(a) 4.03 s
The initial angular velocity of the wheel is
[tex]\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s[/tex]
The angular acceleration of the wheel is
[tex]\alpha = -3.40 rad/s^2[/tex]
negative since it is a deceleration.
The angular acceleration can be also written as
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 0[/tex] is the final angular velocity (the wheel comes to a stop)
t is the time it takes for the wheel to stop
Solving for t, we find
[tex]t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s[/tex]
(b) 27.6 rad
The angular displacement of the wheel in angular accelerated motion is given by
[tex]\theta= \omega_i t + \frac{1}{2}\alpha t^2[/tex]
where we have
[tex]\omega_i=13.7 rad/s[/tex] is the initial angular velocity
[tex]\alpha = -3.40 rad/s^2[/tex] is the angular acceleration
t = 4.03 s is the total time of the motion
Substituting numbers, we find
[tex]\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad[/tex]
