Respuesta :
Answer:
[tex] x_f = 15 m/s *0.8 s + \frac{1}{2} 2.5 m/s^2 (0.8 s)^2 = 12.8 m[/tex]
And we have the following relation between the angular displacement and the linear displacement:
[tex] x = r \theta[/tex]
Where x represent the linear displacement and [tex]\theta[/tex] the angular displacement, if we solve for [tex] \theta[/tex] we got:
[tex] \theta= \frac{x}{R}= \frac{12.8 m}{0.34 m}=37.65 rad[/tex]
Explanation:
For this case we have the following data given:
[tex] v_i = 15 m/s[/tex] represent the initial speed
[tex] a = 2.5 m/s^2[/tex] represent the acceleration
[tex] t = 0.8 s[/tex] represent the time
[tex] R = 34 cm = 0.34 m[/tex] represent the radius
First we can calculate the linear displacement with the following formula from kinematics:
[tex] x_f = x_i + v_i t + \frac{1}{2} at^2[/tex]
And replacing we have:
[tex] x_f = 15 m/s *0.8 s + \frac{1}{2} 2.5 m/s^2 (0.8 s)^2 = 12.8 m[/tex]
And we have the following relation between the angular displacement and the linear displacement:
[tex] x = r \theta[/tex]
Where x represent the linear displacement and [tex]\theta[/tex] the angular displacement, if we solve for [tex] \theta[/tex] we got:
[tex] \theta= \frac{x}{R}= \frac{12.8 m}{0.34 m}=37.65 rad[/tex]
And that would be the angular displacement during the period of acceleration.
