IF the equation for the displancement of a particle moving on a circular path is given by `theta=2t^3+0.5`, where `theta` is in radius and t is in seconds, then the angular velocity of the particle at t=2 s is

Explanation

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

### Step 1: Write down the given equation for angular displacement
The equation for the angular displacement \( \theta \) is given as:
\[ \theta = 2t^3 + 0.5 \]
where \( \theta \) is in radians and \( t \) is in seconds.

### Step 2: Differentiate the displacement equation to find angular velocity
Angular velocity \( \omega \) is defined as the rate of change of angular displacement with respect to time. Therefore, we differentiate \( \theta \) with respect to \( t \):
\[ \omega = \frac{d\theta}{dt} \]
Differentiating \( \theta = 2t^3 + 0.5 \):
\[ \frac{d\theta}{dt} = \frac{d}{dt}(2t^3) + \frac{d}{dt}(0.5) \]
The derivative of \( 0.5 \) is \( 0 \), and the derivative of \( 2t^3 \) is:
\[ \frac{d}{dt}(2t^3) = 6t^2 \]
Thus, we have:
\[ \omega = 6t^2 \]

### Step 3: Substitute \( t = 2 \) seconds into the angular velocity equation
Now, we will substitute \( t = 2 \) seconds into the angular velocity equation:
\[ \omega = 6(2^2) = 6(4) = 24 \text{ radians per second} \]

### Conclusion
The angular velocity of the particle at \( t = 2 \) seconds is:
\[ \omega = 24 \text{ radians per second} \]

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