Calculate the magnitude of linear acceleration of a particle moving in a circle of radius 0.5 m at the instant when its angular velocity is 2.5 rad s–1 and its angular acceleration is `6 rad s^(-2)`.

Options

A. `40sqrt26 m/s^2`
B. `40 m/s^2`
C. Zero
D. `20 m/s^2`

Answer

To calculate the magnitude of linear acceleration of a particle moving in a circle, we need to consider both the tangential acceleration and the centripetal acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the circle (r) = 0.5 m
- Angular velocity (ω) = 2.5 rad/s
- Angular acceleration (α) = 6 rad/s²

2. **Calculate Tangential Acceleration (At):**
- The formula for tangential acceleration is:
\[ A_t = r \cdot \alpha \]
- Substituting the values:
\[ A_t = 0.5 \, \text{m} \cdot 6 \, \text{rad/s}^2 = 3 \, \text{m/s}^2 \]

3. **Calculate Centripetal Acceleration (Ac):**
- The formula for centripetal acceleration is:
\[ A_c = \omega^2 \cdot r \]
- First, calculate ω²:
\[ \omega^2 = (2.5 \, \text{rad/s})^2 = 6.25 \, \text{rad}^2/\text{s}^2 \]
- Now substitute into the centripetal acceleration formula:
\[ A_c = 6.25 \, \text{rad}^2/\text{s}^2 \cdot 0.5 \, \text{m} = 3.125 \, \text{m/s}^2 \]

4. **Calculate the Magnitude of Total Linear Acceleration (A):**
- Since tangential acceleration and centripetal acceleration are perpendicular to each other, we can use the Pythagorean theorem to find the resultant acceleration:
\[ A = \sqrt{A_t^2 + A_c^2} \]
- Substituting the values:
\[ A = \sqrt{(3 \, \text{m/s}^2)^2 + (3.125 \, \text{m/s}^2)^2} \]
- Calculate the squares:
\[ A = \sqrt{9 + 9.765625} = \sqrt{18.765625} \]
- Finally, calculate the square root:
\[ A \approx 4.33 \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the linear acceleration of the particle is approximately **4.33 m/s²**.