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Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 15.1...

Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 15.1 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.560. What is the speed of the automobile after 1.33 s have elapsed? Ignore the effects of air resistance.

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Answer #1

It is given that the automobile is moving with 15.1m/s so its initial speed u is 15.1 m/s. And finally it stops so its final speed v is zero. Here, we apply the following equation of motion

v^2=u^2-2 as...(1)

Here, s is the distance traversed and a represents the acceleration that needs to be determined by considering the effect of frictional force. We know that the frictional force f acts opposite to the external force F and the automobile will stop if the two forces balance each other so

F=f

ma=mu mg

Here, m is the mass and mu is the coefficient of friction.

a = ng (2)

Putting all the values in Eq. (1)

0=u^2-2 mu g s

Solving for s

s=rac{u^2}{2 mu g}

15.20.77m 20.560 x 9.8

The speed vt at time t=1.33s can be given as

20.77 = 15.62m/s t 1.33 2

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