A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant acceleration a. Show that the pickpocket will be caught it `vgesqrt(2ad)`.

Explanation

Let us assume that it takes time t to catch the pickpocket. Distance travelled by the motorcycle in this time interval can be written as follows:
`s=ut+(1)/(2)at^(2)rArr s=(1)/(2)at^(2)`
Distance travelled by jeep in same time interval should be d+s, hence we can write the following:
`d+s=vt`
`rArr d+(1)/(2)at^(2)=vt`
`rArr at^(2)-2vt+2d=0`
`t=(2vpmsqrt(4v^(2)-8ad))/(2a)`
`rArr t=(vpmsqrt(v^(2)-2ad))/(a)`
For real solution of time we can write the following:
`v^(2)-2adge0`
`rArr v^(2) ge 2ad`
or `v ge sqrt(2ad)`