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Four particles A, B, C and D are situate...

Four particles `A, B, C and D` are situated at the cornerst of a square `ABCD` of side `a at t-0`. Each of particles moves with constant speed (v). A always has its velocity along `AB, B` along `BC, C` along `CB~ and D` along `DA`. At what time will these particles meet each other ?

A

`2d//3v`

B

`d//v`

C

`3d//2v`

D

`4d//3v`

Text Solution

Verified by Experts

The correct Answer is:
B

`(v_(A//B))_(x)=v-0=v`
`t=d/v`
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