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An athlete completes one round of a circular track of diameter $200 m$ in $40 s$ . At the end of $2 \min 20 s$ the distance covered by the athlete is $m$ and displacement is $m$ .

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Detailed Solution

An athlete completes one round of a circular track of diameter

200 m

40 s

. At the end of

2 \min 20 s

the distance covered by the athlete is

2200 m

and displacement is

200 m

.
The diameter of the circular track is

200 m

which means that the radius of the track is

100 m

.
If an athlete in

40 s

completes one round then in

2 \min 20 s

the athlete will complete,

2 \times 60 + 20 = 120 + 20

= 140 s

\frac{140}{40} = 3 \frac{1}{2} rounds

So, athlete completes three rounds and half more so, the distance covered will be,

= 2 πr \times 3 + \frac{1}{2} \times 2 πr

= 6 πr + πr

= 7 πr

= 7 πr

= 7 \times π \times 100

= 7 \times \frac{22}{7} \times 100

= 2200 m

The distance covered by athletes is

2200 m

.
Displacement is the shortest distance between initial and final position which is

2 r

as the athlete covers three rounds and half rounds more.

Displacement = 2 r

= 2 \times 100

= 200 m

The distance covered by the athlete is

2200 m

and the displacement is

200 m

at the end of

2 \min 20 s

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