A particle starts moving from rest at `t=0` with a tangential acceleration of constant magnitude of `pi m//s^(2)` along a circle of radius 6 m. The value of average acceleration, average velocity and average speed during the first `2 sqrt(3)` s of motion, are respectively :

Options

A. `3sqrt(2)m//s^(2),pi m//s,pisqrt(3)m//s`
B. `pi m//s^(2),2 sqrt(3) m//s, pi sqrt(3) m//s`
C. `pi sqrt(3) m//s^(2),2 sqrt(3) m//s, pi m//s`
D. None of the above

Answer

To solve the problem, we need to find the average acceleration, average velocity, and average speed of a particle moving in circular motion under constant tangential acceleration. Here are the steps to arrive at the solution:

### Step 1: Calculate Average Acceleration

1. **Given Data**:
- Tangential acceleration \( a_t = \pi \, \text{m/s}^2 \)
- Time \( t = 2\sqrt{3} \, \text{s} \)
- Initial velocity \( u = 0 \) (starts from rest)

2. **Final Velocity Calculation**:
Using the formula for final velocity under constant acceleration:
\[ v = u + a_t t \]
Substituting the values:
\[ v = 0 + \pi \cdot (2\sqrt{3}) = 2\pi\sqrt{3} \, \text{m/s} \]

3. **Average Acceleration Calculation**:
Average acceleration \( a_{avg} \) is defined as the change in velocity over time:
\[ a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v - u}{t} = \frac{2\pi\sqrt{3} - 0}{2\sqrt{3}} = \pi \, \text{m/s}^2 \]

### Step 2: Calculate Average Velocity

1. **Angular Displacement Calculation**:
The angular acceleration \( \alpha \) is given by:
\[ \alpha = \frac{a_t}{r} = \frac{\pi}{6} \, \text{rad/s}^2 \]
The angular displacement \( \theta \) can be calculated using:
\[ \theta = \frac{1}{2} \alpha t^2 = \frac{1}{2} \cdot \frac{\pi}{6} \cdot (2\sqrt{3})^2 = \frac{1}{2} \cdot \frac{\pi}{6} \cdot 12 = \pi \, \text{rad} \]

2. **Linear Displacement Calculation**:
The linear displacement \( s \) for a circular path can be calculated as:
\[ s = r \theta = 6 \cdot \pi = 6\pi \, \text{m} \]

3. **Average Velocity Calculation**:
Average velocity \( v_{avg} \) is defined as total displacement divided by total time:
\[ v_{avg} = \frac{s}{t} = \frac{6\pi}{2\sqrt{3}} = \frac{3\pi}{\sqrt{3}} = \sqrt{3}\pi \, \text{m/s} \]

### Step 3: Calculate Average Speed

1. **Total Distance Calculation**:
The total distance traveled during the time \( t \) can be calculated using:
\[ d = \frac{1}{2} a_t t^2 = \frac{1}{2} \cdot \pi \cdot (2\sqrt{3})^2 = \frac{1}{2} \cdot \pi \cdot 12 = 6\pi \, \text{m} \]

2. **Average Speed Calculation**:
Average speed is defined as total distance divided by total time:
\[ v_{avg, speed} = \frac{d}{t} = \frac{6\pi}{2\sqrt{3}} = \sqrt{3}\pi \, \text{m/s} \]

### Final Results

- Average acceleration: \( \pi \, \text{m/s}^2 \)
- Average velocity: \( \sqrt{3}\pi \, \text{m/s} \)
- Average speed: \( \sqrt{3}\pi \, \text{m/s} \)