Starting from rest, a particle rotates in a circle of radius R = `sqrt 2 m` with an angular acceleration = `pi/4` rad/s2.The magnitude of average velocity of the particle over the time it rotates quarter circle is:

Options

A. 1.5 m/s
B. 2 m/s
C. 1 m/s
D. 1.25 m/s

Answer

To solve the problem, we need to find the average velocity of a particle rotating in a circle of radius \( R = \sqrt{2} \, \text{m} \) with an angular acceleration \( \alpha = \frac{\pi}{4} \, \text{rad/s}^2 \) over the time it takes to rotate a quarter circle.

### Step-by-Step Solution:

1. **Identify the Displacement**:
- The particle moves from point A to point B, covering a quarter of the circle. The displacement is the straight line distance between the initial and final points.
- For a quarter circle, the displacement can be calculated using the Pythagorean theorem. If both the initial and final points are at a distance \( R \) from the center, the displacement \( d \) is given by:
\[ d = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

2. **Substitute the Radius**:
- Given \( R = \sqrt{2} \, \text{m} \):
\[ d = \sqrt{2} \cdot \sqrt{2} = 2 \, \text{m} \]

3. **Calculate Angular Displacement**:
- The angular displacement \( \theta \) for a quarter circle is:
\[ \theta = \frac{\pi}{2} \, \text{rad} \]

4. **Use the Angular Motion Equation**:
- The equation for angular displacement when starting from rest is:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
- Since the initial angular velocity \( \omega_0 = 0 \):
\[ \frac{\pi}{2} = 0 + \frac{1}{2} \left(\frac{\pi}{4}\right) t^2 \]
- Simplifying this gives:
\[ \frac{\pi}{2} = \frac{\pi}{8} t^2 \]
- Canceling \( \pi \) from both sides:
\[ 2 = \frac{1}{8} t^2 \]
- Multiplying both sides by 8:
\[ t^2 = 16 \implies t = 4 \, \text{s} \]

5. **Calculate Average Velocity**:
- The average velocity \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\text{Displacement}}{\text{Time}} = \frac{d}{t} \]
- Substituting the values:
\[ v_{avg} = \frac{2 \, \text{m}}{4 \, \text{s}} = 0.5 \, \text{m/s} \]

### Final Answer:
The magnitude of the average velocity of the particle over the time it rotates a quarter circle is \( 0.5 \, \text{m/s} \).