A-LevelMathematicsMechanics9709/42 · Oct/Nov 2018 · Q58 Marks

A particle moves in a straight line starting from a point O with initial velocity 1 ms¯¹. The acceleration of the particle at time t s after leaving O is a m s¯², where a = 1.2t^(1/2) – 0.6t.

(i)At time T s after leaving O the particle reaches its maximum velocity. Find the value of T.[2]
(ii)Find the velocity of the particle when its acceleration is maximum (you do not need to verify that the acceleration is a maximum rather than a minimum).[6]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[M1] [1.2T^(1/2) – 0.6T = 0] Attempt to find time of maximum v, set a = 0 and solve for T
[A1] T^(1/2) = 2 → T = 4
[M1] [da/dt = 0.6t^(1/2) – 0.6] Attempt to differentiate a
[A1] Solve da/dt = 0 and find t
[M1] [v = 0.8t^(3/2) – 0.3t² (+ C)] Attempt to integrate a to find v
[A1] Correct integration
[M1] [C = 1] Use v = 1 at t = 0 either finding C or by using limits as v(1) – v(0) = [0.8(1)^(3/2) – 0.3(1)²] – [0.8(0)^(3/2) – 0.3(0)²]
[A1] Velocity when acceleration is max is 1.5 ms⁻¹

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Topic: Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.
Source: A-Level Mathematics, Oct/Nov 2018 Paper 4 Variant 2, question 5.

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