A swimmer crosses the river along the line making an angle of `45^@` with the direction of flow. Velocity of the river water is `5(m)/(s)`. Swimmer takes 12 seconds to cross the river of width 60 m. The velocity of the swimmer with respect to water will be:

Options

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

Explanation

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

### Step 1: Understand the problem
The swimmer is crossing a river that is 60 m wide, making an angle of 45 degrees with the direction of the river flow. The velocity of the river is given as 5 m/s. The swimmer takes 12 seconds to cross the river.

### Step 2: Calculate the swimmer's velocity component across the river
The width of the river (d) is 60 m, and the time taken (t) is 12 seconds. The velocity of the swimmer in the direction perpendicular to the flow of the river can be calculated using the formula:
\[ \text{Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{d}{t} \]
Substituting the values:
\[ \text{Velocity}_{\text{perpendicular}} = \frac{60 \, \text{m}}{12 \, \text{s}} = 5 \, \text{m/s} \]

### Step 3: Relate the swimmer's velocity to the angle
Since the swimmer is making an angle of 45 degrees with the direction of the flow, we can express the swimmer's velocity (Vm) in terms of its components:
- The component of the swimmer's velocity perpendicular to the flow (across the river) is:
\[ V_m \sin(45^\circ) = V_m \cdot \frac{1}{\sqrt{2}} \]
- The component of the swimmer's velocity in the direction of the flow is:
\[ V_m \cos(45^\circ) = V_m \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Set up the equation for the perpendicular component
Since we know the perpendicular component of the swimmer's velocity must equal the calculated velocity:
\[ V_m \cdot \frac{1}{\sqrt{2}} = 5 \, \text{m/s} \]
Now, we can solve for \( V_m \):
\[ V_m = 5 \sqrt{2} \, \text{m/s} \]

### Step 5: Calculate the swimmer's velocity with respect to the water
To find the velocity of the swimmer with respect to the water, we need to consider both the swimmer's velocity and the river's velocity. The resultant velocity can be found using the Pythagorean theorem:
\[ V_{\text{relative}} = \sqrt{(V_m \cos(45^\circ) + V_r)^2 + (V_m \sin(45^\circ))^2} \]
Where \( V_r = 5 \, \text{m/s} \) (velocity of the river).

Substituting the values:
\[ V_{\text{relative}} = \sqrt{(5 \sqrt{2} \cdot \frac{1}{\sqrt{2}} + 5)^2 + (5 \sqrt{2} \cdot \frac{1}{\sqrt{2}})^2} \]
This simplifies to:
\[ V_{\text{relative}} = \sqrt{(5 + 5)^2 + (5)^2} = \sqrt{(10)^2 + (5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5 \sqrt{5} \, \text{m/s} \]

### Step 6: Conclusion
Thus, the velocity of the swimmer with respect to the water is:
\[ V_{\text{relative}} = 5 \, \text{m/s} \]