A cyclist starts from the centre O of a circu | vedclass

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(D) The net displacement of the cyclist is $0 	ext{ m}$,since the initial position coincides with the final position (the centre $O$).The total dista

Vedclass · Accepted Answer

$0, 21.4$

Vedclass · Answer

(D) The net displacement of the cyclist is $0 	ext{ m}$,since the initial position coincides with the final position (the centre $O$).The total distance travelled is the sum of the paths $OP$,$PQ$ (arc length),and $QO$.$OP = 1 	ext{ km}$,$QO = 1 	ext{ km}$.Since $O$ is the centre and $OP$ and $OQ$ are perpendicular radii,the arc $PQ$ corresponds to a quarter of the circumference: $PQ = \frac{1}{4} 	imes (2\pi r) = \frac{1}{4} 	imes 2 	imes \pi 	imes 1 = \frac{\pi}{2} 	ext{ km}$.Total distance $= 1 + \frac{\pi}{2} + 1 = 2 + \frac{\pi}{2} = \frac{4 + \pi}{2} 	ext{ km}$.Total time $= 10 	ext{ minutes} = \frac{10}{60} 	ext{ hours} = \frac{1}{6} 	ext{ hours}$.Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{(4 + \pi)/2}{1/6} = \frac{4 + \pi}{2} 	imes 6 = 3(4 + \pi) 	ext{ km/h}$.Using $\pi \approx 3.14$,Average speed $= 3(4 + 3.14) = 3(7.14) = 21.42 	ext{ km/h} \approx 21.4 	ext{ km/h}$.Thus,the displacement is $0 	ext{ m}$ and the average speed is $21.4 	ext{ km/h}$.

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