A cyclist starts from the centre $O$ of a circular park of radius $1\; km$, reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along $QO$ as shown in Figure. If the round trip takes $10 \;min$, what is the
$(a)$ net displacement,
$(b)$ average velocity, and
$(c)$ average speed of the cyclist ?
$(a)$ net displacement,
$(b)$ average velocity, and
$(c)$ average speed of the cyclist ?
$(a)$ Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for $10$ minutes. Hence, his net displacement is zero.
$(b)$ Average velocity is given by the relation:
Average velocity $=\frac{\text { Net displacement }}{\text { Total time }}$
since the net displacement of the cyclist is zero, his average velocity will also be zero.
$(c)$ Average speed of the cyclist is given by the relation:
Average speed $=\frac{\text { Total path length }}{\text { Total time }}$ Total path length $= OP + PQ + QO =1+\frac{1}{4}(2 \pi \times 1)+1$
$=2+\frac{1}{2} \pi=3.570\, km$
Time taken $=10\, \min =\frac{10}{60}=\frac{1}{6}\, h$
$\therefore$ Average speed $=\frac{3.570}{1}=21.42\, km / h$
$(b)$ Average velocity is given by the relation:
Average velocity $=\frac{\text { Net displacement }}{\text { Total time }}$
since the net displacement of the cyclist is zero, his average velocity will also be zero.
$(c)$ Average speed of the cyclist is given by the relation:
Average speed $=\frac{\text { Total path length }}{\text { Total time }}$ Total path length $= OP + PQ + QO =1+\frac{1}{4}(2 \pi \times 1)+1$
$=2+\frac{1}{2} \pi=3.570\, km$
Time taken $=10\, \min =\frac{10}{60}=\frac{1}{6}\, h$
$\therefore$ Average speed $=\frac{3.570}{1}=21.42\, km / h$
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