A ball is spun with angular acceleration α = 6 t ^ 2 - 2t where t...

Given: $\alpha = 6{t}^{2} - 2t$

Using relation $\alpha = \dfrac{d \omega}{d t} = 6{t}^{2} - 2t$

Integrating the above,

${∫}_{10}^{\omega} d \omega = {∫}_{0}^{t}\left(6{t}^{2} - 2t\right) d t$

$⇒ \omega - 10 = 2{t}^{3} - {t}^{2}$

Now, $\omega = \dfrac{d \theta}{d t} = 10 + 2{t}^{3} - {t}^{2}$

${∫}_{4}^{\theta} d \theta = {∫}_{0}^{t}\left(10 + 2{t}^{3} - {t}^{2}\right) d t$

Integrating the above relation,

$\theta - 4 = 10t + \dfrac{{t}^{4}}{2} - \dfrac{{t}^{3}}{3}$

Thus, angular position of the ball is $\theta = \dfrac{{t}^{4}}{2} - \dfrac{{t}^{3}}{3} + 10t + 4$.