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The given equation of the angular position of a particle moving in a circle is:

\[
\theta(t) = 3.0t^2 + 2.0t
\]

We are asked to find the angular velocity (\(\omega\)) and angular acceleration (\(\alpha\)) at \( t = 4.00 \, \text{s} \).

### 1. Angular velocity (\(\omega\))
The angular velocity is the time derivative of the angular position:

\[
\omega(t) = \frac{d\theta(t)}{dt}
\]

Differentiate the given function \( \theta(t) = 3.0t^2 + 2.0t \):

\[
\omega(t) = \frac{d}{dt} \left( 3.0t^2 + 2.0t \right) = 6.0t + 2.0
\]

Now, substitute \( t = 4.00 \, \text{s} \):

\[
\omega(4.00) = 6.0(4.00) + 2.0 = 24.0 + 2.0 = 26.0 \, \text{rad/s}
\]

### 2. Angular acceleration (\(\alpha\))
The angular acceleration is the time derivative of the angular velocity:

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

Differentiate the angular velocity \( \omega(t) = 6.0t + 2.0 \):

\[
\alpha(t) = \frac{d}{dt} \left( 6.0t + 2.0 \right) = 6.0 \, \text{rad/s}^2
\]

Since the angular acceleration is constant, we have:

\[
\alpha = 6.0 \, \text{rad/s}^2
\]

### Final answers:
- Angular velocity at \( t = 4.00 \, \text{s} \): \( \omega = 26.0 \, \text{rad/s} \)
- Angular acceleration: \( \alpha = 6.0 \, \text{rad/s}^2 \)

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How does the angular velocity vary with time for this motion?
2. What would the angular position be at \( t = 4.00 \, \text{s} \)?
3. If the equation of motion was different, how would you approach finding \(\omega\) and \(\alpha\)?
4. What is the significance of constant angular acceleration in circular motion?
5. How would you calculate the total angle swept by the particle during the first 4 seconds?

**Tip**: For motion with constant acceleration, you can use the relationships between position, velocity, and acceleration in rotational motion similar to linear kinematics.

Una partícula se está moviendo en un círculo de acuerdo a la siguiente ecuación: θ = 3,0t^2 + 2,0t. Calcule la velocidad angular y la aceleración angular después de 4,00 s.

Learn how to calculate angular velocity and angular acceleration for a particle moving in a circle based on the equation θ = 3.0t^2 + 2.0t. This physics problem covers key concepts of rotational motion and differentiation. Solution suitable for undergraduate physics students.

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