Answer
We find values for $t_{1}$ when the angular displacement (relative to its orientation at t=0) is $\theta_{1}=22 \mathrm{rad}$ (or 22.09 rad if we want to keep track of true values in all in-between steps and only round off on the last answers). Using Eq. $10-13$ and the quadratic formula, we have
$$
\theta_{1}=\omega_{0} t_{1}+\frac{1}{2} \alpha t_{1}^{2} \Rightarrow t_{1}=\frac{-\omega_{0} \pm \sqrt{\omega_{0}^{2}+2 \theta_{1} \alpha}}{\alpha}
$$
which yields the two roots $5.5\ \mathrm{s}$ and $32\ \mathrm{s}$ .
Thus, the first time the reference line will be at
$$\theta_{1}=22 \mathrm{rad} \quad is \quad t=5.5 \mathrm{s} $$
Work Step by Step
We find values for $t_{1}$ when the angular displacement (relative to its orientation at t=0) is $\theta_{1}=22 \mathrm{rad}$ (or 22.09 rad if we want to keep track of true values in all in-between steps and only round off on the last answers). Using Eq. $10-13$ and the quadratic formula, we have
$$
\theta_{1}=\omega_{0} t_{1}+\frac{1}{2} \alpha t_{1}^{2} \Rightarrow t_{1}=\frac{-\omega_{0} \pm \sqrt{\omega_{0}^{2}+2 \theta_{1} \alpha}}{\alpha}
$$
which yields the two roots $5.5\ \mathrm{s}$ and $32\ \mathrm{s}$ .
Thus, the first time the reference line will be at
$$\theta_{1}=22 \mathrm{rad} \quad is \quad t=5.5 \mathrm{s} $$
