A car travels along a level road with a speed of v = 17 m/s (about 38 mi/hr). Th

Question

A car travels along a level road with a speed of v = 17 m/s (about 38 mi/hr). The coefficient of kinetic friction between the tires and the pavement is K = 0.53.

(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?
m

(b) If the same car is traveling downhill, along a road that makes an angle = 15°, how much does the stopping distance increase?
m

(c) What is the stopping distance on an uphill road with = 15°?
m

Explanation / Answer

a) when barkes are applied, kientic friction acts on car in backwards direction.

friction force f =u N = u mg

and f = ma

umg = ma

a = ug = 0.53 x 9.81 = 5.20 m/s^2

so a = - 5.20 m/s^2

using v^2 - u^2 = 2ad

0 - 17^2 = 2 x -5.20 x d

d = 27.79 m

b) now N = mgcos15

friction f = u N = umgcos15

using F = ma

mgsin15 - f = ma

mgsin15 - umgcos15 = ma

a =(9,81 x sin15) - (0.53 x 9.81 x cos15) = - 2.48 m/s^2

using v^2 - u^ 2 = 2ad

0 - 17^2 = 2 x -2.48 x d

d = 58.19 m

c) uphill,

- mgsin15 - umgcos15 = ma

a = - ( 9.81sin15 + 0.53x9.81cos15 ) = - 7.56 m/s^2

0 - 17^2 = 2 x -7.56 x d

d = 19.11 m

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