A thief spots a jeep 250 m away coming at 36 km/h. After 5 s, the thief runs away at 54 km/h. Ten seconds later, the police speed up to 72 km/h and give chase. How long after the thief first saw the jeep is he caught, and how far does the jeep travel to catch him?

Introduction / Context:This problem involves piecewise-constant speeds with reaction delays. We track the separation over successive intervals, switching relative speeds as the chase dynamics change.

Given Data / Assumptions:

• Initial separation = 250 m.
• Jeep: 36 km/h = 10 m/s; later 72 km/h = 20 m/s.
• Thief: starts after 5 s at 54 km/h = 15 m/s away from jeep.
• Police react 10 s after the thief starts running.

Concept / Approach:Compute separation changes interval by interval: (0–5 s) thief stationary; (5–15 s) thief faster than jeep; (≥15 s) jeep faster than thief. Use relative speed to update separation until closure.

Step-by-Step Solution:0–5 s: separation reduces by 10*5 = 50 m → 200 m remains.5–15 s: thief at 15 m/s, jeep at 10 m/s → separation increases 5 m/s for 10 s → +50 m → 250 m.t ≥ 15 s: jeep at 20 m/s vs thief 15 m/s → closing rate 5 m/s.Time to close 250 m at 5 m/s = 50 s → catch at 15 + 50 = 65 s after first sight.Jeep distance = 10*15 + 20*50 = 150 + 1000 = 1150 m.

Verification / Alternative check:At catch, positions match: thief distance after start (from t = 5 s) is 15*(65 − 5) = 900 m from his start point, consistent with the relative closure.

Why Other Options Are Wrong:Other time–distance pairs conflict with the interval-by-interval separation updates.

Common Pitfalls:Assuming the jeep accelerates immediately or forgetting that separation grows during 5–15 s.

Final Answer:65 s, 1150 m