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A cyclist starts from the centre $O$ of a circular park of radius one kilometre, reaches the edge $P$ of the partk. Then cycles along the circumference and returns to the centre along $QO$ as shown in the figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour respectively) is
$0, 1$
$\frac{{\pi + 4}}{2},0$
$21.4,\frac{{\pi + 4}}{2}$
$0, 21.4$
Solution
Net displacement of the cyclist is zero, since the initial position coincides with the final position.
Average speed of the cyclist $=\frac{\text { total distance travelled }}{\text { total time taken }}$
${=\frac{\mathrm{OP}+\mathrm{PQ}+\mathrm{QO}}{10} \mathrm{km} / \mathrm{min}}$
${=\frac{1+\frac{\pi}{2} \times 1+1}{10}=\frac{\pi+4}{20} \mathrm{km} / \mathrm{min}}$
${=\frac{\pi+4}{20} \times 60 \mathrm{km} / \mathrm{hr}=21.4 \mathrm{km} / \mathrm{hr}}$
