How far will the car slide once the brakes are applied and it comes to a stop in the following case:
A 14,700 N car is traveling at 25 m/s and then it comes to a stop when the brakes are applied. The average braking force due to friction between the tires and the road is 7100 N.
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The car is initially traveling at 25 m/s. When the brakes are applied there is a resistive force that leads to a decrease in the velocity of the car and it eventually comes to a stop. The weight of the car is given as 14700 N. The average braking force between the tires and the road is 7100 N.
The mass of the car is 14700/9.8. The deceleration due to the braking force is 7100/ (14700/9.8) = (7100*9.8)/14700 m/s^2 = 71/15 m/s^2.
The distance traveled by the car while its velocity decreases from 25 m/s to 0 m/s due the deceleration of 4.73 m/s^2 is given by :
d = (v^2 - u^2)/2a
=> (0 - 25^2)/2*(-71/15)
=> 66.02 m
The car slides a distance of 66.02 m before it comes to a halt.
the brake are applied to a rightward moving car and it skids to a stop label the forces
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