Answer:
The answer is below
Explanation:
a) Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\\\theta=angular\ distance\\\\Given\ that:\\\\initial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\\final/ velocity(v)=0(stop)\\\\\omega=v/r=\frac{28.78m/s}{0.25m} =115.12\ rad/s,\omega_o=0,\theta=s/r=\frac{50\ m}{0.25\ m}=200\ rad\\ \\\omega^2=\omega_o^2+2\alpha \theta\\\\115.12^2=0^2+2\alpha(200)\\\\2\alpha(200)=13252.6144\\\\\alpha=33.13\ rad/s^2[/tex]
b)
[tex]\theta=200\ rad=200\ rad*\frac{1\ rev}{2\pi\ rad}=31.83\ rev[/tex]
