Question

A man can row a boat

4 km h^{- 1}

in still water. He is crossing a river where the current is

2 km h^{- 1}

.
(a) In what direction will his boat be headed if he wants to reach a point on the other bank, directly opposite to starting point?
(b) If width of the river is

4 km

, how long will it take him to cross the river with the condition in part

a

?
(c) In what direction should he head the boat if he wants to cross river in shortest time. Find the time taken by him to cross the river? Also find the location of the point where he lands on the other side of the river.

Accepted Answer

We know ${\vec{v}}_{man,water}$ , the velocity of the boat relative to the river and ${\vec{v}}_{water}$ , the velocity of the river water relative to the earth. What we must find is ${\vec{v}}_{man}$ , the velocity of the boat/man relative to the earth. The relationship between these three quantities is ${\vec{v}}_{man} = {\vec{v}}_{man,water} + {\vec{v}}_{water}$ . The terms in the equation must be manipulated as vector quantities; the vectors are shown in the figure. The angle at which man should swim is order to reach just opposite end

$sinθ = \frac{|v_{water}|}{|v_{man,water}|} = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = 30 °$

Hence, to reach the point directly opposite to starting point, he should head the boat an angle $β = (90 ° + 30 °) = 120 °$ with river flow.

(b) Use the Pythagorean theorem to find the net velocity of man.
$v_{m} = \sqrt{v^{2} - u^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{12} = 2 \sqrt{3} km h^{- 1}$
Time taken to cross the river
$t = \frac{d}{v_{man}} = \frac{4}{2 \sqrt{3}} h = \frac{2}{\sqrt{3}} h .$

(c) If man swims across the river, the water current pushes the man down the river. The man will not be able to move directly across the river, but will end up downstream with drift. The analysis now involves the new triangle shown in the figure.

Time taken to cross the river, $t_{\min} = \frac{d}{v_{man,water}} = \frac{4}{4} = 1 h$
Drifting in this case $= v_{water} \times t_{\min} = 2 \times 1 = 2 km$

Hence, man will cross the river in shortest time of $1 h$ and will end up $2 km$ downstream on the other side of river.

B M Sharma Solutions for Chapter: Kinematics II, Exercise 53: ILLUSTRATION 5.49

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 53: ILLUSTRATION 5.49

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 53: ILLUSTRATION 5.49 with Hints & Solutions

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