A walks at a uniform speed of 2 km/h. Two hours later, B starts cycling behind A at 5 km/h on the same straight route. At what distance from the starting point will B catch A?

Introduction / Context:
This is a classic pursuit problem on a straight line with a head start. We compare a walker (2 km/h) who starts first with a cyclist (5 km/h) who starts later. The key is to use the initial lead and relative speed to find the catch-up time and then the meeting position from the start.

Given Data / Assumptions:

• A's speed = 2 km/h (walking).
• B's speed = 5 km/h (cycling).
• B starts 2 hours after A.
• Straight route; speeds are constant.

Concept / Approach:
When one chases another along the same line, the catch-up time equals head start distance divided by relative speed. Meeting distance from the start equals the leader's speed multiplied by total time the leader has been traveling at meeting.

Step-by-Step Solution:
A's head start distance = 2 km/h * 2 h = 4 km.Relative speed (B w.r.t. A) = 5 − 2 = 3 km/h.Catch-up time after B starts = 4 / 3 h = 1 h 20 min.Total time A has moved at meet = 2 + 4/3 = 10/3 h.Meeting distance from start = 2 * (10/3) = 20/3 km = 6 2/3 km.

Verification / Alternative check:
Position of B at meet: B travels (4/3) h at 5 km/h = 20/3 km, same as A's position, confirming the meeting point.

Why Other Options Are Wrong:
6 1/3 km, 6 3/7 km, 6 5/7 km come from incorrect relative-speed or head-start handling.

Common Pitfalls:
Using total time 4/3 h to compute distance for A without adding A's initial 2-hour lead time.

Final Answer:
6 2/3 km