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1.Car A experienced engine failure the moment it enters a 2-km highway that has a constant grade of 4 deg throughout its span (4 deg. inclination). Point C is the l...

Question

1.Car A experienced engine failure the moment it enters a 2-km highway that has a constant grade of 4 deg throughout its span (4 deg. inclination). Point C is the location where car A experienced failure. When car A is towed by the truck it gains an acceleration of 3 m/s^2, at the same instance, car B is passing the same highway with a constant velocity of 40 km/hr and is 100m away from point C (measured along the pavement). a) At what time will car A overtake car B? answer in unit of seconds.b)

1.Car A experienced engine failure the moment it enters a 2-km highway that has a constant grade of 4 deg throughout its span (4 deg. inclination). Point C is the location where car A experienced failure. When car A is towed by the truck it gains an acceleration of 3 m/s^2, at the same instance, car B is passing the same highway with a constant velocity of 40 km/hr and is 100m away from point C (measured along the pavement). a) At what time will car A overtake car B? answer in unit of seconds. b) How far from point C (along the pavement) will the two cars meet? answer in unit of meters. c) What is the velocity of car A at the instance it overtakes car B? answer in unit of km/hr.



Answers

Two cars cover the same distance in a straight line. Car $\mathrm{A}$ covers the distance at a constant velocity. Car B starts from rest, maintains a constant acceleration, and covers the distance in the same time as car A. (a) Is car B's final velocity smaller than, equal to, or greater than car A's constant velocity? Explain. (b) Is car B's average velocity smaller than, equal to, or greater than car A's constant velocity? Why? (c) Given a value for the time to cover the distance and the fact that car B starts from rest, what other variable must be known before car B's acceleration can be determined? Provide a reason for your answer. Problem Both cars cover a distance of $460 \mathrm{~m}$ in $210 \mathrm{~s}$. Assume that they are moving in the $+x$ direction. Determine (a) the constant velocity of car $\mathrm{A},$ (b) the final velocity of car $\mathrm{B},$ and $(\mathrm{c})$ the acceleration of car $\mathrm{B}$. Verify that your answers are consistent with your answers to the Concept Questions.

So first we have that the velocity of for party the velocity of car eh is equaling 12 over six. So this is equaling two meters per second. We can then say that with an initial X value of 20 meters, we can say that the exposition of car, eh when tea is equaling, four seconds would be equaling 28 meters. So we can say then that 28 meters equals 12 meters per second multiplied by t Uh, plus 1/2 times the acceleration of carby times t squared and here T is equaling 4.0 seconds where the acceleration of B is an equaling negative 2.5 meters per second squared. So that'll be four part eh Now, for part B, we can say here the question is that using the value obtained for the acceleration for being part A are there other values where the position of car A is equaling the position of carby? And so what this essentially means is that 20 plus two tea must be equaling 12 t plus 1/2 the 1/2 times acceleration of B times T squared. Now we know that if tea is equaling 4.4 point zero seconds. This is gonna hold true now, Are there any other? Are there any other values of tea where this statement holds true? Given that again, the acceleration of B is negative. 2.50 meters per second squared. Now here there are two distinct roots. Unless the discriminative is zero. And here the discriminatory would be the square root of 10 squared minus two multiplied by negative, 20 times the acceleration of B and this is actually equaling zero. So that means that there only is one route because our discriminative equaling zero and that one route would then be, of course, t equaling 4.0 seconds. Now for part C, they want you to sketch. And this would be the sketch of the graph theory of her ex axes honor we ever X position under y axis in her time in our X axis. Now, this would be for part C for part D. Now again, we only care about riel roots. So that means that 10 squared minus two multiplied by negative 20 multiplied by the acceleration of B s has to be greater than or equal to zero. Now if the absolute value of the acceleration of beat is greater than five over two, then there are no real solutions to the equation. So if this occurs no riel solutions Ah, the cars are never sign by side and weaken Safer part E here, part E 10 squared minus two times negative 20 multiplied by the acceleration of be here This is equaling. It is is greater than zero rather So we have to really roots. Or we can say the cars our side by side at two different times. And so that would be what essentially, what that means for part E. That is the end of the solution. Thank you for watching.

