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Police investigator observes skid marks 30 𝑚 long left by a 1200𝑘𝑔 car. The car skidded to stop on a concrete highway having acoefficient o...

Question

Police investigator observes skid marks 30 𝑚 long left by a 1200𝑘𝑔 car. The car skidded to stop on a concrete highway having acoefficient of kinetic friction with the tires of 0.80. Estimatethe cars speed at the beginning of the skid.

Police investigator observes skid marks 30 𝑚 long left by a 1200 𝑘𝑔 car. The car skidded to stop on a concrete highway having a coefficient of kinetic friction with the tires of 0.80. Estimate the cars speed at the beginning of the skid.



Answers

Police investigators, examining the scene of an accident involving two cars, measure 72 -m-long skid marks of one of the cars, which nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pavement is about $0.80 .$ Estimate the initial speed of that car assuming a level road.

For this problem on the topic of Newton's laws of motion, we are told that at the scene of an accident that involves two cars, We can measure a set of skid marks, which is 72 m long, for one of the cars. This car nearly comes to rest before the collision. If we know the coefficient of Kinetic friction between the rubber of the tires and the pavement to be 0.8, we want to estimate the initial speed of the car, assuming that it's traveling on a level road. Now, we'll assume that the kinetic friction is the net force which is causing the deceleration of the first car. So if we draw a free body diagram for the car, we can first see that the normal force exerted by the road in the car ex vertically upward. This normal force we'll call F. N. They forced due to friction. If the car is moving to the right acts in the opposite direction to the motion, and we'll call this force F subscript F. R friction and the weight of the car X vertically downward to the center of the earth. And that's M times G U M. Is the mass of the car. Now, obviously assuming the right direction to be positive and the direction of motion of the skidding car, we also know there's no acceleration in the vertical direction, so therefore the normal force has the same magnitude as the weight of the car. F N is equal to MG. We can apply Newton's second law to the X. Direction or the horizontal direction, which gives us the following. The resultant force in the horizontal direction is only the friction which is opposite to the direction of motion, so that's minus if friction. And by Newton's 2nd law, this must equal to M. Times A. So this tells us that the frictional force, which is the coefficient of kinetic friction times the normal force F. N. Which we know is equal to M. U k minus mu k MG, which is the weight of the car. By Newton's 2nd law must equal to M. A. So we can simply see then the masses cancel, and we are left with the acceleration A. To B minus UK times G. So then we can use the kind of magic equation. The squid minus the north squared Is equal to two a. Into X minus ex not. And we can find the initial speed of the car. We know the final speed will be zero. So rearranging, we solve for the initial speed V note and we know what is the square root of the final speed. We all squared -2 times the acceleration a, multiplied by the displacement or the distance over which the car takes to come to rest. All of these values are known. So we can find this initial speed. This is the square root. The final speed is zero, so zero square zero minus two times acceleration a, which we showed above is minus mu k times G times the displacement x minus X, not. And so now if we substitute values into this equation, this becomes the square root of two Times The coefficient of kinetic friction between the tire and the road, 0.8 And the acceleration due to gravity 9.8 m/km2, Multiplied by the displacement of the car before it comes to rest 72 m. And so calculating, we get the initial velocity of the car To be 34 meters the second.

I am driven. Here it is, given that car sliding towards stop. But in the states off 200 ft, given coefficient of friction. Toby 5 to 7. We have to find initially speed, mhm and a time office Skip. Mhm. Submission off effects will be Puerto and force off. Friction has got a minus Boom and tow energy, That is I m a X. So from here. Yeah. Hey, X will be minus mu into G, which is to find us d v X, upon duty. Yeah, so D v X will be minus newsy into duty integrating it. So be X is called minus nudity plus constant off integrations. Even eight, please call 20 VX is called Toby. Not so. You will get the eggs, Toby. Minus nudity plus Bill. Not yeah. Come on now. Did it Calibrated eyes. Do you accept one dated Toby Cross. Multiply it. Integrity. I mean, after the second distance. His ex ex bill B minus mu deity square by two. Plus we not in turkey say a question too. Yeah, we have final state ERT. Which access got to l on the X. Yeah. Who? The years of point it is called zero Substitute the value. Any question But we will get Jiro. Is cult o minus music into thine plus B not well, that is time off. Stopping well. We were not upon duty substituting in equation toe X. You will get minus mu ge by two in tow. We not by music who is squared. Plus we not upon me not by music and X is given any. So l is cultured Be not mhm square by cool music. So finally, initial velocity of that car is Rudolph twice Mutlu substitute developed you will too. 179.8 and took length is giving 200 ft. Mhm. Three point 25 16 m. So converting into it my forever it becomes 64.7. It is to me five point 234 m per second or it will be 64 point My poor lover. Yeah, Uh huh. And time, Peter. Time off. Stopping building. Be not my muting. That is 5.234 upon 0.7 to 9.8. It is to be 4.21 2nd. That's all. Thanks for watching

