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Two particle are moing along X and Y axe...

Two particle are moing along `X` and `Y` axes towards the origin with constant speeds `u` and `v` respectively. At time `t=0`, their repective distance from the origin are `x` and `y`. The time instant at which the particles will be closest to each other is

A

`sqrt(x^(2)+y^(2))/sqrt(u^(2)+v^(2))`

B

`(vx+uy)/(u^(2)+v^(2))`

C

`(ux+vy)/(u^(2)+v^(2))`

D

`(2sqrt(x^(2)+y^(2)))/(u+v)`

Text Solution

Verified by Experts

The correct Answer is:
C


`sqrt(x^(2)+y^(2))cos(theta_(1)-theta_(2))=(sqrt(v^(2)+u^(2)))t`
`rArrt=(sqrt(x^(2)+y^(2)))/(sqrt(v^(2)+u^(2)))[(vy+ux)/(sqrt(x^(2)+y^(2'))sqrt(v^(2)+u^(2)))]=(ux+vy)/(u^(2)+v^(2))`
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