A man swims at a speed of 5 km/hour. He wants to cross a canal, 120 meters wide, in a direction perpendicular to the direction of flow. If the canal flows at 4 km/hours , the direction and the time taken by the man to cross the canal are

Options

A. `tan^(-1) "" (3)/(4), 2 . 4 m `
B. `pi - tan^(-1)"" (3)/(4), 144 sec`
C. `tan^(-1) "" (1)/(2) , 100 sec`
D. none

Answer

To solve the problem, we need to analyze the situation using relative velocity concepts. The man swims across a canal that is flowing, and we need to determine both the direction he swims and the time it takes him to cross the canal.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Speed of the man swimming, \( v_m = 5 \) km/h
- Speed of the canal (current), \( v_c = 4 \) km/h
- Width of the canal, \( d = 120 \) meters = 0.12 km (since 1 km = 1000 m)

2. **Determine the Effective Velocity:**
The man swims perpendicular to the flow of the canal. However, the current will affect his actual path. To find the resultant velocity, we need to consider both the man’s swimming speed and the speed of the current.

The man’s swimming velocity is directed straight across the canal, while the current flows downstream. We can represent these velocities as vectors:
- The man’s velocity vector \( \vec{v_m} \) is \( 5 \) km/h across the canal.
- The current’s velocity vector \( \vec{v_c} \) is \( 4 \) km/h downstream.

3. **Calculate the Resultant Velocity:**
To find the resultant velocity \( \vec{v_r} \), we can use the Pythagorean theorem since the two velocities are perpendicular to each other:
\[ v_r = \sqrt{v_m^2 + v_c^2} = \sqrt{(5)^2 + (4)^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4 \text{ km/h} \]

4. **Determine the Angle of Swimming:**
The angle \( \theta \) at which the man swims can be found using the tangent function:
\[ \tan(\theta) = \frac{v_c}{v_m} = \frac{4}{5} \]
Therefore,
\[ \theta = \tan^{-1}\left(\frac{4}{5}\right) \approx 38.66^\circ \]

5. **Calculate the Time Taken to Cross the Canal:**
The time \( t \) taken to cross the canal can be calculated using the width of the canal and the man’s swimming speed:
\[ t = \frac{d}{v_m} = \frac{0.12 \text{ km}}{5 \text{ km/h}} = 0.024 \text{ hours} = 0.024 \times 60 \text{ minutes} = 1.44 \text{ minutes} \]

### Final Results:
- The direction of swimming is approximately \( 38.66^\circ \) upstream from the perpendicular.
- The time taken to cross the canal is approximately \( 1.44 \) minutes.