Pursuit – Overtaking distance with head start: A thief is spotted 200 m ahead. The thief runs at 16 km/h while a policeman chases at 20 km/h. Assuming constant speeds, how far will the thief run before being caught?

Introduction / Context:
With a fixed lead and higher chaser speed, the catch occurs when the chaser closes the head start at the relative speed. The distance run by the thief is his speed multiplied by the catch-up time.

Given Data / Assumptions:

• Lead = 200 m = 0.2 km.
• Thief speed = 16 km/h.
• Policeman speed = 20 km/h.

Concept / Approach:
Relative speed = 20 − 16 = 4 km/h. Time to catch = 0.2 / 4 hours. Distance thief runs = 16 * time.

Step-by-Step Solution:
Time = 0.2 / 4 = 0.05 h = 3 min.Distance thief runs = 16 * 0.05 = 0.8 km = 800 m.

Verification / Alternative check:
Chaser covers 1.0 km in 3 min more than thief's 0.8 km; the extra 0.2 km equals the head start.

Why Other Options Are Wrong:
650, 700, 850, 900 m do not match the computed product speed × time.

Common Pitfalls:
Forgetting to convert 200 m to km or computing the relative speed in the wrong direction.

Final Answer:
800 m