Question

A person reaches a point directly opposite on the other bank of a river. The velocity of the water in the river is $4{ms}^{-1}$ and the velocity of the person in still water is $5{ms}^{-1}$. If the width of the river is $84.6m$, time taken to cross the river in seconds is:

• A. $9.4$
• B. $2$
• C. $84.6$
• D. $28.2$

Answer 1

The correct option is D $28.2$

Let the angle between the path traversed by person and the direction in which he tries to swim with respect to river be $\theta$.

Since the net path traveled is a straight line directly towards the opposite end, the component of net velocity does not exist in the direction of river's flow.

Thus $vsin\theta =v_{river}$

$⟹sin\theta =\frac{4}{5}$

Thus net velocity of man in the direction of crossing=$vcos\theta =5m/s\times \frac{3}{5}=3m/s$

Thus total time taken to cross the river=$\frac{84.6m}{3m/s}=28.2s$