An automobile engine slows down from 4500rpm…

Question:

“an automobile engine slows down from 4500rpm to 1200rpm in2.5seconds..Calculate..
A) Angular Acceleration assumed constant, and
B) The total number of revolutions the engine makes in this time”

Answer:

α=(ω2-ωo2)/t, so first convert the rpm’s to
rad/s then you get 472.5rad/s and 126rad/s and
the acceleration is -138.6rad/s2. The total revolutions that are made are
Δθ=1/2Δωt=748rads, so748rads/2π=1.2x102revolutions

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