An object initially rotating at an angular speed of 1.6 rad/sec turns through 30 revolutions during the time it experienced an angular acceleration of 0.32 rad/s^2

a) For how much time did the acceleration last?
b) What was the final angular speed?

Help please. Use the equation and then step by step and plug the numbers into that equation

Here is the equaton

displcaement=wi*t+1/2*alpha*t^2

displacement=30*2PI rad
wi=1.6 rad/sec
alpha=.32 rad/s^2
solve for time

wf=wi +alpha*t
wf = final angular speed in rad/sec

I still don't get it. Can you explain more and do the problem step by step to get the answer

To find the answers to these questions, we can use the following equation:

θ = ωi * t + (1/2) * α * t^2

Where:
- θ is the total angular displacement in radians
- ωi is the initial angular speed in rad/s
- α is the angular acceleration in rad/s^2
- t is the time in seconds

a) For how much time did the acceleration last?
First, let's express the total angular displacement in terms of revolutions. We know that 30 revolutions is equal to 2π * 30 = 60π radians.

Now we can rearrange the equation to solve for time (t):
θ = ωi * t + (1/2) * α * t^2
60π = 1.6 * t + (1/2) * 0.32 * t^2

This is a quadratic equation, so we can solve it by rearranging and using the quadratic formula:

0.16t^2 + 1.6t - 60π = 0

Using the quadratic formula, we get:
t = (-1.6 ± √((1.6^2)-(4*0.16*(-60π)))) / (2*0.16)

We only need the positive value, so we discard the negative solution:
t ≈ 9.671 seconds

b) What was the final angular speed?
To find the final angular speed (ωf), we can use the following equation:

ωf = ωi + α * t

Substituting the known values:
ωf = 1.6 + 0.32 * 9.671

ωf ≈ 4.963 rad/s

Therefore, the acceleration lasted approximately 9.671 seconds, and the final angular speed was approximately 4.963 rad/s.