 900 + 300a, giving a = −3 m s⁻², a deceleration of 3 m s⁻². Then v = u + at gives 0 = 30 − 3t, so t = 10 s. M1 A1 for the acceleration, M1 A1 for the time. The negative sign is doing real work here, since the acceleration opposes the motion, and an answer of a = −3 quoted as a deceleration of −3 m s⁻² says the car is speeding up.A stone is dropped from rest down a well and hits the water 2.5 s later. Taking g = 9.8 m s⁻², find the depth of the well and the speed of the stone as it hits the water.
Worked answer
Taking downwards as positive, u = 0 and a = 9.8. Depth: s = ½gt2 = ½ × 9.8 × 6.25 = 30.625, so 30.6 m to 3 significant figures. Impact speed: v = gt = 9.8 × 2.5 = 24.5 m s⁻¹. M1 A1 for the depth, M1 A1 for the speed. 'Dropped' supplies u = 0 without saying so, and reading the starting conditions out of the wording is half of every kinematics question.A ball is thrown vertically upwards from ground level with speed 14 m s⁻¹. Taking g = 9.8 m s⁻², find the greatest height it reaches and the total time it spends in the air before returning to the ground.
Worked answer
Take up as positive, so a = −9.8 throughout. At the highest point v = 0, and v2 = u2 + 2as gives 0 = 196 − 19.6s, so s = 10 m. Time to the top is 14/9.8 = 1.43 s, and by symmetry the whole flight takes twice that, 2.86 s. M1 A1 for the height, M1 A1 for the time. Fixing the positive direction once at the start and never revisiting it is what keeps the signs consistent, and mixing a positive g with an upward positive direction is the classic route to a negative height.A ball is thrown vertically upwards from ground level with speed 14 m s⁻¹. Taking g = 9.8 m s⁻², find the times at which the ball is 5 m above the ground, and explain why there are two of them.
Worked answer
With up positive, 5 = 14t − 4.9t2, which rearranges to 4.9t2 − 14t + 5 = 0. The discriminant is 196 − 98 = 98, so t = (14 ± √98)/9.8, giving t = 0.418 s and t = 2.44 s. M1 for forming the quadratic, M1 for solving it, A1 A1 for the two roots, B1 for the explanation. Two roots mean two visits, since the ball passes 5 m on the way up and again on the way down, and a candidate who discards the second root as 'extra' loses two marks for a physically real answer.A ball is projected vertically upwards from ground level with speed 14 m s⁻¹. Exactly 1 s later a second ball is projected vertically upwards from the same point with the same speed. Taking g = 9.8 m s⁻², find the time after the first projection at which the two balls are at the same height, and find that height.
Worked answer
Measure t from the first projection, with up positive. The first ball is at h1 = 14t − 4.9t2. The second has been in the air for t − 1 seconds, so h2 = 14(t − 1) − 4.9(t − 1)2. Setting them equal, the t2 terms cancel and expanding leaves 0 = −14 + 9.8t − 4.9, so 9.8t = 18.9 and t = 1.93 s. Substituting back, h = 14(1.9286) − 4.9(1.9286)2 = 8.78 m. M1 for the first height, M1 for using t − 1 in the second, M1 for equating, A1 for the linear equation, A1 for t, A1 for the height. The whole question turns on the shifted time. Writing 14t for the second ball as well makes the two expressions identical and the equation unsolvable. Both balls land at t = 2.86 s and t = 3.86 s respectively, so 1.93 s is genuinely during the flight of both.A car travelling along a straight road at a constant 20 m s⁻¹ passes a stationary motorcycle. The motorcycle sets off immediately, accelerating uniformly at 2.5 m s⁻² until it reaches 30 m s⁻¹, after which it maintains that speed. Find the time, from the moment the car passed, at which the motorcycle draws level with the car, and the distance it has travelled by then.
Worked answer
Deal with the two phases separately. Phase one: v = u + at gives 30 = 0 + 2.5t, so the motorcycle reaches 30 m s⁻¹ after 12 s, having covered s = ½ × 2.5 × 122 = 180 m. In the same 12 s the car has covered 20 × 12 = 240 m, so it is still 60 m ahead. Phase two: both now travel at constant speed, and the motorcycle closes at 30 − 20 = 10 m s⁻¹, so it takes a further 60/10 = 6 s. The motorcycle draws level 18 s after the car passed, having travelled 180 + 30 × 6 = 360 m, which matches the car's 20 × 18 = 360 m. M1 A1 for the 12 s, M1 A1 for the two distances, M1 for the closing speed, A1 for 18 s, A1 for 360 m. Suvat applies to each phase separately but never across the join, since the acceleration changes at t = 12. Treating the whole 18 s as one uniformly accelerated stretch is the mistake this question is built to expose.
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