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Consult Multiple-Concept Example 10 in preparation for this problem. Traveling at a specd of $16.1 \mathrm{m} / \mathrm{s},$ 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 $0.720 .$ What is the speed of the automobile after 1.30 $\mathrm{s}$ have clapsed? Ignore the effects of air resistance.

6.93 $\mathrm{m} / \mathrm{s}$

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in this question. There are three forces acting on the car to wait force than normal force on the frictional forces, and the initial velocity off the car is 16 16.1 meters per second. Given that information, we have to calculate what is the velocity off the car after 1.3 seconds, given the frictional force. So we have to begin this question by calculating the acceleration off the car. And how can we calculate acceleration, given some forces we can use Newton's second law, Newton's second low, tells us that the net force that acts on the current is equals to the mass off the car times its acceleration. We will apply Newton's second law on the horizontal axis, which I would choose to be pointing to the right so everything that is pointing in the right will be positive and everything there is pointing to the left will be negative. As a consequence, then the net force on that horizontal axis is given by the frictional force only so minus the fictional force is the course, the mass off the car times next generation off the car, but the frictional force in the situation where the object is already moving is given by the kinetic frictional coefficient times the normal force. These is he goes to the mass off the car times in six of the relation. It happens that in this situation, the normal force, it's exactly counterbalanced by the weight force. And then we have that the normal force is equals to the weight force so that here we get my nose different, you know, more efficient times, the weight forced. Is it close to the mask off the car? Times acceleration off the car and the weight force is given by the mass off the car, times acceleration off gravity. So we get these equations. We have mass on the left side of the equation on on the right side of the equation. And it's multiplying both sites so we can simplify it out to get minus me. Okay, times acceleration of gravity is the coast acceleration off the car. The acceleration off gravity near the surface off the earth is ecos to 9.8 meters per second squared, then bringing the values that were given in the problem. We get minor zero point Saturday two times 9.8 is the acceleration off the car and these results in an acceleration off minus 7.0 56 meters per second squared. Now we can calculate what is the velocity off that car? Given its initial velocity on its acceleration we get, we have that the velocity off the car after an interval of time don t is he goes to the initial velocity off that car plus the acceleration off the color times that time interval that blogging in the violence that were given by the problem, we get that the velocity off the car after this 1.3 seconds is the close to the initial velocity of the car, which is 16.1 meters per second minus 7.0 56 which is the acceleration times. Think of off time off 1.3 seconds. This minor sign is because the acceleration is negative. Then we get that the velocity after 1.3 seconds is approximately 6.9 meters per second and this is the final answer off the problem

Brazilian Center for Research in Physics