A swimmer crosses the river along the line making an angle of `45^(@)` with the direction of flow. Velocity of the river is 5 m/s. Swimmer takes 6 seconds to cross the river of width 60 m. The velocity of the swimmer with respect to water will be :

Options

A. `10m//s`
B. 12m/s
C. `5sqrt(5)m//s`
D. `10sqrt(2)m//s`

Answer

To find the velocity of the swimmer with respect to water, we can follow these steps:

### Step 1: Identify the given information
- Width of the river (d) = 60 m
- Velocity of the river (Vw) = 5 m/s
- Time taken to cross the river (t) = 6 s
- Angle made with the direction of flow (θ) = 45°

### Step 2: Calculate the swimmer's effective velocity across the river
The swimmer crosses the river in 6 seconds, so we can calculate the effective velocity (Vx) of the swimmer across the river:
\[ V_x = \frac{d}{t} = \frac{60 \, \text{m}}{6 \, \text{s}} = 10 \, \text{m/s} \]

### Step 3: Relate swimmer's velocity components
Since the swimmer makes an angle of 45° with the direction of the river flow, we can express the swimmer's velocity (Vs) in terms of its components:
- The horizontal component (Vx) and vertical component (Vy) of the swimmer's velocity are equal due to the angle of 45°.
- Therefore, we have:
\[ V_x = V \cos(45°) \]
\[ V_y = V \sin(45°) \]

Since \( \cos(45°) = \sin(45°) = \frac{1}{\sqrt{2}} \), we can say:
\[ V_x = V_y = \frac{V}{\sqrt{2}} \]

### Step 4: Set up the equations
From Step 2, we know \( V_x = 10 \, \text{m/s} \). Therefore:
\[ \frac{V}{\sqrt{2}} = 10 \]
\[ V = 10\sqrt{2} \, \text{m/s} \]

### Step 5: Calculate the swimmer's velocity with respect to water
The swimmer's velocity with respect to water (Vs) can be expressed as:
\[ V_s = V_x \hat{i} + V_y \hat{j} \]
Where:
- \( V_x = 10 \, \text{m/s} \)
- \( V_y = 10 \, \text{m/s} \)

Thus:
\[ V_s = 10 \hat{i} + 10 \hat{j} \]

### Step 6: Adjust for the river's velocity
The velocity of the river (Vw) is in the negative j direction:
\[ V_w = 5 \hat{j} \]

The swimmer's velocity with respect to water is:
\[ V_{swimmer\_water} = V_s - V_w = (10 \hat{i} + 10 \hat{j}) - (0 \hat{i} + 5 \hat{j}) \]
\[ V_{swimmer\_water} = 10 \hat{i} + (10 - 5) \hat{j} = 10 \hat{i} + 5 \hat{j} \]

### Step 7: Calculate the magnitude of the swimmer's velocity with respect to water
The magnitude of the swimmer's velocity with respect to water can be calculated using the Pythagorean theorem:
\[ |V_{swimmer\_water}| = \sqrt{(10)^2 + (5)^2} \]
\[ |V_{swimmer\_water}| = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \, \text{m/s} \]

### Final Answer
The velocity of the swimmer with respect to water is \( 5\sqrt{5} \, \text{m/s} \).

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