Solution:
To cross the river along the minimum distance (i.e., directly perpendicular to the riverbank), the boatman must row with a velocity component equal and opposite to the river flow to cancel the drift.
Given:
- Speed of the boat in still water = 10 km/h
- Speed of the river flow = 5 km/h
- Width of the river = 2 km
The boat's velocity perpendicular to the riverbank is:
\[
v_{\text{perpendicular}} = \sqrt{(10^2 - 5^2)} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \, \text{km/h}
\]
Time to cross the river:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed perpendicular}} = \frac{2}{5\sqrt{3}} \, \text{hours}
\]
Simplifying:
\[{\frac{2}{5\sqrt{3}} \, \text{hours}}
\]
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