Respuesta :
The question is not complete. The complete question is :
During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.
Solution :
Given :
Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]
∴ Angular speed is [tex]$\omega = \frac{d \theta}{dt}$[/tex]
ω = 10 + 4t
And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]
α = 4
a). At time, t = 0.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(0)^2 +10(0)+5$[/tex]
= 5 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(0)
= 10 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]
b). At time, t = 3.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(3)^2 +10(3)+5$[/tex]
= 53 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(3)
= 22 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]
