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A ball is spun with angular acceleration $\alpha=6 t ^{2}-2 t$ where $t$ is in second and $\alpha$ is in $rads$ $^{-2}$. At $t=0$, the ball has angular velocity of $10\,rads$ $^{-1}$ and angular position of $4\,rad$. The most appropriate expression for the angular position of the ball is
$\frac{3}{2} t^{4}-t^{2}+10 t$
$\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4$
$\frac{2 t ^{4}}{3}-\frac{ t ^{3}}{6}+10 t +12$
$2 t^{4}-\frac{t^{3}}{2}+5 t+4$
Solution
$\frac{d \omega}{d t}=6 t^{2}-2 t$
$\int \limits_{10}^{m} d \omega=2 t^{5}-t^{2}$
$\omega=10+2 t^{5}-t^{2}$
$\frac{d \theta}{d t}=10+2 t^{5}-t^{2}$
$\int \limits_{4}^{\theta} d \theta=10+2 t^{3}-t^{2}$
$\int \limits_{4}^{\theta} d \theta=10 t+\frac{t^{4}}{2}-\frac{t^{3}}{3}$
$\theta=4+10 t+\frac{t^{4}}{2}-\frac{t^{5}}{3}$
