Chrome. MasteringPhysics: Ch 8.38.4 Rolling & Torque Secure | https://session.ma

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Chrome. MasteringPhysics: Ch 8.38.4 Rolling & Torque Secure | https://session.masteringphysics.com/myct/itemview?assignmentProblemiD-92538435&offset;=next Constants Part A The tires of a car make 66 revolutions as the car reduces its speed uniformly from 86.0 km/h to 55.0 km/h. The tires have a diameter of 0.90 m. What was the angular acceleration of the tires? Express your answer using two significant figures. Submit Part B If the car continues to decelerate at this rate, how much more time is required for it to stop? Express your answer to two significant figures and include the appropriate units. 1 Value Units Submit Request Answer

Explanation / Answer

Vf = 55/3.6 = 15.28 m/sec

Vi = 88/3.6 = 23.89 m/sec

d = 66*0.90*PI = 186.61 m

t = 2d/(Vf+Vi) = 9.528 sec

a = V/t = (23.89 - 15.28)/9.528 = 0.9036 m/sec^2

1) What was the angular acceleration of the tires?

= a/r = 0.9036/0.45 = 2.0 rad/sec^2

2) If the car continues to decelerate at this rate, how much more time is required for it to stop?

ts = Vf/a = 15.28/0.9036 = 16.91 = 17 sec

3)

slowing space = 66*0.90*PI = 186.61 m

total stopping space = 1/2*Vi*(9.528+16.91) = 320 m

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