Question

The angular speed of a rotor in a centrifuge increases from 428 to 1400 rad/s in a time of 6.26 s. (a) Obtain the angle through which the rotor turns. (b) What is the magnitude of the angular acceleration?

Answer 1

i figured it out,
[v(initial)*time]+[[acceleration*time^2]/2]

Answer 2

d omega / dt = alpha = (1400-428)/6.26
= 155 rad/s^2 which is angular acceleration part b

theta = 428 t + (1/2)(155)t^2
but t is 6.26
so
theta = 5722 radians which is part a