An athlete runs along a circular track of dia | vedclass

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(A) Given,diameter,$d = 80 \ m$.Therefore,radius,$r = d/2 = 40 \ m$.Distance travelled after completion of $3/4$ revolution is given by:Distance $= (3

Vedclass · Accepted Answer

$60 \pi, 40 \sqrt{2}$

Vedclass · Answer

(A) Given,diameter,$d = 80 \ m$.Therefore,radius,$r = d/2 = 40 \ m$.Distance travelled after completion of $3/4$ revolution is given by:Distance $= (3/4) 	imes (2 \pi r) = (3/2) 	imes \pi 	imes 40 = 60 \pi \ m$.Displacement is the shortest distance between the initial point $A$ and the final point $B$. Since the athlete covers $3/4$ of the circle,the angle between the initial and final position vectors is $90^{\circ}$.Using the Pythagorean theorem for the right-angled triangle formed by the two radii:Displacement $= \sqrt{r^2 + r^2} = r \sqrt{2} = 40 \sqrt{2} \ m$.

An athlete runs along a circular track of diameter $80 \ m$. The distance travelled and the magnitude of displacement of the athlete when he covers $3/4$th of the circle is (in $m$):

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