Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time `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?

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`