8.22 The tires of a car make 78 revolutions as the car reduces its speed uniform

ID: 1544982 • Letter: 8

Question

8.22

The tires of a car make 78 revolutions as the car reduces its speed uniformly from 86.0 km/h to 58.0 km/h . The tires have a diameter of 0.90 m .

A) What was the angular acceleration of the tires?

Express your answer using two significant figures.

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.

C) How far does the car go? Find the total distance.

Express your answer to three significant figures and include the appropriate units.

vo = 86 km/h = 23.89 m/s

vf = 58 km/h = 16.11 m/s

r = 0.90/2 = 0.45 m

wo = vo/r = 53.086 rad/s

wf = 35.8 rad/s

wf^2 = wo^2 + 2*alpha*theta

theta = 78 rev = 78 * 2pi rad

alpha = (wf^2 - wo^2)/2*theta

alpha = -1.568 rad/s^2

part b )

wf = wi + alpha*t

t = 35.8 / 1.568

t = 22.84 s

part c )

s = ut - at^2/2

a = alpha*r = 0.705 m/s^2

u = 23.89 m/s

s = 23.89 * 22.84 - 0.705 * 22.84^2/2

s = 361.76 m

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