Speed Drill

A dentist causes the bit of a high-speed drill to accelerate from an angular speed...?

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.42 x 104 rad/s to an angular speed of 4.55 x 104 rad/s. In the process, the bit turns through 3.38 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 9.40 x 104 rad/s, starting from rest?

Thanks for any help!!! 10 points will be awarded to who helps best!

thus test angular path is p1=u1*t1 +0.5*w*t1^2 =3.38e4 rad,
where u1=1.42e4 rad/s, w is angular acceleration,
t1 is time to accelerate from u1 to u2, u2=4.66e4 rad/s;
also w=(u2-u1)/t1, hence t1=(u2-u1)/w;
thus p1= u1*(u2-u1)/w +0.5*w*((u2-u1)/w)^2, hence
w= (u1*(u2-u1) +0.5*(u2-u1)^2)/p1 = 0.5*(u2^2 –u1^2)/p1;
time in question is t=v/w =v*2*p1/(u2^2 –u1^2) =
=9.4*2*3.38 / (4.55^2 -1.42^2) =3.4s; check me though!

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