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

Options

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

Explanation

To solve the problem, we will use the equation of motion for rotational dynamics:

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

where:
- \(\theta\) is the angle rotated (in radians),
- \(\omega_0\) is the initial angular speed (in radians per second),
- \(\alpha\) is the angular acceleration (in radians per second squared),
- \(t\) is the time (in seconds).

Given:
- \(\alpha = 3.0 \, \text{rad/s}^2\)
- \(\omega_0 = 2.0 \, \text{rad/s}\)
- \(t = 2 \, \text{s}\)

Now, we can substitute the values into the equation.

1. Calculate the first term: \(\omega_0 t\)
\[ \omega_0 t = 2.0 \, \text{rad/s} \times 2 \, \text{s} = 4.0 \, \text{rad} \]

2. Calculate the second term: \(\frac{1}{2} \alpha t^2\)
\[ \frac{1}{2} \alpha t^2 = \frac{1}{2} \times 3.0 \, \text{rad/s}^2 \times (2 \, \text{s})^2 \]
\[ = \frac{1}{2} \times 3.0 \times 4 = \frac{12.0}{2} = 6.0 \, \text{rad} \]

3. Now, add both terms to find \(\theta\):
\[ \theta = 4.0 \, \text{rad} + 6.0 \, \text{rad} = 10.0 \, \text{rad} \]

Thus, the wheel has rotated through an angle of \(10.0\) radians in \(2\) seconds.

### Final Answer:
The angle rotated through is \(10.0 \, \text{rad}\).

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