The angular displacement of a particle is given by `theta = t^3 + 2t +1`, where `t` is time in seconds. Its angular acceleration at `t=2s` is

Options

A. `14 rads^(-2)`
B. `17 rads^(-2)`
C. `12 rads^(-2)`
D. `9 rads^(-2)`

Explanation

To find the angular acceleration of the particle at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Find the angular displacement function
The angular displacement is given by:
\[ \theta(t) = t^3 + 2t + 1 \]

### Step 2: Differentiate to find angular velocity
Angular velocity \( \omega \) is the first derivative of angular displacement with respect to time:
\[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(t^3 + 2t + 1) \]
Calculating the derivative:
\[ \omega(t) = 3t^2 + 2 \]

### Step 3: Differentiate to find angular acceleration
Angular acceleration \( \alpha \) is the derivative of angular velocity with respect to time:
\[ \alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(3t^2 + 2) \]
Calculating the derivative:
\[ \alpha(t) = 6t \]

### Step 4: Substitute \( t = 2 \) seconds to find angular acceleration
Now, we will substitute \( t = 2 \) into the angular acceleration equation:
\[ \alpha(2) = 6 \times 2 = 12 \, \text{radians per second squared} \]

### Final Answer
The angular acceleration at \( t = 2 \) seconds is:
\[ \alpha = 12 \, \text{radians per second squared} \]

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