Question

Three particles $A,B$ and $C$ are situated at the vertices of an equilateral triangle of side $r$ at $t=0$. The particle $A$ heads towards $B$, $B$ towards $C$, $C$ towards $A$ with constant speeds $v$. Find the time of their meeting.

Answer 1

The motion of the particles is roughly sketched in Fig. By symmetry they will meet at the centroid $O$ of the triangle. As the speed of all the particles is equal they will cover equal distance in any given interval of time.

If we join the instantaneous position of the particle at any time, the particles will form an equilateral triangle of same centroid as initial triangle.

Let us consider the motion of any one particle say $A$. At any instant, its velocity makes an angle ${30}^o$ with line $AO$.

The component of the velocity along $AO$ is $v\cos {30}^o$. This component will be equal to the rate of change of distance between $A$ and $O$.

At $t=0$, distance between $A$ and $O$,

$AO=\frac{2}{3}AD=\frac{2}{3}\sqrt{l^2-{\left(\frac{l}{2}\right)}^2}=\frac{l}{\sqrt{3}}$

At time $t=T$, the separation between $A$ and $O$ is zero. Hence, time taken for $AO$ to become zero.

$T=\frac{\left(l/\sqrt{3}\right)}{v\cos {30}^o}=\frac{\left(l/\sqrt{3}\right)}{v\left(\sqrt{3}/2\right)}=\frac{2d}{3v}$

After time $t$, let the distance of separation between the insect be $r$. The velocity of approach (component of relative velocity $v_{rel}$ between the insects along the line of their separation) is = $v+vcos{60}^o=3v/3$

Since $(v_{rel}{)}_{II}dt=-dr$ substituting $(v_{rel}{)}_{II}=\frac{3v}{2}$, we obtain

$\frac{3v}{2}dt=-dr$

As at time of meeting, integrating both sides, we have

$\frac{3v}{2}{\int }_0^{\tau }dt=-{\int }_l^{0}dr$

This gives $t=\frac{2l}{3v}$