Q 11-02-026NEETJEE MainMedium
A police jeep moving at a constant $20$ m/s spots a thief on a scooter at rest $64$ m ahead. At that instant the thief starts with a constant acceleration of $2\ \text{m/s}^2$ in the same direction. After how much time will the jeep first reach the thief?
Answer: (C) $4$ s
Measure positions from the jeep's starting point.
Jeep: $x_J = 20t$. Thief: $x_T = 64 + \dfrac{1}{2}(2)t^2 = 64 + t^2$.
They meet when $20t = 64 + t^2$:
$$t^2 - 20t + 64 = 0 \;\Rightarrow\; (t - 4)(t - 16) = 0$$
The jeep first reaches the thief at $t = 4$ s. (If it did not stop him, the faster-accelerating scooter would draw level again at $t = 16$ s.)
Solution by Sreeraj P, M.Sc Physics