If the equation for the displacement of a particle moving on a circle path is given by, `theta = 2t^(2)+0.5` where, `theta` is in radian and `t` is in second, then the angular velocity of the particle after 2 s is

Options

A. `8 rad s^(-1)`
B. `12 rad s^(-1)`
C. `24 rad s^(-1)`
D. `36 rad s^(-1)`

Explanation

To find the angular velocity of the particle after 2 seconds, we can follow these steps:

### Step 1: Understand the given equation
The displacement of the particle moving in a circular path is given by the equation:
\[ \theta = 2t^2 + 0.5 \]
where \(\theta\) is in radians and \(t\) is in seconds.

### Step 2: Recall the formula for angular velocity
Angular velocity (\(\omega\)) is defined as the rate of change of angular displacement with respect to time. Mathematically, it can be expressed as:
\[ \omega = \frac{d\theta}{dt} \]

### Step 3: Differentiate the displacement equation
We need to differentiate the given equation for \(\theta\) with respect to \(t\):
\[ \omega = \frac{d\theta}{dt} = \frac{d}{dt}(2t^2 + 0.5) \]
Using the power rule of differentiation:
\[ \frac{d}{dt}(t^n) = n \cdot t^{n-1} \]
we find:
\[ \frac{d}{dt}(2t^2) = 2 \cdot 2t^{2-1} = 4t \]
The derivative of the constant \(0.5\) is \(0\). Therefore, we have:
\[ \omega = 4t + 0 = 4t \]

### Step 4: Calculate angular velocity at \(t = 2\) seconds
Now we substitute \(t = 2\) seconds into the equation for \(\omega\):
\[ \omega = 4(2) = 8 \text{ radians per second} \]

### Step 5: Conclusion
Thus, the angular velocity of the particle after 2 seconds is:
\[ \omega = 8 \text{ radians per second} \]

### Final Answer
The correct answer is option 1: 8 radians per second.

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