Secure https temView offset-next D-79568533 Tine Remaining 01:45 Question 9 Part

Question

Secure https temView offset-next D-79568533 Tine Remaining 01:45 Question 9 Part A A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the distance been the point t te dri slammed on S brakes (hereby locking the wheels) and the point shere the car came to a stop was 28.0m. From a reference manual he delermines hat the coeficient of kinetic icon beween S tires and the road under the prevailing conditions was 0 300.How fast was the car going when the driver applied the brakes? 21 Amis 32.9ms 12.8 mis y Answers Type here to search

Explanation / Answer

Let the mass of car be 'm' kg and the velocity of car be 'v' m/sec. So the Kinetic energy is K.E = 1/2 x m x v2.

The frictional force on the car is Fk =(0.3 x m x g) and the distance through which it acted is S = 28 m.

To make the car stop work done by friction should equal Kinetic Energy.

So(1/2 x m x v2) = (0.3 x m x g x 28) => v2 = 16.8 x g

So the velocity of car was v = (16.8 x 9.8)1/2 = 12.8 m/sec   (4th option)

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