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)?

May 22, 2017

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
$s = u t + \frac{1}{2} a {t}^{2}$
we get
$s = \frac{1}{2} a {t}^{2}$ ......(1)
During this time interval distance traveled by police jeep $= s + d$
Which is also $= v t$

Equating two we get
$s + d = v t$
$\implies s = v t - d$ ......(2)
From (1) and (2) we get
$\frac{1}{2} a {t}^{2} = v t - d$
$\implies a {t}^{2} - 2 v t + 2 d = 0$

Solving quadratic for $t$
$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$t = \frac{2 v \pm \sqrt{4 {v}^{2} - 4 a \times 2 d}}{2 a}$
$\implies t = \frac{v \pm \sqrt{{v}^{2} - 2 a d}}{a}$
For a real time $t$, the discriminant must be $\ge 0$ and $t$ can not be negative. From the first condition we get
${v}^{2} - 2 a d \ge 0$
$\implies {v}^{2} \ge 2 a d$
$\implies v \ge \sqrt{2 a d}$