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Question: Four particles are situated at four corners of a square of side 'a'. All particles start moving with...

Four particles are situated at four corners of a square of side 'a'. All particles start moving with constant speed 'v' such that first headed second, second headed third, third headed fourth and fourth headed first. When all particles will meet ?

A

3va\frac{3v}{a}

B

2a3v\frac{2a}{3v}

C

a3v\frac{a}{3v}

D

av\frac{a}{v}

Answer

av\frac{a}{v}

Explanation

Solution

 All particles will meet at centre of square.

∴\thereforeTime

= Relative 6muDisplacementRelative 6muvelocity\frac{{Re}lative\mspace{6mu} Displacement}{{Re}lative\mspace{6mu} velocity}

= av+vcos⁡900\frac{a}{v + v\cos 90^{0}} = av\frac{a}{v}