A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of

−4.60 rad/s2.
During a 2.60-s time interval, the wheel rotates through 85.4 rad. What is the angular speed of the wheel at the end of the 2.60-s interval?

To find the angular speed of the wheel at the end of the 2.60-s interval, we can use the equation for angular acceleration:

angular acceleration = (change in angular velocity) / (change in time)

In this case, the angular acceleration is given as -4.60 rad/s^2, and the change in time is 2.60 s. We need to find the change in angular velocity, which we can calculate using the formula:

change in angular velocity = angular acceleration * change in time

Plugging in the values,

change in angular velocity = -4.60 rad/s^2 * 2.60 s = -11.96 rad/s

Now, we need to find the initial angular velocity. We know that the wheel rotates through 85.4 rad in 2.60 s.

initial angular velocity = (change in angular displacement) / (change in time)

initial angular velocity = 85.4 rad / 2.60 s = 32.85 rad/s

Finally, to find the angular speed at the end of the interval, we add the initial angular velocity and the change in angular velocity:

angular speed = initial angular velocity + change in angular velocity

angular speed = 32.85 rad/s - 11.96 rad/s = 20.89 rad/s

Therefore, the angular speed of the wheel at the end of the 2.60-s interval is 20.89 rad/s.