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.