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]