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.