A man is swimming in a lake in a direction `30^(@)` east of north with a speed of 5 km/hour and a cyclist is going on the road along the lake shore towards east at a speed of 10 km/hour. The direction of the swimmer , relative to the cyclist is

Options

A. `30^(@)` west of north
B. west-north
C. `60^(@)` west of north
D. none of these

Answer

To solve the problem of finding the direction of the swimmer relative to the cyclist, we can follow these steps:

### Step 1: Define the velocities of the swimmer and the cyclist
- The swimmer's velocity (\( \vec{V_s} \)) is 5 km/h at an angle of 30 degrees east of north.
- The cyclist's velocity (\( \vec{V_c} \)) is 10 km/h towards the east.

### Step 2: Break down the swimmer's velocity into components
The swimmer's velocity can be resolved into its northward and eastward components using trigonometric functions:
- Northward component (\( V_{s_y} \)):
\[ V_{s_y} = 5 \cos(30^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.33 \text{ km/h} \]
- Eastward component (\( V_{s_x} \)):
\[ V_{s_x} = 5 \sin(30^\circ) = 5 \times \frac{1}{2} = 2.5 \text{ km/h} \]

### Step 3: Define the cyclist's velocity components
The cyclist is moving only in the east direction, so:
- Northward component (\( V_{c_y} \)):
\[ V_{c_y} = 0 \text{ km/h} \]
- Eastward component (\( V_{c_x} \)):
\[ V_{c_x} = 10 \text{ km/h} \]

### Step 4: Calculate the relative velocity of the swimmer with respect to the cyclist
The relative velocity (\( \vec{V_{rel}} \)) of the swimmer with respect to the cyclist is given by:
\[ \vec{V_{rel}} = \vec{V_s} - \vec{V_c} \]
This means we subtract the cyclist's velocity components from the swimmer's:
- Northward component of relative velocity:
\[ V_{rel_y} = V_{s_y} - V_{c_y} = 4.33 - 0 = 4.33 \text{ km/h} \]
- Eastward component of relative velocity:
\[ V_{rel_x} = V_{s_x} - V_{c_x} = 2.5 - 10 = -7.5 \text{ km/h} \]

### Step 5: Determine the direction of the relative velocity
To find the direction of the relative velocity, we can use the arctangent function:
\[ \tan(\theta) = \frac{V_{rel_y}}{V_{rel_x}} = \frac{4.33}{-7.5} \]
Calculating the angle:
\[ \theta = \tan^{-1}\left(\frac{4.33}{-7.5}\right) \approx -30.5^\circ \]
This angle is measured from the negative x-axis (westward), which means the direction is 30.5 degrees north of west.

### Step 6: Finalize the direction
Since the angle is negative, we can express the direction as:
- 30.5 degrees north of west or equivalently, 90 - 30.5 = 59.5 degrees south of east.

### Conclusion
The direction of the swimmer relative to the cyclist is approximately 30.5 degrees north of west.

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