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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a 9.0 -km trip in two situations: $(a)$ the stations at which the trains must stop are 1.8 $\mathrm{km}$ apart $(a$ total of 6 stations, including those at the ends); and $(b)$ the stations are 3.0 $\mathrm{km}$ apart $(4$ stations total). Assume that at each station the train accelerates at a rate of 1.1 $\mathrm{m} / \mathrm{s}^{2}$ until it reaches 95 $\mathrm{km} / \mathrm{h}$ stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at $-2.0 \mathrm{m} / \mathrm{s}^{2}$ . Assume it stops at each intermediate station for 22 $\mathrm{s}$ .

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Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

University of Sheffield

Lectures

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Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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So here we need to separate it into three different sections of the problem. Essentially Ah, we have the acceleration part, the constant speed. Second, the second part would be the constant speed. And the third part would be the deceleration, so we can find the maximum speed of the train. Ah, this is gonna be 95 kilometers per hour. Let's just convert real quick. One multiplied by one meter per second for every 3.6 kilometers per hour. Ah, this is going to be equal to 26.39 meters per second. At this point, we're going to we're going to, um um evaluate the acceleration phase so we can say that VI final. So let's say part one, uh, the final equals the initial plus 80. We know that the initial velocity is going to be equal to zero, So T is going to be equal to 26.39 our final velocity and our maximum velocity of the train divided by our acceleration of the train. So 1.1 meters per second squared. Ah, this is giving us 23.99 seconds. So let's not round until the very end. Let's find the displacement during this part. So we can say that Delta X would be equal to be ex initial t plus 1/2 A T squared. We know that the ex initial is zero, so we can eliminate that term. And this is gonna be equal to 1/2 times 1.1 meters per seconds squared, multiplied by 23.99 seconds. Quantity squared. This is gonna equal 316.5 meters. Now we need to find the ah me to evaluate now part uh, the deceleration phase. So here we can say that for the deceleration phase, well, say acceleration for part one and then, um, for part three, essentially, because this comes after the constant speed face, we can say deceleration. Let's find how long it takes to decelerate. So we can say that the final equals V initial plus 80. We know that here now the final is zero because they're coming to a stop and t would be equal to negative 26.39 meters per second, divided by our maximum deceleration. So this is not the ste acceleration. This would be negative. 2.0 meters per second squared. This is equaling approximately 13.20 seconds. And then let's find the distance traveled during this phase so Delta X would be equal to be ex initial T plus 1/2 A T spared. This would be equal to 26.39 meters per second, multiplied by 13.20 seconds plus 1/2 times negative, 2.0 meters per second squared multiplied by 13.20 seconds. Quantity squared, and we find that here Delta X is gonna equal 174.1 meters. Now, let's get a new workbook. Um, we want to find the total elapsed time, so t of acceleration plus t of deceleration would be simply equal to 23.99 seconds plus 13.20 seconds and this is equaling 37.19 seconds and then for the total distance, this would simply be able to 316.5 plus 174.1. This is equal in 491 meters now. Four part, eh uh, we want to find rather we know that the stations are spaced 1800 meters apart or 1.8 kilometres s o. We know that there's a total of 9000. We can say we could say Ah, stations are 1800 meters apart so we can say 9000 meters divided by 1800 meters. This would equal five inter station segments. We'll call it and at this point we know that a train making the entire trip would have five of these segments. Um and we know that we have four stops of 22 seconds each at the stations, so we can say four stops of 22 seconds each. Um, 491 meters again is traveled during acceleration and deceleration. So 18 weaken, say 1800 meters minus 491 meters. Uh, this would equal 13 09 meters at an average speed. The average speed where we can say average velocity. So at an average velocity equal to 26.39 meters per second, the maximum velocity off the train s so we can solve for the time and we can say that ah, time would then be equal to 13 09 meters divided by 26.39 meters per second. Ah, this is giving us 49.60 seconds. Um so the total inter station segment so we can say tee total would be equal to 37 0.19 seconds plus 20 rather 49.6. Ah, this is giving us 86.79 seconds. At this point, we can say that we have five of these segments and then four stops of 22 seconds each. Therefore, the tow, the time we can say for 0.8 kilometers. All right. Sorry. The time for 1.8 kilometres. My homes kilograms, one penny corners would be equal to five times 86.79 seconds plus four times 22 seconds. This is giving us 522 seconds. So, uh, essentially that would be your answer for part eight. This would be the total time now for Part B. However, we have this, we only have three stations because now they're 3000 or three kilometers apart or 3000 meters apart. So again, 9000 meters divided by 3000 meters. This would be three inter station segments and we can say that we have two stops 22 seconds each. So essentially doing the exact same thing however we're going to we're trying to solve for the 3000 meter apart stations. Um, again, we say we can say that 3002 meters minus 491 meters. This is giving us 25 09 meters at a constant velocity. Eso let's find that tea so T would then be equal to 25 09 meters, divided by the maximum speed 26.39 meters per seconds on then this would be equal to 95.7 seconds. And so once one segment would be equal to 37.19 seconds plus 95.7 seconds. This is giving us 132 0.3 seconds. And at this point, we can solve for the total time, so would be equal to three times 132.3 seconds. Given that we have three inter station segments and then we have two stops at 22nd at 22 22 seconds each. So our total time here would be 441 seconds. This would be your total time for part B. That is the end of the solution. Thank you for watching.

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