Find the value of \(t\) when the particle is instantaneously at rest.

The particle returns to its initial position at \(t = T\).

Find the value of T.

\(3 - \frac{t}{2} = 0 \Rightarrow t = 6{\text{ (s)}}\)     (M1)A1

Note:     Award A0 if either \(t =  - 0.236\) or \(t = 4.24\) or both are stated with \(t = 6\).

[2 marks]

let \(d\) be the distance travelled before coming to rest

\(d = \int_0^4 {5 - {{(t - 2)}^2}{\text{d}}t + \int_4^6 {3 - \frac{t}{2}{\text{d}}t} } \)     (M1)(A1)

Note:     Award M1 for two correct integrals even if the integration limits are incorrect. The second integral can be specified as the area of a triangle.

\(d = \frac{{47}}{3}\;\;\;( = 15.7){\text{ (m)}}\)     (A1)

attempting to solve \(\int_6^T {\left( {\frac{t}{2} - 3} \right){\text{d}}t = \frac{{47}}{3}} \) (or equivalent) for \(T\)     M1

\(T = 13.9{\text{ (s)}}\)     A1

[5 marks]

Total [7 marks]

Part (a) was not done as well as expected. A large number of candidates attempted to solve \(5 - {(t - 2)^2} = 0\) for \(t\). Some candidates attempted to find when the particle’s acceleration was zero.

Most candidates had difficulty with part (b) with a variety of errors committed. A significant proportion of candidates did not understand what was required. Many candidates worked with indefinite integrals rather than with definite integrals. Only a small percentage of candidates started by correctly finding the distance travelled by the particle before coming to rest. The occasional candidate made adroit use of a GDC and found the correct value of \(t\) by finding where the graph of \(\int_0^4 {5 - {{(t - 2)}^2}{\text{d}}t + \int_4^x {3 - \frac{t}{2}{\text{d}}t} } \) crossed the horizontal axis.