Answer:
4.25 s
Explanation:
Given:
angular acceleration'α'= 1.8 rad/s²
angle 'θ'= 45 rad
time 't'= 4s
initial angular velocity '[tex]w_{o}[/tex]'=0[tex]rad^{-1}[/tex]
as we know that,
θ= [tex]w_{1}[/tex]t + [tex]\frac{1}{2} \alpha t^{2}[/tex]
45 = 4 [tex]w_{1}[/tex] + (0.5 x 1.8 x 16)
45- 14.4 = 4 [tex]w_{1}[/tex]
30.6 = 4 [tex]w_{1}[/tex]
[tex]w_{1}[/tex]= 7.65 [tex]rad^{-1}[/tex]
Next is to find t by using the equation
[tex]w_{1}[/tex] = [tex]w_{o}[/tex] + [tex]\alpha t_{1}[/tex]
7.65= 0 + (1.8)[tex]t_{1}[/tex]
[tex]t_{1}[/tex]= 7.65/1.8
[tex]t_{1}[/tex]= 4.25 s
Therefore, At the start of 4s interval the motion is at 4.25 second
