A particle is moving on a circular track of radius 30 cm with a constant speed of `6ms^(-1)`. Its acceleration is

Options

A. zero
B. 120`ms^(-2)`
C. `1.2ms^(-2)`
D. 36`ms^(-2)`

Explanation

To find the acceleration of a particle moving in a circular path with a constant speed, we can use the concept of centripetal acceleration. Here's the step-by-step solution:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the circular track (r) = 30 cm = 0.3 m (convert cm to m)
- Speed of the particle (v) = 6 m/s

2. **Understand Centripetal Acceleration**:
- When an object moves in a circular path with constant speed, it experiences centripetal acceleration directed towards the center of the circle. This acceleration can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
where \( a_c \) is the centripetal acceleration, \( v \) is the speed, and \( r \) is the radius.

3. **Substitute the Values into the Formula**:
- Substitute \( v = 6 \, \text{m/s} \) and \( r = 0.3 \, \text{m} \) into the formula:
\[ a_c = \frac{(6 \, \text{m/s})^2}{0.3 \, \text{m}} \]

4. **Calculate the Numerator**:
- Calculate \( v^2 \):
\[ v^2 = 6^2 = 36 \, \text{m}^2/\text{s}^2 \]

5. **Calculate the Centripetal Acceleration**:
- Now, substitute \( v^2 \) into the formula:
\[ a_c = \frac{36 \, \text{m}^2/\text{s}^2}{0.3 \, \text{m}} = 120 \, \text{m/s}^2 \]

6. **Conclusion**:
- The centripetal acceleration of the particle is \( 120 \, \text{m/s}^2 \).

### Final Answer:
The acceleration of the particle is \( 120 \, \text{m/s}^2 \).

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