Question

A boat crosses a river from one bank A to another bank B which is opposite. The distance between them is D. The speed of water is \(V_W\) and that of boat relative to water is \(V_B\). If \(V_B = 2V_W\), the time taken by the boat to cross the river directly along AB is \((sin30^{\circ} = \frac{1}{2})\), \((cos30^{\circ} = \frac{\sqrt{3}}{2})\)

• A. \(\frac{D}{V_B\sqrt{2}}\)
• B. \(\frac{D\sqrt{2}}{V_B}\)
• C. \(\frac{2D}{V_B\sqrt{3}}\)
• D. \(\frac{\sqrt{3}D}{2V_B}\)

Hint:

To go straight across, the upstream component of boat velocity must cancel the river current.

Answer 1

The Correct Option is C

Step 1: Use velocity addition:
The resultant velocity of the boat is along AB, with magnitude $\sqrt{V_B^2 - V_W^2}$, because the water velocity is perpendicular to this resultant.

Step 2: Insert V_W = V_B/2:
$\sqrt{V_B^2 - V_B^2/4} = V_B\sqrt{3/4} = \frac{\sqrt3}{2}V_B$.

Step 3: Time:
$t = D \big/ \left(\frac{\sqrt3}{2}V_B\right) = \frac{2D}{\sqrt3V_B}$.

Final Answer:
The time is 2D/(V_B root 3), option (C). \[ \boxed{\frac{2D}{V_B\sqrt{3}}} \]