Write down the first value of \(t\) at which P changes direction.

Find the total distance travelled by P, for \(0 \leqslant t \leqslant 8\).

A second particle Q also moves along a straight line. Its velocity, \({v_{\text{Q}}}{\text{ m}}\,{{\text{s}}^{ - 1}}\) after \(t\) seconds is given by \({v_{\text{Q}}} = \sqrt t \) for \(0 \leqslant t \leqslant 8\). After \(k\) seconds Q has travelled the same total distance as P.

Find \(k\).

\(t = 2\)     A1     N1

[1 mark]

substitution of limits or function into formula or correct sum     (A1)

eg\(\,\,\,\,\,\)\(\int_0^8 {\left| v \right|{\text{d}}t,{\text{ }}\int {\left| {{v_Q}} \right|{\text{d}}t,{\text{ }}\int_0^2 {v{\text{d}}t - \int_2^4 {v{\text{d}}t + \int_4^6 {v{\text{d}}t - \int_6^8 {v{\text{d}}t} } } } } } \)

9.64782

distance \( = 9.65{\text{ (metres)}}\)     A1     N2

[2 marks]

correct approach     (A1)

eg\(\,\,\,\,\,\)\(s = \int {\sqrt t ,{\text{ }}\int_0^k {\sqrt t } } {\text{d}}t,{\text{ }}\int_0^k {\left| {{v_{\text{Q}}}} \right|{\text{d}}t} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\int {\sqrt t  = \frac{2}{3}{t^{\frac{3}{2}}} + c,{\text{ }}\left[ {\frac{2}{3}{x^{\frac{3}{2}}}} \right]_0^k,{\text{ }}\frac{2}{3}{k^{\frac{3}{2}}}} \)

equating their expression to the distance travelled by their P     (M1)

eg\(\,\,\,\,\,\)\(\frac{2}{3}{k^{\frac{3}{2}}} = 9.65,{\text{ }}\int_0^k {\sqrt t {\text{d}}t = 9.65} \)

5.93855

5.94 (seconds)     A1     N3

[4 marks]