Answer
(a) The angular acceleration is positive.
(b) The speed of of the wheel is decreasing between t = 0 and t = 4.20 s.
The speed of of the wheel is increasing between t = 4.20 s and t = 7.00 s.
(c) $\theta = -6.97~rad$
Work Step by Step
(a) We can find the angular acceleration.
$\alpha = \frac{\omega_2-\omega_1}{t}$
$\alpha = \frac{4.00~rad-(-6.00~rad/s)}{7.00~s}$
$\alpha = 1.43~rad/s^2$
The angular acceleration is positive.
(b) We can find $t$ when $\omega = 0$
$\omega_0+\alpha ~t = 0$
$t = \frac{-\omega_0}{\alpha} = \frac{-(-6.00~rad/s)}{1.43~rad/s^2}$
$t = 4.20~s$
The speed of of the wheel is decreasing between t = 0 and t = 4.20 s.
The speed of of the wheel is increasing between t = 4.20 s and t = 7.00 s.
(c) $\theta = \omega_0 ~t + \frac{1}{2}~\alpha ~t^2$
$\theta = (-6.00~rad/s)(7.00~s) + \frac{1}{2}(1.43~rad/s^2) (7.00~s)^2$
$\theta = -6.97~rad$
