A particle moves along a circle if radius (`20 //pi`) m with constant tangential acceleration. If the velocity of the particle is ` 80 m//s` at the end of the second revolution after motion has begun the tangential acceleration is .

Options

A. `40 m//s^2`
B. ` 640 m //s^2`
C. `160 m//s^(2)`
D. `40 m//s^2`

Explanation

To solve the problem, we need to find the tangential acceleration of a particle moving along a circular path with a given radius and final velocity after completing two revolutions.

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the circle, \( r = \frac{20}{\pi} \) m
- Final velocity after two revolutions, \( v = 80 \) m/s
- The particle starts from rest, so initial velocity \( u = 0 \) m/s.

2. **Calculate the total angular displacement for two revolutions:**
- One complete revolution corresponds to an angular displacement of \( 2\pi \) radians.
- Therefore, for two revolutions, the angular displacement \( \theta = 2 \times 2\pi = 4\pi \) radians.

3. **Relate linear velocity to angular velocity:**
- The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
- Thus, the final angular velocity \( \omega_f \) can be expressed as:
\[ \omega_f = \frac{v}{r} \]

4. **Substitute the values to find \( \omega_f \):**
- Substitute \( v = 80 \) m/s and \( r = \frac{20}{\pi} \):
\[ \omega_f = \frac{80}{\frac{20}{\pi}} = 80 \times \frac{\pi}{20} = 4\pi \text{ rad/s} \]

5. **Use the equation of motion for angular motion:**
- The equation relating final angular velocity, initial angular velocity, angular acceleration, and angular displacement is:
\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]
- Here, \( \omega_i = 0 \) (initial angular velocity), \( \alpha \) is the angular acceleration, and \( \theta = 4\pi \).

6. **Substituting the known values into the equation:**
- Plugging in the values:
\[ (4\pi)^2 = 0 + 2\alpha (4\pi) \]
- Simplifying gives:
\[ 16\pi^2 = 8\pi \alpha \]

7. **Solve for angular acceleration \( \alpha \):**
- Dividing both sides by \( 8\pi \):
\[ \alpha = \frac{16\pi^2}{8\pi} = 2\pi \text{ rad/s}^2 \]

8. **Relate angular acceleration to tangential acceleration:**
- The tangential acceleration \( a_t \) is related to angular acceleration by:
\[ a_t = r \alpha \]
- Substituting \( r = \frac{20}{\pi} \) and \( \alpha = 2\pi \):
\[ a_t = \frac{20}{\pi} \times 2\pi = 40 \text{ m/s}^2 \]

### Final Answer:
The tangential acceleration \( a_t \) is \( 40 \, \text{m/s}^2 \).