A spot light `S` rotates in a horizontal plane with a constant angular velocity of `0.1 rad//s`. The spot of light `P` move along the wall at a disatnce `3 m`. What is the velocity of the spot `P` when `theta = 45^(@)` ?

Explanation

To solve the problem, we need to find the velocity of the spot of light \( P \) on the wall when the angle \( \theta \) is \( 45^\circ \). The spotlight rotates with a constant angular velocity \( \omega = 0.1 \, \text{rad/s} \) and the distance from the spotlight to the wall is \( D = 3 \, \text{m} \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
The spotlight \( S \) rotates in a horizontal plane, and the light spot \( P \) moves along the wall at a distance \( D = 3 \, \text{m} \). The angle \( \theta \) is the angle between the line from the spotlight to the wall and the vertical line.

2. **Finding the Radius \( r \)**:
The radius \( r \) from the spotlight to the point \( P \) on the wall can be calculated using the formula:
\[ r = \frac{D}{\cos(\theta)} \]
For \( \theta = 45^\circ \):
\[ r = \frac{3}{\cos(45^\circ)} = \frac{3}{\frac{1}{\sqrt{2}}} = 3\sqrt{2} \, \text{m} \]

3. **Calculating the Linear Velocity \( v \)**:
The linear velocity \( v \) of the spot of light \( P \) can be calculated using the formula:
\[ v = \omega \cdot r \cdot \sin(\theta) \]
Substituting the known values:
\[ v = 0.1 \, \text{rad/s} \cdot (3\sqrt{2}) \cdot \sin(45^\circ) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ v = 0.1 \cdot (3\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 0.1 \cdot 3 = 0.3 \, \text{m/s} \]

4. **Final Calculation**:
Therefore, the velocity of the spot \( P \) when \( \theta = 45^\circ \) is:
\[ v = 0.3 \, \text{m/s} \]

### Final Answer:
The velocity of the spot \( P \) when \( \theta = 45^\circ \) is \( 0.3 \, \text{m/s} \).