A-LevelPhysicsKinematics9702/21 · Oct/Nov 2017 · Q28 Marks

The variation with time t of the velocity v of two cars P and Q is shown in Fig. 2.1. The cars travel in the same direction along a straight road. Car P passes car Q at time t = 0. [Figure 2.1]

(a)The speed limit for cars on the road is 100kmh⁻¹. State and explain whether car Q exceeds the speed limit.[1]
(b)Calculate the acceleration of car P.[2]
(c)Determine the distance between the two cars at time t = 12s.[3]
(d)From time t = 12s, the velocity of each car remains constant at its value at t = 12s. Determine the time t at which car Q passes car P.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[B1] 30 ms⁻¹ = 108kmh⁻¹ or 100kmh⁻¹ = 28 ms⁻¹ and so exceeds speed limit
[C1] acceleration = gradient or Δv/(Δ)t or (v – u)/t
[A1] e.g. acceleration = (24 – 20) / 12 [other points on graph line may be used] = 0.33 ms⁻²
[C1] distance travelled by Q = ½ × 12 × 30 (= 180 m)
[C1] distance travelled by P = ½ × (20 + 24) × 12 (= 264 m)
[A1] distance between cars = 264 – 180 = 84 m
[C1] 30 – 24 = 6ms⁻¹ 'extra' time T = 84/6 (= 14 s) or 180 + 30T = 264 + 24T 'extra' time T = 84/6 (= 14 s)
[A1] t = 12 + 14 = 26 s

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