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During a very quick stop, a car decelerates at 7.8 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).

Randomized Variablesat = 7.8 m/s2
r = 0.29 m
ω0 = 93 rad/s

Part (a) What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.29 m and do not slip on the pavement? sig.gif?tid=8O75-6C-7E-4A-961B-22885
Part (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 93 rad/s ? sig.gif?tid=8O75-6C-7E-4A-961B-22885

Part (c) How long does the car take to stop completely in seconds? sig.gif?tid=8O75-6C-7E-4A-961B-22885
Part (d) What distance does the car travel in this time in meters? sig.gif?tid=8O75-6C-7E-4A-961B-22885
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Answer #1

Os Solaat= 7.8 m/s? 8= 0.29m Wo= 93 rod/ s a n alinear = QR q=170 8 ) = 726.896 od of motion eqr w = Wotak I 0= 93 + (-261896

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