Angle rotated by accelerating dentist's drill in 4.95 s

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hoseA
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Angular Acceleration/ Find angle in period -- PLEASE help!

I'm having trouble with part "B" please help.

A.)A dentist's drill starts from rest. After 4.95 s
of constant angular acceleration, it turns at a
rate of 28600 rev/min.
Find the drill's angular acceleration. An-
swer in units of rad/s^2.

B.)Find the angle through which the drill rotates
during this period. Answer in units of rad.

A.) [(28600*2pi)/60]/4.95 = 605.047 rad/s^2

B.) [(28600*2pi)/60] * 4.95 = 14825 rad. (but this is not the right answer?)

Am I using the wrong formula? Any help appreciated.
 

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hoseA said:
B.) [(28600*2pi)/60] * 4.95 = 14825 rad. (but this is not the right answer?)
This assumes that the speed is constant at (28600*2pi)/60 rad/s. But the speed is not constant; the drill starts from rest and accelerates.
 
Thanks so much. Didn't realize that. So it's theta = .5 (605.047)(4.95^2)

= 7412.58 radians.

Thanks again. :)