Question

A boat moves relative to water with a velocity 'v' which is n times less than the river flow velocity u. At what angle to the north direction must the boat move to minimize drifting if the river flows from west to east?

• A. $\theta =si{n}^{-1}n$
• B. $\theta =si{n}^{-1}\frac{1}{n}$
• C. $\theta =co{s}^{-1}\frac{1}{n}$
• D. $noneofthese$

Answer 1

The correct option is B

$\theta =si{n}^{-1}\frac{1}{n}$

In this case, the velocity of boat is less than the river flow velocity. Hence boat cannot reach the point directly opposite to its starting point. i.e. drift can never be zero.) Suppose boat starts at an angle $\theta$ from the normal direction up stream as shown. Component of velocity of boat along the river, $v_x$ = $u-vsin\theta$

and velocity perpendicular to the river, $v_y$ = v cos$\theta$

Time taken to cross the river is t =$\frac{d}{v_y}=\frac{d}{vcos\theta }$

Drift x = ($v_x$)t = $(u-vsin\theta )\frac{d}{vcos\theta }=\frac{ud}{v}sec\theta -dtan\theta$

The drift x is minimum, when $\frac{dx}{d\theta }$= 0, or $\left(\frac{ud}{v}\right)(sec\theta .tan\theta )-dse{c}^2\theta =0or\frac{u}{v}sin\theta =1\Rightarrow sin\theta =\frac{v}{u}$

i.e., for minimum drift, the boat must move at an angle $\theta =si{n}^{-1}\left(\frac{v}{u}\right)=si{n}^{-1}\frac{1}{n}$ from normal

direction.