Acar is traveling at 50.0 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?

(b) what is the stopping distance when the surface is dry and the coefficient of friction is 0.600?

Answer

given,

speed of the car = 50 Km/h

                            = 50 x 0.278 = 13.9 m/s

a) normal force acting on the car

   force along x-direction

friction act along x-direction in opposite direction

a) using equation of motion

     v² = u² + 2 a s

final speed, v = 0 m/s

     0² = 13.9² - 2 x μg x s

      2 x 0.1 x 9.8 x s = 13.9²

      s = 98.57 m

minimum distance in which the car will stop is equal to 98.57 m

b) When coefficient of friction is 0.6

 v² = u² + 2 a s

final speed, v = 0 m/s

     0² = 13.9² - 2 x μg x s

      2 x 0.6 x 9.8 x s = 13.9²

      s = 16.43 m

stopping distance when coefficient of friction is 0.600 is equal to 16.43 m

explanation:

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