Problem 8.22 The tires of a car make 77 revolutions as the car reduces its speed

Question

Problem 8.22

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 93.0 km/h to 57.0 km/h . The tires have a diameter of 0.82 m .

Part A

What was the angular acceleration of the tires?

Express your answer using two significant figures.

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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.

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Part C

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

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

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Problem 8.22

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 93.0 km/h to 57.0 km/h . The tires have a diameter of 0.82 m .

Part A

What was the angular acceleration of the tires?

Express your answer using two significant figures.

=   rad/s2  

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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.

t =

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Part C

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

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

s =

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Explanation / Answer

vi = 93 km/h = 25.83 m/s

vf = 57 km/h = 15.83 m/s

r = diameter/2 = 0.82/2 = 0.41 m

we know

w = v/r

wi = 63 rad/s

wf = 38.6 rad/s

theta = 77 rev = 77 * 2pi = 483.8 rad

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

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

alpha = -2.56 rad/s^2 = -2.6 rad/s^2

part b )

wi = 38.6 rad/s

wf = 0

alpha = -2.56 rad/s^2

t = ??

wf = wi + alpha*t

t = 15 s

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