17M.2.SL.TZ1.S_7

pestleMathematicsAISLPaper 217M· ahl-5-13-kinematic-problemssource ↗

A particle P moves along a straight line. Its velocity ${v_{\text{P}}}{\text{ m}}\,{{\text{s}}^{ - 1}}$ after $t$ seconds is given by ${v_{\text{P}}} = \sqrt t \sin \left( {\frac{\pi }{2}t} \right)$ , for $0 \leqslant t \leqslant 8$ . The following diagram shows the graph of ${v_{\text{P}}}$ .

M17/5/MATME/SP2/ENG/TZ1/07

Write down the first value of $t$ at which P changes direction.

[1]

a.i.

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

[2]

a.ii.

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 $k$ .

[4]

Markscheme / solution

$t = 2$ A1 N1

[1 mark]

a.i.

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]

a.ii.

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]

Examiners’ report

[N/A]

a.i.

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a.ii.

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