Question

The ratio of the distance carried away by the water current downstream in crossing a river by a person making same angle with downstream and upstream are, respectively,

2 : 1

. Show that the ratio of the speed of the person to water current cannot be less than

\frac{1}{3}

.

Accepted Answer

Case I: Motion of the person making an angle (say $α$ ) with the downstream.

The time taken to cross the river, $t = \frac{d}{v \sin α}$ .

The distance carried away downstream in the same time,

$x_{1} = (u + v \cos α) \frac{d}{v \sin α}$ .

Case II: Motion of the person, making an angle (say $α$ ) with the upstream.

The time taken to cross the river, $t = \frac{d}{v \sin α}$ .

The distance carried away downstream in the same time, $x_{2} = (u - v \cos α) \frac{d}{v \sin α}$ .

Given,

$\frac{x_{1}}{x_{2}} = \frac{2}{1}$

$\Rightarrow \frac{(u + v \cos α) \frac{d}{v \sin α}}{(u - v \cos α) \frac{d}{v \sin α}} = \frac{2}{1} \Rightarrow \frac{(u + v \cos α)}{(u - v \cos α)} = \frac{2}{1}$

$\Rightarrow u + v \cos α = 2 u - 2 v \cos α$
$\Rightarrow 3 v \cos α = u \Rightarrow \cos α = \frac{u}{3 v}$ .

But, $\cos α \leq 1$

$\Rightarrow \frac{u}{3 v} \leq 1 \Rightarrow \frac{v}{u} \geq \frac{1}{3}$

Hence, $\frac{v}{u}$ cannot be less than $\frac{1}{3}$ .

B M Sharma Solutions for Chapter: Kinematics II, Exercise 56: ILLUSTRATION 5.52

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 56: ILLUSTRATION 5.52

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