Three boys A,B and C are situated at the vertices of an equilateral triangle of side d at t=0. Each of the boys moves with constant speed v. A always moves towards C and C towards A, At what time and where they meet each other?

Explanation

This can be solved by considering reference frame attached to one of the particles. (By relative motion) Velocity of A w.r.t. B `" "vec(v)_(AB)=vec(v)_(A)-vec(v)_(B)`
`=(sqrt(2)hat(i)+sqrt(2)hat(j))-(-7sqrt(2)hat(i)+7sqrt(2)hat(j))`
`=8sqrt(2)hat(i)-6sqrt(2)hat(j)`
Dir. of `vec(v)_(AB)" "tan alpha=(-6sqrt(2))/(8sqrt(2))=(-3)/4`
`alpha=37^(@)` (downward from horizontal)
from
`tan 37^(@)=(AC)/(CD)=3/4`
`CD=(9xx4)/3=12 m" " :' " "BD=27-12=15m`
From
`(BP)/(BD)=sin 37^(@)=3/5`
`(BP)/15=3/5 rArr BP=9 m`