Further Mathematics Paper 2, May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength

Question 17

A particle moves from point 0 along a straight line such that its acceleration at any time, t seconds is a = (4 - 2 t) ms". At t = 0, its distance from 0 is 18 metres while its velocity is 5 ms'.
(a) At what time will the velocity be greatest?
(b) Calculate the:
(i) time;
(ii) distance of the particle from 0
when the particle is momentarily at rest.

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Observation

Majority of the candidates were also reported to have performed poorly in this question. Most of them failed to recognize the problem as that of integration.

In part (a), many candidates failed to understand that the velocity will be greatest when the acceleration is 0 i.e. 4 - 2t = O. Hence t = 2. In part (b), candidates were expected to obtain the velocity by integrating the acceleration with respect to t i.e. v = J(4 - 2t) dt = 4t - t2 + c. When t = 0, v = 5 i.e. 4(0) - (0) + C = 5.
Thus C = 5, hence v = 4t - t2 + 5. The particle will be momentarily at rest when v = 0 i.e. 4t - e + 5 = 0. Solving this quadratic equation gave t = 5 seconds. Integrating v with respect to t gave the displacement, s = J(4t - t2 + 5) dt = 2t2 - ~t3 + 5t + C. At t = 0, s = 18 = 2(0) - 1/3(0)+ C. 3
Thus c = 18. Hence s = 2e - ~ e + 5t + 18. When t = 5, s = 2(5)2 - J 13 (5i + 5(5) + 18 = 51.33 m