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Three particles A, B and C are situated ...

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?

A

`2d//3v`

B

`d//3v`

C

`3d//2v`

D

`4d//3v`

Text Solution

Verified by Experts

The correct Answer is:
A

`(v_(A//B))` along `x`-axis `=v+vcos 60^(@)=(3v)/2`
`t=d/(3v//2)=(2d)/(3v)`
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