Three particles A, B and C are at the vertices of an equilateral triangle of side a. They start simultaneously with the same constant speed v, each always moving toward the next particle in cyclic order. Time after which they meet is:
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Question

Three particles A, B and C are at the vertices of an equilateral triangle of side a. They start simultaneously with the same constant speed v, each always moving toward the next particle in cyclic order. Time after which they meet is:

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G1 Physics · Accepted Answer

Consider particles A and B. A moves directly toward B with speed v. B's velocity makes 60° with BA, so the component of B's velocity away from A along AB is v cos60°=v/2. Thus the side AB decreases at rate vᵣ = v + v/2 =3v/2. Initial separation is a. Therefore t = a/(3v/2) =2a/(3v).