Three particles A, B and C and situated at the vertices of an equilateral triangle ABC of side of length $l$ at time $t = 0$ , Each of the particles move with constant speed $u$ . A always has its velocity along AB, B along BC and C along CA.

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The time after which they meet is $\frac{2 l}{3 u}$

Total distance travelled by each particle before they meet is $\frac{2 l}{3}$

Relative velocity of approach between any two particles is $\frac{3 u}{2}$

Average velocity during the motion is $\frac{\sqrt{3} u}{2}$

answer is A, B, C, D.

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Detailed Solution

$t = \frac{l}{u_{r e l}} = \frac{l}{u + \frac{u}{2}} = \frac{2 l}{3 u}$

Distance $= \int v d t = u \times \frac{2 l}{3 u} = \frac{2 l}{3}$

$v_{a v g} = \frac{total disp}{total time} = \frac{\frac{l}{\sqrt{3}}}{\frac{2 l}{3 u}} = \frac{\sqrt{3} u}{2}$

$V_{A B} = u + u \cos 60 = \frac{3 u}{2}$

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