Catch-up times from staggered starts and different speeds: A, B, C walk 1 km in 5 min, 8 min, and 10 min, respectively. C starts at t = 0, B starts 1 min later, and A starts 2 min later than C. After A starts, when does A catch B and C (times in minutes)?

Introduction / Context:
With staggered starts, catch-up times use relative speeds and the head start each person has when A begins. We measure times from A's start instant.

Given Data / Assumptions:

• Speeds: vA = 1/5 km/min = 0.2; vB = 1/8 = 0.125; vC = 1/10 = 0.1.
• Start times: C at 0, B at 1, A at 2 (minutes).
• Uniform speeds, straight line.

Concept / Approach:
Let x be minutes after A starts. Distances from the common start are equated to find catch-up times: 0.2x = 0.125(x + 1) for B; 0.2x = 0.1(x + 2) for C.

Step-by-Step Solution:

Verification / Alternative check:
Convert to metres and seconds; the same proportions hold, confirming the times.

Why Other Options Are Wrong:
Other time pairs do not solve both linear equations simultaneously.

Common Pitfalls:
Measuring from absolute time instead of from A’s start; mixing minutes and hours.

Final Answer:
5/3 min, 2 min