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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

`(d)/(v)`

B

`(sqrt(2)d)/(v)`

C

`(d)/(2v)`

D

`(d)/(sqrt(2)v)`

Text Solution

Verified by Experts

The correct Answer is:
A
`T = (a)/(2vsin^(2)((pi)/(n))), (n=4)`
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