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Question
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?
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Solution
(a) The net displacement is zero because the initial and final positions of the cyclist are the same.
(b) Average velocity is given by the relation:
Average velocity = `"Net displacement"/"Total time"`
As net displacement is zero, the average velocity of the cyclist is also zero.
(c) Average speed of the cyclist is given by the relation:
Average speed = `"Total path length"/"Total time"`
= `(OP + PQ + QO)/t`
Now, OP = QO = 1 km;
`PQ = 1 + 1/4(2 pi r)`
= `1/4(2pi xx 1)`
= 1.571 km
Time taken = `10 min = 10/60 = 1/6 h`
∴ Average speed = `(1 + 3.570 + 1)/(1/6)`
= 3.571 × 6
= 21.42 km/h
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