A phonograph record initially moving at 33 1/3 rpm accelerates uniformly to 45 rpm in 2.5 s. What is the angular distance traveled by the record during this time?

Answer 1

Vo = 33.3333rev/min * 6.28rad/rev 1mim/60s = 3.49rad/s.

Vf = (45/33.3333) * 3.49rad/s=4.71rad/s

a = (Vf - Vo) / t,
a = (4.71 - 3.49) / 2.5 = 0.488.

d = Vo*t + 0.5a*t^2,
d = 3.49*2.5 +.244*(2.5)^2 = 10.25rad.