A car travels at 54 mi/h when the brakes are suddenly applied. Consider how the

Question

A car travels at 54 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect to the road; otherwise, when the tires roll, normally the point where the tire contacts the road is stationary. Assume the coefficients of friction between the tires and the road are

?xskid ? ?xno skid = m ?xskid ?xno skid = %

A car travels at 54 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect to the road; otherwise, when the tires roll, normally the point where the tire contacts the road is stationary. Assume the coefficients of friction between the tires and the road are ?K = 0.80 and ?S = 0.90. Calculate the distance required to bring the car to a full stop when the car is skidding. Calculate the distance required to bring the car to a full stop when the wheels are not locked up. How much farther does the car go if the wheels lock into a skidding stop? Give your answer as a distance in meters and as a percent of the nonskid stopping distance.

Explanation / Answer

Part A)
First 54 mph converts to 24.1 m/s

F = ma

F = uFn = ma

umg = ma (mass cancels)

a = ug = (.8)(9.8) = 7.84 m/^2

vf^2 = vo^2 + 2ad

0 = 24.1^2 + (2)(-7.84)(d)

d = 37.0 m

Part B)

a = (.9)(9.8) = 8.82 m/s^2

0 = (24.1)^2 + 2(-8.82)(d)

d = 32.9 m

Part C

The difference is 37 - 32.9 = 4.1 m

37/32.9 = 112.5%

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