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 start with the given angular displacement equation:

\[ \theta(t) = t^4 + t^3 + t^2 + 1 \]

### Step 1: Differentiate the angular displacement to find angular velocity

The angular velocity \(\omega\) is defined as the rate of change of angular displacement with respect to time. Mathematically, this is expressed as:

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

We need to differentiate \(\theta(t)\) with respect to \(t\).

### Step 2: Differentiate the function

Using the power rule of differentiation, we differentiate each term in the equation:

\[ \frac{d\theta}{dt} = \frac{d}{dt}(t^4) + \frac{d}{dt}(t^3) + \frac{d}{dt}(t^2) + \frac{d}{dt}(1) \]

Calculating each derivative:

- \(\frac{d}{dt}(t^4) = 4t^3\)
- \(\frac{d}{dt}(t^3) = 3t^2\)
- \(\frac{d}{dt}(t^2) = 2t\)
- \(\frac{d}{dt}(1) = 0\)

Combining these results, we get:

\[ \omega(t) = 4t^3 + 3t^2 + 2t \]

### Step 3: Substitute \(t = 2\) seconds into the angular velocity equation

Now we need to find the angular velocity at \(t = 2\) seconds:

\[ \omega(2) = 4(2^3) + 3(2^2) + 2(2) \]

Calculating each term:

- \(4(2^3) = 4 \times 8 = 32\)
- \(3(2^2) = 3 \times 4 = 12\)
- \(2(2) = 4\)

Now, add these values together:

\[ \omega(2) = 32 + 12 + 4 = 48 \text{ radians per second} \]

### Final Answer

The angular velocity after 2 seconds is:

\[ \omega = 48 \text{ radians per second} \]

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