A particle moves in a circle of radius 5 cm with constant speed and time period `0.2pis`. The acceleration of the particle is

Options

A. `25m//s^(2)`
B. `36m//s^(2)`
C. `5m//s^(2)`
D. `15m//s^(2)`

Explanation

To find the acceleration of a particle moving in a circle with constant speed, we can follow these steps:

### Step 1: Identify the Given Values
- Radius of the circle, \( R = 5 \, \text{cm} = 0.05 \, \text{m} \) (convert to meters for standard SI units)
- Time period, \( T = 0.25 \, \text{s} \)

### Step 2: Calculate the Speed of the Particle
The speed \( V \) of the particle can be calculated using the formula for the circumference of a circle and the time period:
\[ \text{Circumference} = 2\pi R \]
\[ V = \frac{\text{Circumference}}{T} = \frac{2\pi R}{T} \]
Substituting the values:
\[ V = \frac{2\pi \times 0.05}{0.25} \]
Calculating this gives:
\[ V = \frac{0.1\pi}{0.25} = 0.4\pi \, \text{m/s} \]

### Step 3: Calculate the Centripetal Acceleration
Centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{V^2}{R} \]
Substituting \( V = 0.4\pi \, \text{m/s} \) and \( R = 0.05 \, \text{m} \):
\[ a_c = \frac{(0.4\pi)^2}{0.05} \]
Calculating \( (0.4\pi)^2 \):
\[ (0.4\pi)^2 = 0.16\pi^2 \]
Now substituting this back:
\[ a_c = \frac{0.16\pi^2}{0.05} = 3.2\pi^2 \, \text{m/s}^2 \]

### Step 4: Final Calculation
Using the approximate value of \( \pi \approx 3.14 \):
\[ a_c \approx 3.2 \times (3.14)^2 \approx 3.2 \times 9.8596 \approx 31.55 \, \text{m/s}^2 \]

Thus, the acceleration of the particle is approximately \( 31.55 \, \text{m/s}^2 \).

### Summary of Steps:
1. Identify the radius and time period.
2. Calculate the speed using the circumference and time period.
3. Use the speed to find the centripetal acceleration.
4. Substitute values and calculate the final result.