Pursuit problem with start delay A thief steals a car at 1:30 p.m. driving 45 km/h. The theft is discovered at 2:00 p.m., and the owner starts at 50 km/h. At what time will the owner catch the thief?

Introduction / Context:
Classic chase problems compare the initial head start to the relative speed. The pursuer closes the gap at the difference of speeds, provided the pursuer is faster.

Given Data / Assumptions:

• Thief speed = 45 km/h, starts at 1:30 p.m.
• Owner speed = 50 km/h, starts at 2:00 p.m.
• Uniform speeds; straight route.

Concept / Approach:
Head start distance = thief speed * start gap (0.5 h). Catch-up time after the owner starts equals head start / relative speed.

Step-by-Step Solution:

Verification / Alternative check:
Distances at 6:30 p.m.: thief drives 5 h from 1:30 ⇒ 225 km; owner drives 4.5 h from 2:00 ⇒ 225 km.

Why Other Options Are Wrong:
Earlier times ignore the 22.5 km head start or use the wrong relative speed.

Common Pitfalls:
Adding speeds (for opposite directions) or forgetting the 30-minute start delay.

Final Answer:
6:30 p.m.