Question

A man can swim at a rate of

5 km h^{- 1}

in still water. A

1 km

wide river flows at the rate of

3 km h^{- 1}

. The man wishes to swim across the river directly opposite to the starting point.
(a) Along which direction must the man swim?
(b) What should be his resultant velocity?
(c) How much time will he take to cross the river?

Accepted Answer

Given, the velocity of man with respect to river water, $v = 5 km h^{- 1}$ ,

width of river, $d = 1 km$ ,
velocity of river water, $u = 3 km h^{- 1}$ .

(a) The velocity of man with respect to river water, $v = 5 km h^{- 1}$ . This is greater than the river flow velocity. Therefore, he can cross the river directly (along the shortest path or no drift condition from flow velocity). The angle of swim,

$θ = \frac{π}{2} + \sin^{- 1} (\frac{u}{v}) = 90 ° + \sin^{- 1} (\frac{u}{v})$

$= 90 ° + \sin^{- 1} (\frac{3}{5}) = 90 ° + 37 ° = 127 °$

w.r.t. the river flow.

or $37 °$ w.r.t. perpendicular in upstream direction.

(b) The resultant velocity or velocity of man will be,
$v_{m} = \sqrt{v^{2} - u^{2}} = \sqrt{5^{2} - 3^{2}} = 4 km h^{- 1}$

in the direction perpendicular to the river flow.

(c) The time taken to cross the river,
$t = \frac{d}{\sqrt{v^{2} - u^{2}}} = \frac{1 km}{4 km h^{- 1}} = \frac{1}{4} h = 15 \min$ .

B M Sharma Solutions for Chapter: Kinematics II, Exercise 55: ILLUSTRATION 5.51

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 55: ILLUSTRATION 5.51

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