#### Answer

(a) t = 6.32 s
(b) The motor turns through an angle of 84.3 radians.

#### Work Step by Step

(a) At the moment when the motor changes direction, the angular velocity will be zero. We can find the time when the angular velocity is zero.
$\omega(t) = (20-\frac{1}{2}t^2)~rad/s = 0$
$t^2 = 40$
$t = \sqrt{40}$
$t = 6.32~s$
(b) The angle that the motor turns is equal to the area under the omega versus time graph. We can integrate to find this area.
$\Delta \theta = \int_{0}^{\sqrt{40}}\omega(t)~dt$
$\Delta \theta = \int_{0}^{\sqrt{40}}(20-\frac{1}{2}t^2)~dt$
$\Delta \theta = (20t-\frac{1}{6}t^3)\vert_{0}^{\sqrt{40}}$
$\Delta \theta = (20)(\sqrt{40})-\frac{1}{6}(\sqrt{40})^3$
$\Delta \theta = 84.3~rad$
The motor turns through an angle of 84.3 radians.