Angular displacement (`theta`) of a flywheel varies with time as `theta=2t+3t^2` radian. The angular acceleration at t=2s is given by

Options

A. 14`rad//s^2`
B. 18`rad//s^2`
C. 6`rad//s^2`
D. 16`rad//s^2`

Answer

To find the angular acceleration of the flywheel at \( t = 2 \) seconds, we start with the given angular displacement equation:

\[ \theta = 2t + 3t^2 \]

### Step 1: Differentiate the angular displacement to find angular velocity.

The angular velocity \( \omega \) is given by the first derivative of angular displacement \( \theta \) with respect to time \( t \):

\[ \omega = \frac{d\theta}{dt} = \frac{d}{dt}(2t + 3t^2) \]

Calculating the derivative:

\[ \omega = 2 + 6t \]

### Step 2: Differentiate the angular velocity to find angular acceleration.

The angular acceleration \( \alpha \) is given by the derivative of angular velocity \( \omega \) with respect to time \( t \):

\[ \alpha = \frac{d\omega}{dt} = \frac{d}{dt}(2 + 6t) \]

Calculating the derivative:

\[ \alpha = 6 \]

### Step 3: Evaluate angular acceleration at \( t = 2 \) seconds.

Since the angular acceleration \( \alpha \) is a constant value (independent of time), we can conclude:

\[ \alpha = 6 \, \text{rad/s}^2 \]

Thus, the angular acceleration at \( t = 2 \) seconds is:

\[ \alpha = 6 \, \text{rad/s}^2 \]

### Final Answer:
The angular acceleration at \( t = 2 \) seconds is \( 6 \, \text{rad/s}^2 \).

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