A ball is spun with angular acceleration `alpha=6t^(2)-2t` where t is in second and `alpha` is in `rads^(-2)`. At t = 0, the ball has angular velocity of 10 `rads^(-1)` and angulra position of 4 rad. The most appropriate expression for the angular position of the ball is :

Options

A. `(3)/(2)t^(4)-t^(2)+10t`
B. `(t^(4))/(2)-(t^(3))/(3)+10t+4`
C. `(2t^(4))/(3)-(t^(3))/(6)+10t+12`
D. `2t^(4)-(t^(3))/(2)+5t+4`

Explanation

To find the angular position of the ball as a function of time, we will follow these steps:

### Step 1: Understand the given data
- The angular acceleration is given by:
\[ \alpha(t) = 6t^2 - 2t \]
- Initial angular velocity at \( t = 0 \):
\[ \omega(0) = 10 \, \text{rads}^{-1} \]
- Initial angular position at \( t = 0 \):
\[ \theta(0) = 4 \, \text{rad} \]

### Step 2: Relate angular acceleration to angular velocity
We know that angular acceleration is the rate of change of angular velocity:
\[ \alpha = \frac{d\omega}{dt} \]
Thus, we can write:
\[ d\omega = (6t^2 - 2t) dt \]

### Step 3: Integrate to find angular velocity
Integrate both sides with respect to \( t \):
\[ \int d\omega = \int (6t^2 - 2t) dt \]
This gives:
\[ \omega - \omega(0) = 2t^3 - t^2 + C \]
Where \( C \) is the constant of integration. Since \( \omega(0) = 10 \):
\[ \omega - 10 = 2t^3 - t^2 \]
Thus,
\[ \omega(t) = 2t^3 - t^2 + 10 \]

### Step 4: Relate angular velocity to angular position
Now, we know that:
\[ \omega = \frac{d\theta}{dt} \]
Thus, we can write:
\[ d\theta = (2t^3 - t^2 + 10) dt \]

### Step 5: Integrate to find angular position
Integrate both sides with respect to \( t \):
\[ \int d\theta = \int (2t^3 - t^2 + 10) dt \]
This gives:
\[ \theta - \theta(0) = \left( \frac{2t^4}{4} - \frac{t^3}{3} + 10t \right) + C \]
Where \( C \) is the constant of integration. Since \( \theta(0) = 4 \):
\[ \theta - 4 = \frac{t^4}{2} - \frac{t^3}{3} + 10t \]
Thus,
\[ \theta(t) = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4 \]

### Final Expression
The most appropriate expression for the angular position of the ball is:
\[ \theta(t) = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4 \]

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