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A cyclist starts from the centre $O$ of a circular park of radius $1 \mathrm{~km}$, reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $Q O$ as shwon in Fig. If the round trip takes $10 \mathrm{~min}$, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist.
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(a) Since both the initial and final positions are the same therefore the net displacement is zero.
(b) Average velocity is the ratio of net displacement and total time taken. Since the net displacement is zero therefore the average velocity is also zero.
(c) Average speed $=\frac{\text { distance covered }}{\text { time taken }}$ $=\frac{O P+\text { Actual distance } P Q+Q O}{10 \text { minute }}$
$=\frac{1 \mathrm{~km}+\frac{1}{4} \times 2 \pi \times 1 \mathrm{~km}+1 \mathrm{~km}}{\frac{10}{60 h}}$
$=6\left(2+\frac{22}{14}\right) \mathrm{kmh}^{-1}=6 \times \frac{50}{14}=21.43 \mathrm{~km} \mathrm{~h}^{-1}$.
(b) Average velocity is the ratio of net displacement and total time taken. Since the net displacement is zero therefore the average velocity is also zero.
(c) Average speed $=\frac{\text { distance covered }}{\text { time taken }}$ $=\frac{O P+\text { Actual distance } P Q+Q O}{10 \text { minute }}$
$=\frac{1 \mathrm{~km}+\frac{1}{4} \times 2 \pi \times 1 \mathrm{~km}+1 \mathrm{~km}}{\frac{10}{60 h}}$
$=6\left(2+\frac{22}{14}\right) \mathrm{kmh}^{-1}=6 \times \frac{50}{14}=21.43 \mathrm{~km} \mathrm{~h}^{-1}$.
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