A car travels along a level road with a speed of 25m/s. The coefficient of kinet

Question

A car travels along a level road with a speed of 25m/s. The coefficient of kinetic friction between the tires and the pavement is 0.55.  a.) If the driver applies the brakes and the tires "lock up" so that they skid along the road, how far does the car travel before it comes to a stop?  b.) Is the same car is traveling downhill along a road that makes an angle of 12 degrees, how much does the car's stopping distance increase?  c.) What is the stopping distance on an uphill road with an angle of 12 degrees?

Explanation / Answer

A car travels along a level road with a speed of 25m/s. Thecoefficient of kinetic friction between the tires and the pavementis 0.55. a.) If the driver applies the brakes and the tires"lock up" so that they skid along the road, how far does the cartravel before it comes to a stop? 1/2m*25*25=0.55*9.8*m*s calculating s=stopping distance=57.97m b.) Is the same car is traveling downhill along a road thatmakes an angle of 12 degrees, how much does the car's stoppingdistance increase? acceleration down the hill=a=g(sin12-0.55*cos12)=-3.233m/sec2 stopping distance=-v2/2a=-625/-3.233=193.31m c.) What is the stopping distance on an uphill road with anangle of 12 degrees? acceleration up thehill=a=g(sin12+0.55*cos12)=7.3096m/sec2 stopping distance = -v2/2a=-625/7.3096=-85.50m=magnitudeof stopping distance=85.50m

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