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(B) Let $r = 10 \ m$ be the radius of the circular track.The speed of the man along the track is $v = r \frac{d	heta}{dt} = 10 \ m/sec$.Since $r = 10

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let $r = 10 \ m$ be the radius of the circular track.The speed of the man along the track is $v = r \frac{d	heta}{dt} = 10 \ m/sec$.Since $r = 10 \ m$,we have $10 \frac{d	heta}{dt} = 10$,which implies $\frac{d	heta}{dt} = 1 \ rad/sec$.Let $y$ be the position of the shadow on the wall at an angular distance $	heta$ from $P$.From the geometry,we have $	an 	heta = \frac{y}{r}$,so $y = r 	an 	heta$.Differentiating with respect to time $t$,we get $\frac{dy}{dt} = r \sec^2 	heta \cdot \frac{d	heta}{dt}$.At $	heta = 60^{\circ}$,we have $\sec(60^{\circ}) = 2$,so $\sec^2(60^{\circ}) = 4$.Substituting the values,$\frac{dy}{dt} = 10 	imes 4 	imes 1 = 40 \ m/sec$.Thus,the speed of the shadow is $40 \ m/sec$.

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