Answer:
(a) t=3s
(b) β=18.9rad
Explanation:
Given data
[tex]wo_{angular-velocity}=12.60 rad/s\\ \alpha_{angular-acceleration}=-4.20rad/s^{2}[/tex]
Part(a)
From the equation of angular motion to find time
So:
[tex]w=w_{o}+\alpha t\\0=12.60rad/s+(-4.20rad/s^{2} ) t\\-4.20 t=-12.60\\t=3s[/tex]
Part(b)
From the equation of simple motion to find angle β
So
[tex]\beta=w_{o}t+(1/2)\alpha t^{2}\\\beta=(12.60rad/s)(3)+(1/2)(-4.20rad/s^{2} ) (3s)^{2}\\ \beta =18.9rad[/tex]
