Question

Three particles $A,B,C$ are situated at the vertices of an equilateral triangle of side $l$. Each of the particle starts moving with a constant velocity $v$ such that $A$ is always directed towards $B$, $B$ towards $C$ and $C$ towards $A$. Find the time when they meet.

• A. $\frac{l}{\sqrt{3}v}$
• B. $\frac{2l}{v}$
• C. $\frac{2l}{3v}$
• D. none

Answer 1

The correct option is C $\frac{2l}{3v}$

We look into it as a problem of relative velocity and find $v_{BA}$ in the direction of $B$.
Thus,
$t=\frac{l}{v_A-v_B\cos 120}$
$=\frac{l}{v+v/2}$
$=\frac{2l}{3v}$