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?

Options

A. `(d)/(u)` seconds
B. `(2d)/(3u)` seconds
C. `(2d)/sqrt(3u)` seconds
D. `dsqrt(3u)` seconds

Answer

B

Explanation

`d_("rel")=d`
`u_("rel")=u+u cos 60^(@)=(3u)/(2)`
time
For general `d_("rel")=` side of polygon
`u_("rel")=u+u cos theta [theta= "Interior angle"]`
`t=("side of polygon")/(u+ucos theta)`