A particle moves in circle of radius R with a constant speed v. Then, find the magnitude of average
acceleration during a time interval `(pi R)/(2v)`.

Options

A. `3sqrt(2)V^(2)// pi R `
B. `2sqrt(2)V^(2)// pi R `
C. `3V^(2)// pi R `
D. `2V^(2)// pi R `

Answer

To find the magnitude of the average acceleration of a particle moving in a circle of radius \( R \) with a constant speed \( v \) during a time interval of \( \frac{\pi R}{2v} \), we can follow these steps:

### Step 1: Determine the distance traveled
The distance \( d \) traveled by the particle in the given time interval can be calculated using the formula:
\[ d = v \cdot t \]
Substituting the time interval \( t = \frac{\pi R}{2v} \):
\[ d = v \cdot \left(\frac{\pi R}{2v}\right) = \frac{\pi R}{2} \]

### Step 2: Find the angle covered
The distance \( \frac{\pi R}{2} \) corresponds to an angle \( \theta \) in radians. The relationship between the arc length \( s \), radius \( R \), and angle \( \theta \) is given by:
\[ s = R \theta \]
Thus, we can find \( \theta \):
\[ \frac{\pi R}{2} = R \theta \implies \theta = \frac{\pi}{2} \text{ radians} \]
This means the particle moves through a quarter of a circle.

### Step 3: Determine the initial and final velocity vectors
Assuming the particle starts at point A (0, R) and moves to point B (R, 0) in the quarter circle:
- **Initial velocity \( \vec{v_i} \)** at point A (0, R) is directed tangentially to the circle, which is horizontally to the right:
\[ \vec{v_i} = (v, 0) \]
- **Final velocity \( \vec{v_f} \)** at point B (R, 0) is directed tangentially to the circle, which is vertically downwards:
\[ \vec{v_f} = (0, -v) \]

### Step 4: Calculate the change in velocity
The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = (0, -v) - (v, 0) = (-v, -v) \]

### Step 5: Find the magnitude of the change in velocity
The magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{2v^2} = v\sqrt{2} \]

### Step 6: Calculate the average acceleration
The average acceleration \( \vec{a_{avg}} \) is defined as the change in velocity divided by the time interval:
\[ \vec{a_{avg}} = \frac{\Delta \vec{v}}{t} = \frac{v\sqrt{2}}{\frac{\pi R}{2v}} = \frac{2v^2\sqrt{2}}{\pi R} \]
Thus, the magnitude of the average acceleration is:
\[ |\vec{a_{avg}}| = \frac{2\sqrt{2}v^2}{\pi R} \]

### Final Answer
The magnitude of the average acceleration during the time interval \( \frac{\pi R}{2v} \) is:
\[ |\vec{a_{avg}}| = \frac{2\sqrt{2}v^2}{\pi R} \]

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