The coefficient of static friction between the tires of a car and a horizontal r

Question

The coefficient of static friction between the tires of a car and a horizontal road is ?s = 0.315. If the net force on the car is the force of static friction exerted by the road, what is the magnitude of the maximum acceleration of the car when it is braked? What is the least distance in which the car can stop if it is initially traveling at 44.0 m/s?

Explanation / Answer

A car is traveling at 53.0 km/h on a flat highway. a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? ______m (b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.600? we can use vf^2=v0^2 + 2ad to find stopping distance where vf = final velocity =0 v0 = initial velocity = 14.72m/s a = acceleration d = distance traveled = stopping distance we can find the acceleration knowing that F = ma, and in the case of friction, f = u mg, so that m a = u m g or a = u g in this case, a =- 0.1 x 9.8m/s/s = -0.98m/s/s the acceleration will be negative since it opposes motion therefore, 0=(14.72m/s)^2 +2(-0.98m/s/s) d => d = 110.6m if the coefficient of friction were 6 times greater, the accel would be 6 times greater and the stopping distance would be 1/6 as much, so it would be 18.4m

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