A police party is moving in a jeep at a constant speed $v$. They saw a thief at a distance $x$ on a…
A police party is moving in a jeep at a constant speed $v$. They saw a thief at a distance $x$ on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration $a$. Which of the following relations is true if the police is able to catch the thief?
$v^2 < a x$
$v^2 < 2 a x$
$v^2 \geq 2 a x$
$v^2=a x$
Solution
Let the police party catch the thief after $t$ second.
$\therefore$ Distance travelled by police party in $t$ second $=v t$
and distance travelled by thief $=x+\frac{1}{2} \alpha t^2$
So, $\quad x+\frac{1}{2} \alpha t^2 \leq v t$
or $\quad \frac{\alpha t^2}{2}-v t+x=0$
or $\quad t=\frac{v \pm \sqrt{v^2-2 \alpha x}}{\alpha}$
For $t$ to be real, $v^2 \geq 2 \alpha x$
