Question

A man with some passengers in his boat, starts perpendicular to the flow of river $200m$ wide and flowing with $2m/s$. Speed of boat in still water is $4m/s$. When he reaches half the width of river, the passengers asked him that they want to reach the point just opposite to the end from where they have started. If the direction due which he must row to reach the required end is $\alpha =2ta{n}^{-1}\left(\frac{1}{x}\right)$. Find $x$.

Answer 1

Time taken by boat to reach from A to B is
$t=\frac{100}{v_y}=\frac{100}{4}=25s$
Also $AD=v_{x}t=2t=50m$
From B to C
Actual velocity of boat should be along BC. This actual velocity is found by resultant of $v_{rel}$ and $v_r$
$v_{rel}=v_b-v_r$
x component of actual velocity is
$v_x=v_r-v_{rel}sin\alpha =2-4sin\alpha$
y component of actual velocity is
$v_y=v_{rel}cos\alpha$
The time taken from B to C is
$t_o=\frac{BE}{v_y}$
$=\frac{100}{v_{rel}cos\alpha }=\frac{100}{4cos\alpha }$
$=\frac{25}{cos\alpha }$
Also
$EC=-v_{x}t$
$=-(2-4sin\alpha )t_o$
$=(4sin\alpha -2)\frac{25}{cos\alpha }$
But
$EC=AD=50m$
Thus
$EC=(4sin\alpha -2)\frac{25}{cos\alpha }$
$50=(4sin\alpha -2)\frac{25}{cos\alpha }$
$50cos\alpha =100sin\alpha -50$
$cos\alpha =2sin\alpha -1$
$1+cos\alpha =2sin\alpha$
$2co{s}^2\frac{\alpha }{2}=4sin\frac{\alpha }{2}cos\frac{\alpha }{2}$
$tan\frac{\alpha }{2}=\frac{1}{2}$
$\frac{\alpha }{2}=ta{n}^{-1}\frac{1}{2}$
$\alpha =2ta{n}^{-2}\frac{1}{2}$