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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?
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Verified Answer
The correct answer is:
$v^2 \geq 2 a x$
Let the police party catch the thief after $t \mathrm{~s}$
$\therefore$ Distance travelled by police party in $t \mathrm{~s}=v t$
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, $v^2 \geq 2 \alpha x$
$\therefore$ Distance travelled by police party in $t \mathrm{~s}=v t$
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, $v^2 \geq 2 \alpha x$
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