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(D) For a regular polygon with $n$ sides of length $a$,where each person moves with speed $v$ towards the next person,the rate of approach between two

Vedclass · Accepted Answer

$\frac{2 a}{v(2-\sqrt{3})}$

Vedclass · Answer

(D) For a regular polygon with $n$ sides of length $a$,where each person moves with speed $v$ towards the next person,the rate of approach between two adjacent persons is given by $v_{rel} = v - v \cos 	heta$,where $	heta$ is the exterior angle of the polygon.For a regular polygon with $n$ sides,the exterior angle is $	heta = \frac{2 \pi}{n}$.The time taken to meet is $t = \frac{a}{v_{rel}} = \frac{a}{v(1 - \cos \frac{2 \pi}{n})}$.Given $n = 12$,we have:$t = \frac{a}{v(1 - \cos \frac{2 \pi}{12})} = \frac{a}{v(1 - \cos \frac{\pi}{6})}$.Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,we get:$t = \frac{a}{v(1 - \frac{\sqrt{3}}{2})} = \frac{a}{v(\frac{2 - \sqrt{3}}{2})} = \frac{2 a}{v(2 - \sqrt{3})}$.

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