The velocity of a boat in still water is $10 m s^{- 1} .$ If water flows in the river with velocity of $6 m s^{- 1} .$ What is the difference in times taken to cross the river in shortest path and shortest time. The width of river is 80m.

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Detailed Solution

In shortest path
$t_{1} = \frac{d}{\sqrt{V_{b}^{2} - V_{r}^{2}}} = \frac{80}{\sqrt{10^{2} - 6^{2}}} = \frac{80}{8} = 10 \sec$
In shortest time
$t_{2} = \frac{d}{v_{b}} = \frac{80}{10} = 8 \sec$
Difference $Δ t = t_{1} - t_{2} = 10 - 8 = 2 \sec$

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