On the grid below, sketch the graph of v , clearly indicating the maximum point.

(i)     Write down an expression for d .

(ii)    Hence, write down the value of d .

     A1A1A1     N3

Note: Award A1 for approximately correct shape, A1 for right endpoint at \((25{\text{, }}0)\) and A1 for maximum point in circle.

[3 marks]

(i) recognizing that d is the area under the curve     (M1)

e.g. \(\int {v(t)} \)

correct expression in terms of t, with correct limits     A2     N3

e.g. \(d = \int_0^9 {(15\sqrt t } - 3t){\rm{d}}t\) , \(d = \int_0^9 v {\rm{d}}t\)

(ii) \(d = 148.5\) (m) (accept 149 to 3 sf)     A1     N1

[4 marks]

The graph in part (a) was well done. It was pleasing to see many candidates considering the domain as they sketched their graph.

Part (b) (i) asked for an expression which bewildered a great many candidates. However, few had difficulty obtaining the correct answer in (b) (ii).