A boy is moving with constant velocity `v_(0)`on a straight road. When he is at a distance `d` behind the bus, the bus starts from rest and moves with a constant acceleration `alpha`. Find the minimum value of `v_(0)` so that the boy catches the bus.

Explanation

Let the boy catch the bus after time t and at distance s.

Bus: `s=1/2at^(2)=(alphat^(2))/2 …(i)`
Boy: `d+s=v_(0)t …(ii)`
Eliminating t between (i) and (ii) we get
`d+(alpha t^(2))/2=v_(0)t`
`alphat^(2)-2v_(0)t+2d=0`
The boy will catch the bus if time t is real and positive. For time t to be real, discriminant,
`Dgt0`
`(b^(2)-4ac)ge0`
`(-2v_(0))^(2)-4alpha. 2d ge0`
`4v_(0)^(2)-8alpha d gt = 0`
`v_(0)^(2) gt = 2 alpha d implies v_(0) ge sqrt(2alphad)`
`(v_(0))_(min)=sqrt(2alphad)`