A dentist's drill accelerates from rest at 605 rad/s² for 2.15s

and then runs at constant angular velocity for 6.70s.

Through how many total revolutions has the drill turned?

ΔΘ=?

Well 2pi radians equals one full revolution, so you first want to find out how much distance it has traveled, then divide by 2pi to figure out how many revolutions that is.

You can first find the final angular velocity (W = omega = angular velocity; A = alpha = angular acceleration) with the rotational motion equation:
Wf = Wi + at = 0m/s + (605x2.15)
Wf = 1300.75 rad/s
Find the distance of this first part: (Θ = Wi(t) + .5a(t^2))
Θ = 0 + 0.5(605)(2.15)(2.15) = 1398.3 radians. That's for the acceleration

Now for the constant velocity (no acceleration).
Θ = Wi(t) + 0.5a(t^2)
Θ = 1300.75 rad/s(6.70s) + 0 = 8715.0 radians

So in the first part (acceleration), it traveled 1398.3 radians, and in the second part (constant velocity, no acceleration), it traveled 8715.0 radians. Add them together for a total of 10,113.3 radians. Divide that by 2pi for your total revolutions (2pi = 360º or one full revolution/rotation).

YOUR ANSWER:
You get: 1,609.6 rotations (10,113.3 radians).

find the angular displacement for the acceleration phase:

theta 1= 1/2 alpha t^2 = 1/2 x 605rad/s/s x(2.15s)^2

then find the angular displacement in the second phase:

theta2 = angular velocity x time

the angular velocity is the angular velocity at the end of the acceleration phase

ang velociy = alpha x time = 605rad/s/s x 2.15s

and

theta2 = (605rad/s/s x 2.15s) x (6.7s)

add theta1 and theta2 for your final angular displacement

since this is inradians, divide by 2 pi to find revolutions