The acceleration,  $a{\text{ m}}{{\text{s}}^{ - 2}}$, of a particle at time t seconds is given by $$a = \frac{1}{t} + 3\sin 2t {\text{, for }} t \ge 1.$$

The particle is at rest when $t = 1$ .

Find the velocity of the particle when $t = 5$ .

evidence of integrating the acceleration function     (M1)

e.g. $\int {\left( {\frac{1}{t} + 3\sin 2t} \right)} {\rm{d}}t$

correct expression $\ln t - \frac{3}{2}\cos 2t + c$     A1A1

evidence of substituting (1, 0)     (M1)

e.g. $0 = \ln 1 - \frac{3}{2}\cos 2 + c$

$c = - 0.624$ $\left( { = \frac{3}{2}\cos 2 - \ln {\text{1 or }}\frac{{\rm{3}}}{{\rm{2}}}\cos 2} \right)$     (A1)

$v = \ln t - \frac{3}{2}\cos 2t - 0.624$ $\left( { = \ln t - \frac{3}{2}\cos 2t + \frac{3}{2}\cos {\text{2 or ln}}t - \frac{3}{2}\cos 2t + \frac{3}{2}\cos 2 - \ln 1} \right)$     (A1)

$v(5) = 2.24$ (accept the exact answer $\ln 5 - 1.5\cos 10 + 1.5\cos 2$ ) A1     N3

[7 marks]

This problem was not well done. A large number of students failed to recognize that they needed to integrate the acceleration function. Even among those who integrated the function, there were many who integrated incorrectly. A great number of candidates were not able to handle the given initial condition to find the integration constant but incorrectly substituted $t = 5$ directly into their expression.