Answer
$(\frac{6}{5}t^5-\frac{4}{3}t^3+2)\ rad/s$
Work Step by Step
Given:
angular acceleration of a wheel $\alpha = 6t^4-4t^2$
initial angular velocity $\omega_o = 2\ rad/s$
Angular acceleration is rate of change of angular velocity of the object. Therefore;
$\alpha =\frac{d\omega}{dt}$
$dw=\alpha dt$
Integrate with respect to its limits:
$\int_{\omega_o}^{\omega} d\omega =\int_ {0}^{t} \alpha dt $
$\omega-\omega_o = \int_ {0}^{t} ( 6t^4-4t^2) dt $
$\omega -2\ rad/s = 6(\frac{t^5}{5})-4(\frac{t^3}{3})$
$\omega =(\frac{6}{5}t^5-\frac{4}{3}t^3+2)\ rad/s$
