First of all, let's convert the time interval into minutes. Since
[tex]60 s: 1 min = 5 s: x[/tex]
we find
[tex]\Delta t = \frac{5.0 s}{60 s/min}=0.083 min [/tex]
Then we can find the angular acceleration of the flywheel:
[tex]\alpha = \frac{\omega _f - \omega_i}{\Delta t}= \frac{853 rpm-161 rpm}{0.083 min}=8337 rev/min^2 [/tex]
At this point, we can use the law of motion of an uniformly accelerated rotational motion. The angular displacement after a time [tex]\Delta t[/tex] is given by
[tex]\theta (\Delta t)= \omega_i t + \frac{1}{2} \alpha t^2 = [/tex]
[tex]=(161 rpm)(0.083 min)+ \frac{1}{2}(8337 rev/min^2)(0.083 min)^2 =42.1 rev[/tex]
So, the flywheel covers 42.1 revolutions.