So let's start a free body diagram off the car first, and then we can talk about how to solve the problem. So this is the car which skid on the road. Now. If we assume that the kind of dick friction is the Net force that caused the car to decelerate, then we'll see that. Ah, if we consider from left to right direction as the direction or, uh, as the distance the car traveled, then the frictional force will be acting opposite to that. Let's call that if some f s then there were the MG, which is acting vertically downwards and to counterbalance that will have if in, or the normal force, which is acting upwards. And that's it. There is no other force that's acting on the system. No, we see that in the X direction. Let's call that FX. If we apply, um, Newton's second law, the force on FX direction or the porcelain extraction will be negative F off offense, which is equal to M times eight, where we're considering from left to right as positive X and upward direction as positive light. So since we're considering kinetic friction here, then f off F s must be equal to mu k times f n where n is tthe e normal fours. And since we since, uh, there's no motion in the vertical direction, then the normal force must be equal to M G. So f n was illegal tow. Am time. See? No. If we put all these together, we'll see that we have negative mu k times mg, which is equal to m times, eh? On we can get you tough, the EMS, and then we have a which is equal to negative mu K Times, Chief. No, For the initial velocity, we can apply Question 2.12 c and see that, um, Reece squared minus V zero squared, which is equal to eight times x minus X not where x More ex wives ex not, is the displacement and ah ah, we not is the initial velocity. So that's the in nation velocity. These great is the final velocity. And finally, is the acceleration or, in our case, the situation. Excuse me. So we see that the car came to a stop. That means we can get it off the final velocity because, uh, there's no because the final velocity pseudo So we have be known squared. And if we solve for Venus could receive that equals, um, the square where we got rid of this. So we have negative to a times x minus x not. And again explain is excellent is the displacement if we put in here which from the previous page we see that easy, Pollux. Negative. Mewes times. Sorry to nooky nooky times G. So, um, we put that here on DA then. Since there are two native since we got rid of that. Now, if we put all the given values, we see that displacement was 72 meters. Um G's 9.8 meters per second squared uk was pointed and we multiply two with the whole thing. The final velocity becomes 30. Sorry. The initial velocity because 34 meters per second. Thank you.

Hi. In the given problem, there are two cars out of which one off the car is moving. Do you west? These are the directions. This is not This is south here. This is east and this is west one off the car is moving the west Cozma's is Ambon which has given us 1100 kg and it is moving. Suppose with a speed off human which is not given to us. Another car is moving due north and it's the speed is also missing. Let it be you too. And the mass of this car his M two which is given ass 1300 kilograms the collide her at the origin and stuck together. Then the combined mass moves at an angle off 30 degree or we can say in a direction. 30 degree north off west. This is a combined mass which is m one plus m to on the speed of this combined mass. Let it be we No. After the collision, the combined mass moves through a distance as which has given us 17 m and the coefficient of friction over the road is given as 0.80 So the force of friction new into em into G will be equal to I am in tow A where a is the retardation produced due to the friction into this combined mass. As for Newton's second law off motions. So the magnitude off this retardation comes out, Toby New times off G or we can say this is 0.8 multiplied by 9.8 m per second square, which comes out to be 7.84 meter per second square. And as it is our retardation, So we assign a negative sign to it. No, using the third equation of motion which says the final velocity we F square is equal to the initial velocity V I square plus two a s here the final velocity of the combined must should be zero for we I This is not known to us minus to a plus to a and four a. This is minus 7.84 meet or poor. Secondly, square multiplied by the distance trouble which is 17 meter. So the value off this speed which is actually the speech off the combine muss off the cars off Prada collision. It comes out, Toby, we is equal toe Squire route off the product off this two with 7.84 and with 17 and it comes out to be 266 point 56 m per second. Or we can say this is we is equal to 16.3 meter per second, which is the speed off the two cars achieved. Africa Coalition Not using conservation off linear momentum during the collision, First of all, along X axis So along X axis means in a direction from east to west. It is assumed to be X axis, and the direction on our north to south is assumed to be y axis. So along X axis, the first car was moving, so initial momentum will be off. First car and final momentum will be off the combined mass off the two cars, so M one plus M two and the component off final velocity along X axis, which comes out to be we cause 30 degree. Hence, using this you one comes out to be for everyone Placenta. This is 1100 plus 1300 kilogram multiplied by V, which is 16.3 m per second in tow, cause sign off 30 degree divided by M one, the mass of the first car, which is 1100 kilogram. So this value off you one here comes out to be 30.8 meter for second. No in the similar manner. If he use a conservation off linear momentum along, why access as the second car was moving along y axis means due north. So I am too. You do the initial momentum off the second car along y axis visible toe, the component off momentum off the combined mass off the car along by excess means and one plus M two into we sign 30 degree so you, too, comes out to be 1100 plus 1300 into 1.69 30 which is half divided by empty, which is 1300. And finally the magnitude of this uto comes out. Toby. 15 0.0 meter per second. This is the speed. This was the speed. Initially, speed off the second card. No, the maximum allowable speed is given us or speed limit. We can say it has given us just 70 kilometer for art, so if we convert it into meter per second, we multiply it by fired by 18 and it comes out to be 19.44 meter per second. So when we compare this speed limit with the two cars, it is obvious that the first card actually crossing the speed limit before collision hands we can say the lighter car, the car having the less mass, the lighter car Waas increasing. It's a speed beyond limit. Thank you.


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