The angular displacement of a particle is given by `theta=t^(3)+ t^(2)+ t+1` then, its angular velocity at t = 2sec is ……….. rad `s^(-1)`

Options

A. 27
B. 17
C. 15
D. 16

Answer

To find the angular velocity of a particle given its angular displacement, we follow these steps:

1. **Identify the given angular displacement function**:
The angular displacement \( \theta \) is given by:
\[ \theta = t^3 + t^2 + t + 1 \]

2. **Differentiate the angular displacement with respect to time**:
The angular velocity \( \omega \) is the derivative of the angular displacement \( \theta \) with respect to time \( t \):
\[ \omega = \frac{d\theta}{dt} \]
We differentiate \( \theta \):
\[ \frac{d\theta}{dt} = \frac{d}{dt}(t^3 + t^2 + t + 1) \]
Using the power rule of differentiation:
\[ \frac{d\theta}{dt} = 3t^2 + 2t + 1 \]

3. **Substitute \( t = 2 \) seconds into the angular velocity equation**:
Now we need to find \( \omega \) at \( t = 2 \):
\[ \omega = 3(2^2) + 2(2) + 1 \]
Calculate each term:
\[ = 3(4) + 2(2) + 1 \]
\[ = 12 + 4 + 1 \]
\[ = 17 \text{ rad/s} \]

4. **Final answer**:
The angular velocity at \( t = 2 \) seconds is:
\[ \omega = 17 \text{ rad/s} \]