Question

A jeep chases a thief with a constant velocity `v`. When the jeep is at distance d from the thief, he starts to run with a constant acceleration `a`. Show that the police will be able to catch the thief when `v >= sqrt(2ad)`?

Answer 1

Shown

Explanation:

Given is speed of police jeep `= v`

Distance between the thief and the police jeep`=d`

At this point constant acceleration of thief `= a`.

Let the time when both meet `=t`
Let `s` be the distance covered by thief.
Using kinematic expression
`s = ut+1/2 at^2`
we get
`s = 1/2 at^2` ......(1)
During this time interval distance traveled by police jeep `= s + d`
Which is also `= vt`

Equating two we get
`s+d=vt`
`=>s = vt - d` ......(2)
From (1) and (2) we get
`1/2 at^2=vt - d`
`=>at^2 - 2vt + 2d = 0`

Solving quadratic for `t`
`t=(-b+-sqrt(b^2-4ac))/(2a)`
`t=(2v+-sqrt(4v^2-4axx2d))/(2a)`
`=>t=(v+-sqrt(v^2-2ad))/(a)`
For a real time `t`, the discriminant must be `>=0` and `t` can not be negative. From the first condition we get
`v^2-2ad>=0`
`=>v^2>=2ad`
`=>v>=sqrt(2ad)`