So in this given problem there is a car of must 1200 Kg ingen is providing it Um, acceleration force, our faith, we can say and there is tense force of 500. Newton is always present. So it will, it must have to be nullified by this exciting force. So F. A. Now developed the graph is of velocity and for the first two seconds each reaches after the valley for so we can see the next lesson for the first two, second is to city by time. And here velocity is four and at the last part again, velocity by time. So you can write it as -4. The FAA is there for uh that is calculate the fourth, accelerating the car in the first two seconds. So this will be the acceleration that is gained of course will be M. A. Since it is exciting with less are this also must be nullified, which is 1200 in two, two Plus 500. So 2900. So this much amount of force should be applied by the engine. No, the part B is singing about power, average power. The average we can see integration of FBS upon time. So integration FDSFS 2900 integration. Do you simply in the strangle? Is this area under the car of velocity times. So it is simply house into this into high upon time which is to so canceling. We see the value is 5800 jewels or 5.8 kilowatt in the part C. We think it is asked about The force pushing the car forward for the next 10 seconds For the next 10 seconds. F will be I am of a plus alpha. So here is zero. Only effort needs to be recovered cancer. 500 Newton in the part D. Yes, He's asking calculate the power delivered by injuring those 10 seconds of power delivered In those 10 seconds will be f. d. s. upon time which is again forced into area area at this time is time into velocity which is for By time you can see from this about half time is canceled and it comes out with 200. Jeweller took a lot, you can see it now the last part it is calculate the backing force in the last four seconds. So breaking forth at that time. Breaking force. Yeah. Yeah. Mhm. Well played the baking fourth in the last four seconds of the motion. So this party is a tricky one because here uh this value 500 will be subtracted instead of adding because it is helping in breaking so minus off. Or we can say one in 2 12:00- of 500 700 newton. No the last part it is saying described the energy transformation that has taken place in 16 seconds, energies transformation. If you know the exhalation, what has happened is first of all, whenever the velocity is increased it means kinetic energy is increased in the system so it will be. 1st, Kinetic energy was increased, then the velocity was postures and the energy was also constant and then kinetic energy has decreased. Since the potential energy is not changed, kinetic energy, rough potential energy is always zero because the height of the car is not changing is zero all the time. So kinetic in Nigeria the equals two total energy in this case. So this is the stuff time was his energy. Yeah, yeah.

So part of this problem is asking which position is larger for really small values of tea and a way you can do this by plugging in really small value of tea such as point zero one and then just crunching the numbers on your calculator and seeing which one of these guys is larger. Or you could notice that we have larger powers of tea for B. Then we do for a And so if we plugged in something like one one hundredth of a second these numbers, they're going to be really small because one one hundredth square is a really small number and one one hundredth. Cuba's an extremely small hunger. Also, we have an A minus sign here, so we're going to be subtracting too small numbers, whereas for this one we're adding two bigger numbers. And so shortly after Ciro tea is equal to zero. XB is going to be smaller. But then you'LL see if tea is equal to one hundred seconds, then these higher power start toe make B be much larger than a and so I just like to solve this by intuition. And so, by the reasoning, I just said, Hey, is a shorter distance away just after he's equal zero Herbie's asking, At what time are the two distances equal? And so to solve this, we just equate our two formulas for the existence except Annex B. And so for eggs, eh? I have this and I'm equating it to Bixby, which is this for these problems where they give you a bunch of constants. I find it easiest to just leave them in terms of these parameter at Alpha Beta Gamma Delta rather than playing them in. It just makes it go a little quicker. At this point, we can cancel T. And then we have a form for a quadratic equation so that we can use the quadratic formula. The standard form of that equation is this bait a scammer Times t yourself is equal to zero. And so contract formula tells us that tea is equal to game on lines Beta posted, last square room of beta minus, game squared minus four. Don't Alfa. And this is all over two times a which in this case, is too deep. Now, at this point, you do actually have to plug in what thes Constance represented. I wrote written them down here. These four numbers here where you do that and crunch the numbers, numbers their calculator. You get that tea is equal to two point two seven five seconds and he is equal to five point seven two five seconds. And so these are the two times where the two X values for NPR equivalent and that's Barbie. Percy, Ask us at which point is the distance between them not changing. And so the way I didn't this is I wrote down the equation for the distance between them, which is equal to X a minus six B, but accident expert already have forms. And so, playing in those forms like it Al Fatih Oh, spade A T squared minus camel t squared pastel Tha t cute. This is a function of T, and it gives us the distance. If you plug in a value of tea and crunch the numbers, you'LL get the distance between A and B. We won't know when this is not changing, and so we have to take extra motive and set that equal to zero. So the derivative I'm just going to denote by distance prime is equal to Alfa plus to bait a T minus two Gamaty plus three Delta t spurred. And we need to set this equal to zero to find out when it's not changing. And so when we set that equals zero, we once again get a quadratic because we have t squared and then we have a large linear term here and then our constant term. And so we can just plug this into the quadratic formula again. I'm not going to write out the whole steppe at this point, I'll just go with the answer because this is just like part be at this point, but two solutions for tea R t is equal to four point three three seconds and he is equal to one point zero seconds. So these were the two times where the distance between Ambi is not changing and that is the answer to Percy party. Ask us when the accelerations are equivalent and so we have to start taking some directors of our position functions. So just recall the exes. Alfa T possibilities word. To get acceleration, we have to first get the velocity We'LL get the velocity by taking the derivative of the position So the glossy of a is Alva just drew that plus to bait a tea which is the derivative of this trip. And then from that we can get the acceleration of A Which is this the truth about this? He often goes away, and we just have to me to beta, we're going to do something for being We're called at X is equal to gamma t squared minus delta t cubed. Taking her this because it's the velocity. And so it is to gamaty minus three Delta T squared and then taking a tour of this gives a sixth generation of B, which is to gamma minus six Delta T. Now we want to find out at what times? These are equivalent. So we equate them to each other and salt for tea. So equating them gives us this equation. We can cancel two. I'm also going to bring this term over here. I'm gonna bring the beta over here. Whenever I do that, I get three. Delta T is equal to gamma minus beta, so tea is equal to gamma minus beta over three delta and you have to plug in the constant values which were given here. Into this expression, we do that you get a T value of two point six seven seconds and I see a problem


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