The width of a river is 80 m. The velocity of a boat in still water is 10 m/s. If water flows in the river with a velocity of 6 m/s, the difference in times taken to cross the river in the shortest path and the shortest time is

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10 s

1 s

$\frac{\sqrt{3}}{2} s$

2 s

answer is D.

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Detailed Solution

$\begin{array}{l} V_{0} = 10 {ms}^{- 1} V_{w} = 6 {ms}^{- 1} W = 80 m \\ t_{1} = \frac{W}{V_{res}} = \frac{80}{8} = 10 \sec t_{2} = \frac{W}{V_{0}} = \frac{80}{10} = 8 \sec \\ Δt = 10 - 8 = 2 \sec \end{array}$

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