Physics
posted by Rebecca .
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?

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.
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