Answer
Considering the first half of the motion again, Eq. $10-11$ leads to
$$
\omega=\omega_{0}+\alpha t \Rightarrow \alpha=\frac{40 \mathrm{rad} / \mathrm{s}}{20 \mathrm{s}}=2.0 \mathrm{rad} / \mathrm{s}^{2}
$$
Work Step by Step
Considering the first half of the motion again, Eq. $10-11$ leads to
$$
\omega=\omega_{0}+\alpha t \Rightarrow \alpha=\frac{40 \mathrm{rad} / \mathrm{s}}{20 \mathrm{s}}=2.0 \mathrm{rad} / \mathrm{s}^{2}
$$
