Physics Angular Velocity/Acceleration problem.?

A) Find the angular acceleration of the drill

ω = (2PI/60)*n = 0.1047*2.51*10^4 = 2628.0 rad/sec

ώ = Δω/Δt = (2628.0-0)/(3.20-0) = 821.25 rad/sec^2

B) Calculate the angle through which the drill rotates.

Θ = 1/2*ω*t = 2628.0*1.6 = 4204.75 rad

a. First convert rev/min to rad/s, (2.51 * 10^4 * 2pi)/60 = 2628.5 rad/s. Now using the formula w =at;

2628.5 = a * 3.2

therefore a = 821.4 rad/s^2.

b. Now for the second one im not completely sure, but i think you use the equation w = change in theta/change in time, therefore 2628.5 * 3.2 = 8411.2 rad

w of drill after t = 3.2 s:

w = 2π(2.51E4)/60 = 0.263E4 = 2630 rad/s

α = w/t = 2630/3.2 ≈ 822 rad/s² ANS A)

Θ = w•t = 2630(3.2) = 8416 rad ANS B)