A particle 'A' moves along a circle of radius R = 50 cm, so that its radius vetor 'r' relative to the point O rotates with the constant angular velocity `omega = 0.4 rad//s`. Then :

Options

A. linear velocity of particle is `0.2 m//s`
B. linear velocity of particle is `0.4 m//s`
C. magnitude of net acceleration is `0.08 m//s^(2)`
D. acceleration of particle is zero

Answer

To solve the problem, we need to find the linear velocity and the net acceleration of a particle moving in a circular path with a given radius and angular velocity.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the circle, \( R = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Angular velocity, \( \omega = 0.4 \, \text{rad/s} \)

2. **Calculate the Linear Velocity:**
The linear velocity \( v \) of a particle moving in a circle can be calculated using the formula:
\[ v = R \cdot \omega \]
Substituting the values:
\[ v = 0.5 \, \text{m} \cdot 0.4 \, \text{rad/s} = 0.2 \, \text{m/s} \]

3. **Determine the Angular Velocity for the Radius Vector:**
Since the radius vector \( r \) rotates with a constant angular velocity \( \omega \), we can consider that the effective angular velocity for the linear motion is \( \omega_{eff} = 2\omega \) (as derived from the properties of circular motion).

4. **Calculate the Effective Linear Velocity:**
Using the effective angular velocity:
\[ v_{eff} = R \cdot \omega_{eff} = R \cdot (2\omega) \]
Substituting the values:
\[ v_{eff} = 0.5 \, \text{m} \cdot (2 \cdot 0.4 \, \text{rad/s}) = 0.5 \, \text{m} \cdot 0.8 \, \text{rad/s} = 0.4 \, \text{m/s} \]

5. **Calculate the Net Acceleration:**
The net acceleration \( a \) of a particle moving in a circle can be calculated using the formula:
\[ a = \omega^2 \cdot R \]
Substituting the values:
\[ a = (0.4 \, \text{rad/s})^2 \cdot 0.5 \, \text{m} = 0.16 \cdot 0.5 = 0.08 \, \text{m/s}^2 \]

### Final Answers:
- The linear velocity of the particle is \( 0.4 \, \text{m/s} \).
- The magnitude of the net acceleration is \( 0.08 \, \text{m/s}^2 \).