A wheel has angular acceleration of `3.0 rad//s^2` and an initial angular speed of `2.00 rad//s`. In a time of `2 s` it has rotated through an angle (in radian) of

Options

A. 6
B. 10
C. 12
D. 4

Explanation

To find the angle through which the wheel has rotated in a given time under uniform angular acceleration, we can use the following kinematic equation for rotational motion:

\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]

Where:
- \(\theta\) = angular displacement (in radians)
- \(\omega_0\) = initial angular speed (in radians per second)
- \(\alpha\) = angular acceleration (in radians per second squared)
- \(t\) = time (in seconds)

### Step-by-Step Solution:

1. **Identify the given values:**
- Angular acceleration, \(\alpha = 3.0 \, \text{rad/s}^2\)
- Initial angular speed, \(\omega_0 = 2.0 \, \text{rad/s}\)
- Time, \(t = 2.0 \, \text{s}\)

2. **Substitute the values into the formula:**
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
\[ \theta = (2.0 \, \text{rad/s}) \times (2.0 \, \text{s}) + \frac{1}{2} \times (3.0 \, \text{rad/s}^2) \times (2.0 \, \text{s})^2 \]

3. **Calculate each term:**
- First term:
\[ \omega_0 t = 2.0 \times 2.0 = 4.0 \, \text{radians} \]
- Second term:
\[ \frac{1}{2} \times 3.0 \times (2.0)^2 = \frac{1}{2} \times 3.0 \times 4.0 = \frac{12.0}{2} = 6.0 \, \text{radians} \]

4. **Add the two terms together:**
\[ \theta = 4.0 + 6.0 = 10.0 \, \text{radians} \]

5. **Conclusion:**
The angle through which the wheel has rotated in 2 seconds is \(10.0 \, \text{radians}\).

### Final Answer:
The wheel has rotated through an angle of \(10 \, \text{radians}\).

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