A particle is moving along a circle of radius `R` with a uniform speed `v`. At `t = 0` , the particle is moving along the east. Find the average acceleration (magnitude and direction) in `1//4 th` revolution.

Options

A. `(V^(2))/(sqrt(2)piR)S-W`
B. `(2sqrt(2)v^(2))/(piR)S-W`
C. `(2sqrt(2)v^(2))/(piR)S-E`
D. `0`

Explanation

To find the average acceleration of a particle moving in a circular path of radius \( R \) with uniform speed \( v \) during one-fourth of a revolution, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle starts at the east direction (let's denote this as the positive x-direction) at \( t = 0 \).
- After one-fourth of a revolution (which corresponds to \( \frac{\pi}{2} \) radians), the particle will be at the north direction (positive y-direction).

2. **Determine Initial and Final Velocity**:
- The initial velocity \( \vec{V_i} \) at \( t = 0 \) is directed east (positive x-direction):
\[ \vec{V_i} = v \hat{i} \]
- The final velocity \( \vec{V_f} \) after one-fourth of the revolution is directed north (positive y-direction):
\[ \vec{V_f} = v \hat{j} \]

3. **Calculate Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} = v \hat{j} - v \hat{i} = -v \hat{i} + v \hat{j} \]

4. **Calculate the Time Taken for One-Fourth Revolution**:
- The arc length \( s \) for one-fourth of the revolution is:
\[ s = R \theta = R \left(\frac{\pi}{2}\right) = \frac{\pi R}{2} \]
- The time \( t \) taken to cover this arc length at speed \( v \) is:
\[ t = \frac{s}{v} = \frac{\frac{\pi R}{2}}{v} = \frac{\pi R}{2v} \]

5. **Calculate Average Acceleration**:
- Average acceleration \( \vec{a} \) is defined as the change in velocity divided by the time taken:
\[ \vec{a} = \frac{\Delta \vec{V}}{t} = \frac{-v \hat{i} + v \hat{j}}{\frac{\pi R}{2v}} = \frac{-2v^2 \hat{i} + 2v^2 \hat{j}}{\pi R} \]
- Simplifying gives:
\[ \vec{a} = \frac{2v}{\pi R}(-\hat{i} + \hat{j}) \]

6. **Magnitude of Average Acceleration**:
- To find the magnitude of the average acceleration:
\[ |\vec{a}| = \frac{2v}{\pi R} \sqrt{(-1)^2 + (1)^2} = \frac{2v}{\pi R} \sqrt{2} = \frac{2\sqrt{2}v}{\pi R} \]

7. **Direction of Average Acceleration**:
- The direction of the average acceleration is given by the vector \( -\hat{i} + \hat{j} \), which points southwest.

### Final Answer:
- The magnitude of the average acceleration is \( \frac{2\sqrt{2}v}{\pi R} \).
- The direction of the average acceleration is towards the southwest.