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 $\alpha$. Which of the following relations is true if the police is able to catch the thief?
Answer: (C) $v^2 > 2\alpha x$
After time $t$, the police have covered $vt$ and the thief is at $x + \dfrac{1}{2}\alpha t^2$ from the police's starting point.
The police catch the thief if these are equal for some real $t$:
$$vt = x + \frac{1}{2}\alpha t^2 \;\Rightarrow\; \alpha t^2 - 2vt + 2x = 0$$
A real (positive) solution exists only if the discriminant is non-negative:
$$4v^2 - 8\alpha x \ge 0 \;\Rightarrow\; v^2 \ge 2\alpha x$$
Among the options, the police can catch the thief when $v^2 > 2\alpha x$. (At exactly $v^2 = 2\alpha x$ they just reach him.)
Solution by Sreeraj P, M.Sc Physics