Question

A man can row a boat at a speed of $4km/hr$ in still water, if he crosses a river when the speed of the current is $2km/hr$. The width of the river is $4km$. Suppose $t_1$ is the time taken by the boat to reach the other end, directly opposite to the starting point. In the same conditions, $t_2$ is the minimum time taken to reach the other end of the river. Then the value of $4{\left(\frac{t_2}{t_1}\right)}^2$ is

Answer 1

Let the width of the river be $(AB=d)$

Let $\overrightarrow{V_{BR}},\overrightarrow{V_R}$ be the velocity of boat with respect to river and velocity of river respectively
Given: $\overrightarrow{V_{BR}}=4km/hr,\overrightarrow{V_R}=2km/hr,d=4km$For finding the value of $\theta$ the vertical component $(sin\theta$) of the boat with respect to river should be equal but opposite in direction
$\therefore$ $V_{BR}sin\theta =2$, $4sin\theta =2,\theta ={30}^o$
,now to find the time taken for the boat to reach the other end $t_1$ ,we take the horizontal component $(cos\theta$)
$\therefore$ $V_{BR}\cos \left(\theta \right)t_1=4,t_1=\frac{4}{V_{BR}\cos \left(\theta \right)}=\frac{4}{4\cos \left({30}^o\right)}=\frac{2}{\sqrt{3}}$
Now to find the minimum time taken to reach the other end of the river $\left(t_2\right)$, he need to row in such a way that his velocity with respect to river is horizontal

$\overrightarrow{V_{BG}}=V_{BR}\hat{i}+\overrightarrow{V_R}=\overrightarrow{V_{BG}}=V_{BR}\hat{i}+2\hat{j}=4\hat{i}+2\hat{j}$
$V_{BR}\times t_2=d,4\times t_2=4$
$t_2=1$
$\therefore$ $4{\left(\frac{t_2}{t_1}\right)}^2=3$

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