The angular displacement of a particle is given by `theta = t^4 +t^3 +t^2 +1` where ‘t’ is time in seconds. Its angular velocity after 2 sec is

Options

A. ` 20 rad s^(-1)`
B. ` 48 rads ^(-1)`
C. `52 rad s^(-1)`
D. `22 rads ^(-1)`

Explanation

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

### Step 1: Write down the expression for angular displacement.
The angular displacement is given by:
\[ \theta(t) = t^4 + t^3 + t^2 + 1 \]

### Step 2: Differentiate the angular displacement to find angular velocity.
Angular velocity (\(\omega\)) is the derivative of angular displacement with respect to time:
\[ \omega(t) = \frac{d\theta}{dt} \]
Using the power rule of differentiation, we differentiate each term:
- The derivative of \(t^4\) is \(4t^3\).
- The derivative of \(t^3\) is \(3t^2\).
- The derivative of \(t^2\) is \(2t\).
- The derivative of a constant (1) is 0.

Thus, we have:
\[ \omega(t) = 4t^3 + 3t^2 + 2t \]

### Step 3: Substitute \(t = 2\) seconds into the angular velocity equation.
Now, we need to find \(\omega\) at \(t = 2\):
\[ \omega(2) = 4(2^3) + 3(2^2) + 2(2) \]

Calculating each term:
- \(2^3 = 8\) so \(4(2^3) = 4 \times 8 = 32\)
- \(2^2 = 4\) so \(3(2^2) = 3 \times 4 = 12\)
- \(2(2) = 4\)

Now, adding these values together:
\[ \omega(2) = 32 + 12 + 4 = 48 \text{ radian/second} \]

### Final Answer:
The angular velocity after 2 seconds is:
\[ \omega = 48 \text{ radian/second} \]

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