Two particles start simulataneously from the same point and moves along two straight lines inclined at angle `alpha`, one with uniform velocity `v_(0)` and other from the rest with uniform acceleration, `a_(0)`. After how much time their relative velocity is minimum and what is it equal to?

Explanation

After time `t,v_(1)=u+at=a_(0)t=v_(1//g)`
`v_(2)=v_(0)=v_(2//g)`
`v_(1//2)`=resultant of (velocity of `1` and velocity of `2` in opposite direction)

`(v_(1//2))^(2)=v_(0)^(2)+(a_(0)t)^(2)+2(v_(0))(a_(0)t)cos(180-alpha)`
For `(v_(1//2))` to be minimum
`(d(v_(1//2)))/(dt)` or `(d(v_(1//2)^(2)))/(dt)=0`
`implies (2a_(0)t)a_(0)-2v_(0)a_(0)cos alpha=0`
`t=(v_(0)cos alpha)/(a_(0))`
`(v_(1//2))_(min)=sqrt(v_(0)^(2)+(v_(0) cos alpha)^(2)+2v_(0)(v_(0)cos alpha)(-cos alpha))`
`=v_(0)sin alpha`
Relative velocity is minimum after time `(v_(0)cos alpha)/(a_(0))` and equal to `v_(0) sin alpha`.