A particle `A` is at origin and particle `B` is at distance `y=-d` at `t=0.` They move with constant velocity `v`, A towards positive x-axis and `B` towards origin. The time at which distance between them is minimum and minimum distance will be

Options

A. `d/(2v),d`
B. `d/v,d/(sqrt(2))`
C. `d/(2v),d/(sqrt(2))`
D. `d/v,d`

Answer

C

Explanation

After time `t`,

`s^(2)=(vt)^(2)+(d-vt)^(2)`
For `s` to be minimum,
`(ds)/(dt)` or `(d(s^(2)))/(dt)=v^(2).2t+2(d-vt)(-v)=0`
`vt=(d-vt) implies t=d/(2v)`
`s_(min)=sqrt((d/2)^(2)+(d/2)^(2))=d/(sqrt(2))`