By using static friction coefficient Uk, we assume that the car does NOT go into a skid. The maximum static friction amount is applied to the tires even though they are still turning. That force decelerates the car over a distance X
friction work = Uk*M g X = (1/2) M V^2
X = V^2/(2 g Uk)
V = 51 mph = 74.8 ft/s = 22.8 m/s
Solve for X, using V in m/s and g = 9.8 m/s^2, or V in ft/s and g = 32.2 ft/s^2
A car is traveling at 51.0 mi/h on a horizontal highway
If the coefficient of static friction between road and tires on a rainy day is 0.096, what is the minimum distance in which the car will stop
What is the stopping distance when the surface is dry and µs = 0.603?
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