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

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Radius of the circle, \( r = \frac{20}{\pi} \) meters
- Final velocity after the second revolution, \( v = 80 \) m/s
- Initial velocity, \( u = 0 \) m/s (since the motion has just begun)

### Step 2: Calculate the distance traveled in two revolutions
The distance \( s \) covered in two revolutions of a circle can be calculated using the formula:
\[ s = 2 \times \text{Circumference of the circle} \]
The circumference \( C \) of a circle is given by:
\[ C = 2\pi r \]
Substituting the value of \( r \):
\[ C = 2\pi \left(\frac{20}{\pi}\right) = 40 \text{ meters} \]
Thus, the distance for two revolutions is:
\[ s = 2 \times 40 = 80 \text{ meters} \]

### Step 3: Use the kinematic equation to find tangential acceleration
We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
- \( v = 80 \) m/s
- \( u = 0 \) m/s
- \( s = 80 \) m

The equation becomes:
\[ (80)^2 = (0)^2 + 2a \cdot 80 \]
This simplifies to:
\[ 6400 = 160a \]

### Step 4: Solve for tangential acceleration \( a \)
Rearranging the equation to solve for \( a \):
\[ a = \frac{6400}{160} = 40 \text{ m/s}^2 \]

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

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