A man can swim with a speed v relative to water . There is a river of width d which is flowing at a speed u . A is the point on river bank from where he starts and B is point directly opposite to A on the other side of the river .

Options

A. Man can directly reach point B in time d/l.
B. Man can cross the river in minimum time d/v
C. Man can reach point B directly in time `(d)/(sqrt(v^(2)-u^(2)))`
D. If `u gt v` then man cannot reach the point B directly .

Explanation

To solve the problem step by step, we will analyze the situation of a man swimming across a river with a current. We will derive the necessary equations to determine the time taken to cross the river and the conditions for minimum drift and minimum time.

### Step 1: Understand the Problem
We have a river of width \( d \) and a current flowing at speed \( u \). The swimmer has a speed \( v \) relative to the water. The goal is to find the time taken to swim directly across the river to point B, which is directly opposite point A.

### Step 2: Determine the Velocity Components
To swim directly across the river with no drift, the swimmer must angle his swim against the current. We can break down the swimmer's velocity into two components:
- The component perpendicular to the river (across the river) is \( v \sin(\theta) \).
- The component parallel to the river (against the current) is \( v \cos(\theta) \).

### Step 3: Set Up the Equation for Minimum Drift
For the swimmer to reach point B directly across from point A, the downstream drift caused by the river's current must be zero. This means the swimmer's upstream component must equal the river's speed:
\[ v \cos(\theta) = u \]
From this, we can express \( \cos(\theta) \):
\[ \cos(\theta) = \frac{u}{v} \]

### Step 4: Calculate the Time Taken to Cross the River
The time taken to cross the river can be calculated using the width of the river and the swimmer's effective speed across the river:
\[ t = \frac{d}{v \sin(\theta)} \]
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can find \( \sin(\theta) \):
\[ \sin(\theta) = \sqrt{1 - \left(\frac{u}{v}\right)^2} = \frac{\sqrt{v^2 - u^2}}{v} \]
Substituting this back into the time equation gives:
\[ t = \frac{d}{v \cdot \frac{\sqrt{v^2 - u^2}}{v}} = \frac{d}{\sqrt{v^2 - u^2}} \]

### Step 5: Analyze Special Cases
1. **Minimum Drift**: If \( u < v \), the swimmer can reach point B directly with no drift.
2. **Minimum Time**: If the swimmer swims directly across without considering the current, the time taken is:
\[ t = \frac{d}{v} \]
3. **If \( u > v \)**: The swimmer cannot reach point B directly, as the current is too strong.

### Conclusion
The answers to the options provided in the question are:
- Option 2: Time taken to reach point B directly is \( \frac{d}{\sqrt{v^2 - u^2}} \) (correct).
- Option 3: Time taken to reach point B directly is \( \frac{d}{v} \) (correct for minimum time).
- Option 4: If \( u > v \), the swimmer cannot reach point B directly (correct).