A person crossing a road with a certain velocity due north sees a car moving towards east The relative velocity of the car w.r.t the person is `sqrt(2)` times that fo the velocity of the persons. The angle made by the relative velocity with the east is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to analyze the motion of the person and the car in terms of their velocities and the relative velocity between them.

### Step-by-Step Solution:

1. **Define the Velocities**:
- Let the velocity of the person (moving due north) be \( \vec{v_p} = u \hat{j} \).
- Let the velocity of the car (moving due east) be \( \vec{v_c} = v \hat{i} \).

2. **Relative Velocity**:
- The relative velocity of the car with respect to the person is given by:
\[ \vec{v_{cp}} = \vec{v_c} - \vec{v_p} = v \hat{i} - u \hat{j} \]

3. **Magnitude of Relative Velocity**:
- The magnitude of the relative velocity \( |\vec{v_{cp}}| \) can be calculated using the Pythagorean theorem:
\[ |\vec{v_{cp}}| = \sqrt{v^2 + u^2} \]

4. **Given Condition**:
- According to the problem, the magnitude of the relative velocity is \( \sqrt{2} \) times the velocity of the person:
\[ |\vec{v_{cp}}| = \sqrt{2} u \]
- Therefore, we have:
\[ \sqrt{v^2 + u^2} = \sqrt{2} u \]

5. **Square Both Sides**:
- Squaring both sides gives:
\[ v^2 + u^2 = 2u^2 \]
- Simplifying this, we find:
\[ v^2 = u^2 \]

6. **Relationship Between Velocities**:
- From \( v^2 = u^2 \), we conclude that:
\[ v = u \]

7. **Finding the Angle**:
- The relative velocity vector can now be expressed as:
\[ \vec{v_{cp}} = v \hat{i} - v \hat{j} = v(\hat{i} - \hat{j}) \]
- The angle \( \theta \) that this vector makes with the east (the positive x-axis) can be found using the tangent function:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{v}{v} = 1 \]
- Therefore, the angle \( \theta \) is:
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Final Answer:
The angle made by the relative velocity with the east is \( 45^\circ \).

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