A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 8.0-s interval at a rate of 0.040 rad/s2. What is its angular speed after this 8.0-s interval?Select one:a.0.48 rad/sb.0.24 rad/sc.0.32 rad/sd.1.6 rad/se.0.05 rad/s
Question
A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 8.0-s interval at a rate of 0.040 rad/s2. What is its angular speed after this 8.0-s interval?Select one:a.0.48 rad/sb.0.24 rad/sc.0.32 rad/sd.1.6 rad/se.0.05 rad/s
Solution
The final angular speed of an object undergoing angular acceleration can be found using the formula:
ωf = ωi + αt
where: ωf is the final angular speed, ωi is the initial angular speed, α is the angular acceleration, and t is the time interval.
Given in the problem, we have: ωi = 0.16 rad/s, α = 0.040 rad/s², and t = 8.0 s.
Substituting these values into the formula, we get:
ωf = 0.16 rad/s + (0.040 rad/s² * 8.0 s) = 0.16 rad/s + 0.32 rad/s = 0.48 rad/s.
So, the final angular speed of the Ferris wheel after the 8.0-s interval is 0.48 rad/s. Therefore, the correct answer is a. 0.48 rad/s.
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