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

Answer: (A)

$\frac{2D}{v_{B}\sqrt{3}}$

Answer 2

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
$t=\frac{2 D}{v_{B} \sqrt{3}}$