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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 Fig. 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 1

**(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 = `"Net displacement"/"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 =`"Total path length"/"Total time"`

Total path length = OP + PQ + QO

`= 1 + 1/4(2pi xx 1) + 1`

`=2 + 1/2 pi = 3.570 km`

Time taken = `10 min = 10/60 = 1/6 h`

∴ Average speed = `(3.570)/(1/6) = 21.42 "km/h"`

#### Solution 2

(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 = `"distance covered"/"time taken"`

`= (OP + "Actual distance " PQ + QO)/"10 minute"`

`= (1km + 1/4 xx 2pi xx 1 km + 1 km)/("10/60 h")`

`=6(2+22/14) km h^(-1) = 6xx50/14 km h^(-1)`

=21.43 `km h^(-1)`