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
Concept. Write both positions from the jeep's start and solve for when they are equal.
Step 1. Positions. Jeep: $20t$. Thief: $64 + t^2$.
Step 2. Meeting.
$$t^2 - 20t + 64 = 0 \;\Rightarrow\; (t - 4)(t - 16) = 0$$
Step 3. First meeting. $t = 4$ s.
Answer. $4$ s (Option C).
Solution by Sreeraj P, M.Sc Physics