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 pi "m/s"^(-2)`
C. `160 pi"m/s"^(-2)`
D. `40 pi "m/s"^(-2)`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the problem
We are given a particle moving in a circular path with a radius of \( \frac{20}{\pi} \) meters. The particle has a constant tangential acceleration, and we need to find this acceleration after the particle has completed two revolutions, given that its final velocity is \( 80 \, \text{m/s} \).

### Step 2: Calculate the distance covered in two revolutions
The distance \( s \) covered in two revolutions can be calculated using the formula for the circumference of a circle:
\[ \text{Circumference} = 2\pi r \]
For two revolutions, the distance \( s \) is:
\[ s = 2 \times \text{Circumference} = 2 \times 2\pi r = 4\pi r \]
Substituting the radius \( r = \frac{20}{\pi} \):
\[ s = 4\pi \left(\frac{20}{\pi}\right) = 4 \times 20 = 80 \, \text{m} \]

### Step 3: Use the equation of motion to find tangential acceleration
We can use the equation of motion that relates initial velocity \( u \), final velocity \( v \), acceleration \( a \), and distance \( s \):
\[ v^2 - u^2 = 2as \]
Here, the initial velocity \( u = 0 \, \text{m/s} \) (since the motion starts from rest), the final velocity \( v = 80 \, \text{m/s} \), and the distance \( s = 80 \, \text{m} \).

Substituting the known values into the equation:
\[ (80)^2 - (0)^2 = 2a(80) \]
This simplifies to:
\[ 6400 = 160a \]

### Step 4: Solve for \( a \)
Now, we can solve for \( a \):
\[ a = \frac{6400}{160} = 40 \, \text{m/s}^2 \]

### Conclusion
The tangential acceleration of the particle is \( 40 \, \text{m/s}^2 \).

---