At t = 0, three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 1.5d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet?
Solution
Velocity of A along AB = $v$
Velocity of B along BA = $v cos 60^\circ = 0.5v$
Relative velocity of B towards A = $v+0.5v=1.5v$
Displacement covered by B to reach A with respect to A = 1.5d
time = $\frac {d_{rel}}{v_{rel}}=\frac {1.5d}{1.5v}=\frac {d}{v}$
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