Question

Three particles $A,B$ and $C$ are situated at the vertices of an equilateral triangle $ABC$ of side $d$ at $t=0$. Each of the particles moves with constant speed $v$. $A$ always has its velocity along $AB$ i.e. along the line joining the two particles, $B$ along $BC$ and $C$ along $CA$. At what time will the particles meet each other?

• A. $t=\frac{d}{3v}$
• B. $t=\frac{2d}{3v}$
• C. $t=\frac{2d}{v}$
• D. $t=\frac{d}{v}$

Answer 1

The correct option is B $t=\frac{2d}{3v}$


Motion of the particle is shown in the figure. Let they meet at point $O$
Considering motion of particle $B$ w.r.t $C$, the relative velocity of $B$ w.r.t $C$ along the line joining them is $V_{BC}=V_B-V_C=v-(-vcos{60}^{\circ })$
$⟹V_{BC}=\frac{3v}{2}$
$\therefore$ time taken in reducing the separation $BC$ from $d$ to zero is
$t=\frac{d}{V_{BC}}=\frac{d}{\frac{3v}{2}}=\frac{2d}{3v}$