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: Understanding the Concept:
To reach the point directly opposite, the boat is steered upstream at an angle \(\theta\) to the line AB so that the sideways velocity cancels the water flow.

Step 2: Condition for direct crossing:
\(V_B\sin\theta = V_W\). With \(V_B = 2V_W\), \(\sin\theta = \frac12\), so \(\theta = 30^{\circ}\).

Step 3: Speed across the river:
The component perpendicular to the banks is \(V_B\cos 30^{\circ} = V_B\cdot\frac{\sqrt3}{2}\).

Step 4: Time:
\[ t = \frac{D}{V_B\cos 30^{\circ}} = \frac{2D}{\sqrt3\,V_B} \]

Step 5: Why the other options are wrong.
Options (A) and (B) have \(\sqrt2\), which would arise if the angle were \(45^{\circ}\). Option (D) has \(\frac{\sqrt3 D}{2V_B}\), the reciprocal of the correct factor.

Final Answer:
The time is \(\frac{2D}{V_B\sqrt3}\), option (C). \[ \boxed{\frac{2D}{V_B\sqrt{3}}} \]