A man can row at 8 km/h in still water. The river's current is 2 km/h. He rows to a place and back in a total of 32 minutes. How far is that place from the start (one-way distance in km)?

Introduction / Context:Round-trip time with different upstream and downstream speeds can be used to determine one-way distance. Here, the still-water speed and current are given, so both leg speeds are known.

Given Data / Assumptions:

• Still-water speed b = 8 km/h
• Current speed c = 2 km/h
• Downstream speed vd = b + c = 10 km/h
• Upstream speed vu = b − c = 6 km/h
• Total time T = 32 minutes = 32/60 h = 8/15 h
• Let one-way distance be d (km)

Concept / Approach:Total time T = d/vd + d/vu. Solve for d given T, vd, vu.

Step-by-Step Solution:

Verification / Alternative check:Time downstream = 2/10 = 0.2 h; time upstream = 2/6 ≈ 0.333... h; total ≈ 0.533... h = 32 minutes, matches.

Why Other Options Are Wrong:1.5, 2.5, or 3 lead to totals not equal to 32 minutes when substituted into d/vd + d/vu with vd = 10 and vu = 6.

Common Pitfalls:Using average speed over unequal legs or averaging 10 and 6. Average speed cannot be used because times per leg differ.

Final Answer:2 km.