A horse runs along a circle with a speed of 2 | vedclass

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(B) Let $r$ be the radius of the circle and $v = 20 \, km/hr$ be the speed of the horse.Let $	heta$ be the angle subtended by the horse at the centre

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let $r$ be the radius of the circle and $v = 20 \, km/hr$ be the speed of the horse.Let $	heta$ be the angle subtended by the horse at the centre of the circle at any time $t$.The position of the shadow on the fence is given by $x = r 	an 	heta$.Differentiating with respect to $t$,we get $\frac{dx}{dt} = r \sec^2 	heta \frac{d	heta}{dt}$.Since the speed of the horse is $v = r \frac{d	heta}{dt}$,we have $\frac{d	heta}{dt} = \frac{v}{r}$.Substituting this into the expression for $\frac{dx}{dt}$,we get $\frac{dx}{dt} = r \sec^2 	heta \left(\frac{v}{r}ight) = v \sec^2 	heta$.The horse covers $1/8$ of the circle,which corresponds to an angle $	heta = \frac{1}{8} 	imes 360^\circ = 45^\circ$.At $	heta = 45^\circ$,$\sec^2(45^\circ) = (\sqrt{2})^2 = 2$.Therefore,the speed of the shadow is $\frac{dx}{dt} = 20 	imes 2 = 40 \, km/hr$.

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