Question

A boat crosses a river from port $A$ to port $B$, which are just on the opposite sides. The speed of the water is $v_{W}$ and that of boat is $Vg$ relative to water. Assume $v_{B}=2 v_{W}$. What is the time taken by the boat, if it has to cross the river directly on the $A B$ line?

• A. $ \frac{2D}{v_{B}\sqrt{3}} $
• B. $ \frac{\sqrt{3}D}{2v_{B}} $
• C. $ \frac{D}{v_{B}\sqrt{2}} $
• D. $ \frac{D\sqrt{2}}{v_{B}} $

Answer 1

The Correct Option is A

Let the velocity of the boat, if it has to cross the river directly on water the line $A B$ be $v_{A}$ and the angle between and $v_{B}$ be $\theta$.
Then, from the figure $\sin \theta=\frac{v_{w}}{v_{B}}$
Given, $v_{B}=2 v_{w} $
$\therefore \sin \theta=\frac{v_{w}}{2 v_{w}}=\frac{1}{2} $
$\Rightarrow$ $\theta=30^{\circ}$
Now, time taken by the boat to cross the river directly from $A$ to $B$
$t=\frac{D}{v_{A}}=\frac{D}{v_{B} \cos \theta}$
$=\frac{D}{v_{B} \times \cos 30^{\circ}}$ or
$t=\frac{2 D}{v_{B} \sqrt{3}}$