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Q.

Six particles situated at the corners of a regular hexagon of side $d$ move at a constant speed $v$ . Each particle maintains a direction towards the particle at the next corner. The time of the particles will take to meet each other.

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a

$2 d / 3 v$

b

$2 d / v$

c

$3 d / 2 v$

d

$4 d / 3 v$

answer is B.

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$T h e p a r t i c l e s w i l l m e e t a t t h e c e n t r e o f t h e h e x a g o n w h o s e d i s \tan c e f r o m p o i n t A i d d . T h e r e f o r e t = \frac{D i s p l a c e m e n t o f a n y p a r t i c l e}{v \cos 60^{0}} \Rightarrow t = \frac{d}{v / 2} = \frac{2 d}{v}$

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