the angular velocity omega of a particle varies with time t as `omega = 5t^2 + 25 rad/s`. the angular acceleration of the particle at` t=1`s is

Options

A. `10 rad/s^2`
B. `5rad/s^2`
C. Zero
D. `3 rad/s^2 `

Answer

To find the angular acceleration of the particle at \( t = 1 \) second, we start with the given angular velocity equation:

\[ \omega = 5t^2 + 25 \quad \text{(in rad/s)} \]

### Step 1: Understand the relationship between angular velocity and angular acceleration
Angular acceleration (\( \alpha \)) is defined as the rate of change of angular velocity with respect to time. Mathematically, this is expressed as:

\[ \alpha = \frac{d\omega}{dt} \]

### Step 2: Differentiate the angular velocity function
We need to differentiate the given angular velocity function \( \omega \) with respect to time \( t \):

\[ \alpha = \frac{d}{dt}(5t^2 + 25) \]

### Step 3: Apply differentiation
Using the power rule of differentiation, we differentiate each term:

1. The derivative of \( 5t^2 \) is \( 10t \).
2. The derivative of the constant \( 25 \) is \( 0 \).

Thus, we have:

\[ \alpha = 10t \]

### Step 4: Substitute \( t = 1 \) second
Now we substitute \( t = 1 \) second into the expression for angular acceleration:

\[ \alpha = 10(1) = 10 \quad \text{(in rad/s}^2\text{)} \]

### Conclusion
Therefore, the angular acceleration of the particle at \( t = 1 \) second is:

\[ \alpha = 10 \, \text{rad/s}^2 \]

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