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Q.
A wheel has angular acceleration of $ 3.0\,rad/s^{2} $ and an initial angular speed of $2.00 \,rad/s$. In a time of $2 \,s$ it has rotated through an angle (in radian) of
Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement.
By definition $ \alpha=\frac{d \omega}{d t}$
ie, $ d \omega=\alpha d t$
So, if in time $t$ the angular speed of a body changes from $\omega_{0}$ to $\omega$
$\int\limits_{\omega_{0}}^{\omega} d \omega=\int\limits_{0}^{t} \alpha d t$
If $\alpha$ is constant
$\omega-\omega_{0} =\alpha t $
or $ \omega =\omega_{0}+\alpha t$ ... (i)
Now, as by definition
$\omega=\frac{d \theta}{d t}$
Eq. (i) becomes
$\frac{d \theta}{d t}=\omega_{0}+\alpha t $
$d \theta=\left(\omega_{0}+\alpha t\right) d t$
So, if in the time $t$ angular displacement is $\theta$.
$\int\limits_{0}^{\theta} d \theta=\int\limits_{0}^{t}\left(\omega_{0}+\alpha t\right) d t$
or $ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$ ... (ii)
Given, $\alpha = 3.0\, rad / s ^{2}$,
$\omega_{0} = 2.0 \,rad / s , t = 2 \,s$
Hence, $ \theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2}$
or $ \theta=4+6=10 \,rad$
Note Eqs. (i) and (ii) are similar to first and second equations of linear motion.