Question

Six particles situated at the corners of a constant speed $v$. Each particles maintains a direction towards the particle at the next. The time which the particle will take to meet each other is:

• A. $\frac{2a}{v}sec$
• B. $\frac{a}{v}sec$
• C. $\frac{2a}{3v}sec$
• D. $\frac{3a}{2v}sec$

Answer 1

The correct option is A $\frac{2a}{v}sec$

In the given figure$;-$

Particle $A$ moves towards $B$ with a speed $Vm/s$

Let us calculate the speed of $B$

$B$ is also moving with velocity $V$ along $BC$ direction.

Hence velocity component along $BC$:

$Vcos60$

$=V/2$

Relative speed of particle $A$ with respect$B=V-V/2$

$=V/2$

To travel a distance $a,$ time taken by the particle$:$

Time$=$ Distance$/$ Speed

$=a/\left(V/2\right)$

$=2a/V$

Hence,

option $\left(A\right)$ is correct answer.