A boat crosses a river from port A to port B, which are just on the opposite side. The speed of the water is `V_W` and that of boat is `V_B` relative to still water. Assume `V_B = 2V_W`. What is the time taken by the boat, if it has to cross the river directly on the AB line

Options

A. `(2D)/(V_B sqrt3)`
B. `(sqrt3 D)/(2V_B)`
C. `(D)/(V_B sqrt2)`
D. `(Dsqrt2)/(V_B)`

Explanation

To solve the problem of how long it takes for a boat to cross a river directly from port A to port B, we will follow these steps:

### Step 1: Understand the problem
We have a river with a width \( D \) (the distance between ports A and B), a boat with speed \( V_B \) relative to still water, and a current with speed \( V_W \). We know that \( V_B = 2V_W \).

### Step 2: Set up the scenario
The boat must cross the river directly from A to B. To do this, it must angle its path upstream to counteract the current of the river. We denote the angle at which the boat must head upstream as \( \theta \).

### Step 3: Analyze the velocities
- The component of the boat's velocity that acts against the current is \( V_B \sin \theta \).
- The component of the boat's velocity that acts across the river (toward B) is \( V_B \cos \theta \).

To ensure the boat moves directly across the river, the upstream component of the boat's speed must equal the downstream speed of the current:
\[ V_B \sin \theta = V_W \]

### Step 4: Substitute the relationship between \( V_B \) and \( V_W \)
Since \( V_B = 2V_W \), we can substitute this into the equation:
\[ 2V_W \sin \theta = V_W \]
Dividing both sides by \( V_W \) (assuming \( V_W \neq 0 \)):
\[ 2 \sin \theta = 1 \]
Thus,
\[ \sin \theta = \frac{1}{2} \]

### Step 5: Determine the angle \( \theta \)
From trigonometry, we know that:
\[ \sin \theta = \frac{1}{2} \implies \theta = 30^\circ \]

### Step 6: Calculate the effective speed across the river
Now, we need to find the effective speed of the boat across the river:
\[ V_B \cos \theta = 2V_W \cos 30^\circ \]
Using \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\[ V_B \cos \theta = 2V_W \cdot \frac{\sqrt{3}}{2} = V_W \sqrt{3} \]

### Step 7: Calculate the time taken to cross the river
The time \( T \) taken to cross the river can be calculated using the formula:
\[ T = \frac{\text{Distance}}{\text{Speed}} = \frac{D}{V_B \cos \theta} \]
Substituting the effective speed:
\[ T = \frac{D}{V_W \sqrt{3}} \]

### Step 8: Substitute \( V_W \) back in terms of \( V_B \)
Since \( V_W = \frac{V_B}{2} \):
\[ T = \frac{D}{\frac{V_B}{2} \sqrt{3}} = \frac{2D}{V_B \sqrt{3}} \]

### Final Answer
Thus, the time taken by the boat to cross the river directly from A to B is:
\[ T = \frac{2D}{V_B \sqrt{3}} \]