#### Answer

i) In case a), angular acceleration is positive. The magnitude of $\vec{\alpha}$ is $1.94 rad/s^2$ and it is directed counterclockwise.
ii) In case b), angular acceleration is negative. The magnitude of $\vec{\alpha}$ is $1.23 rad/s^2$ and it is directed clockwise.

#### Work Step by Step

We have $$\alpha=\frac{\omega-\omega_0}{\Delta t}$$
i) In the case when the angular speed increases, the final angular speed is greater than the initial one, making $(\omega-\omega_0)\gt0$. Therefore, $\alpha\gt0$, or the angular acceleration is positive.
$$\alpha=\frac{28.5-21.7}{3.5}=+1.94rad/s^2$$
The magnitude of $\vec{\alpha}$ is $1.94 rad/s^2$. The positive sign indicates $\vec{\alpha}$ is directed counterclockwise.
ii) In the case when the angular speed decreases, the final angular speed is less than the initial one, making $(\omega-\omega_0)\lt0$. Therefore, $\alpha\lt0$, or the angular acceleration is negative.
$$\alpha=\frac{15.3-28.5}{10.7}=-1.23rad/s^2$$
The magnitude of $\vec{\alpha}$ is $1.23 rad/s^2$. The negative sign indicates $\vec{\alpha}$ is directed clockwise.