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, B along BC and C along CA. At what time will the particles meet each other?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The motion of the particles is roughly sketched in diagram. By symmetry they will meet at the centroid O of the triangle. At any instant the particles will form an equilateral triangle ABC with the same centroid O. Concentrate on the motion of any one particle, say A. At any instant its velocity makes angle ${30}^{\circ }$ with AO.

The component of this velocity along AO is v cos ${30}^{\circ }$. This component is the rate of decrease of the distance AO. Initially,

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

Therefore, the time taken for AO to become zero

=$\frac{\frac{d}{\sqrt{3}}}{vcos{30}^{\circ }}=\frac{2d}{\sqrt{3}v\times \sqrt{3}}=\frac{2d}{3v}$

Alternative: Velocity of A is v along AB. The velocity of B is along BC. Its component along BA is

$vcos\left({60}^{\circ }\right)=\frac{v}{2}$. Thus, the separation AB decreases at the rate

$v+\frac{v}{2}=\frac{3v}{2}$

Since this rate is constant, the time taken in reducing the separation AB from d to zero is

$t=\frac{d}{\frac{3v}{2}}=\frac{2d}{3v}$