Physics Motion Math. (Read objectives on screen.)

Physics 302 - Motion Math (Read objectives on screen.) Welcome back. When we ended the last program, your teacher gave you some motion graphs to interpret. For each section, you were to describe the motion of the object, as specifically as you could. So let s see how you did. (graph on screen) Let s look at graph one. The very first thing you must do when you are asked to interpret a motion graph is to look at the variable on the y-axis. Time will always be on the x-axis, so the other variable is very important. This graph has position d on the y-axis. And you know that as position changes, it s called delta d or displacement. The slope of each line represents displacement over time or velocity. In section a-b, as time progresses, the object s position increases, so it s moving forward. But the slope changes and gets steeper, so the velocity is increasing. This is a form of accelerated motion. You can t say more than this. You can t know if the motion is uniformly accelerated without a velocity, time graph. In section b-c, the object is still moving forward, but this time the graph is a straight line, which means that the slope or velocity is constant. So the motion is uniform. In c-d, the object is still moving forward, but the graph is a curve, meaning that motion is accelerated. This time, the slope decreases as time progresses. If you have trouble seeing this, imagine trying to climb the hill. It gets easier as you go along on this one. Since the slope, or velocity, decreases, the motion is decelerated. You can also call it negative acceleration. The graph, d-e, is a flat line. This shows that the object does not change position and has no velocity. It is not in motion. From e-f, as time progresses, the object s position decreases. Since the graph is a straight line, the motion is uniform. But it is going backwards. You could also say that velocity is negative. How did you do? Remember: NO SURFING. Don t skim the surface without thinking. Two graphs can look the same, but describe completely different types of motion. Watch that y-axis and decide what the slope represents before you look at each section. (graph on screen) Now we ll switch gears and look at velocity versus time graphs. Since rise over run gives us change in velocity over time, the slope of each graph will tell us what s happening to acceleration. 1

Let s start with section g-h. Velocity increases as time progresses, so the motion is accelerated. And since the slope is a straight line, acceleration is constant, so the best description is uniformly accelerated motion. In h-j, the line is flat. This does not mean that the object is motionless, because it has velocity. It just means the velocity is constant. This is uniform motion. Surfers might think that the object is moving backwards in i-j. But they aren t looking at the y-axis. The only thing that is decreasing is velocity. Our object is slowing down. And since the graph is a straight line, acceleration is constant. It s also negative. So the best answer is uniformly decelerated motion. The graph for section j-k looks very much like section g-h. The motion is uniformly accelerated. The only difference is that the slope or acceleration is less. If you noticed that, you re not a surfer. Good for you! If you had trouble with some of the motion graphs, your local teacher may want to give you a chance to practice more before you take a quiz on interpreting these graphs. Your local teacher will stop the tape now, and we ll see you after the quiz. Good luck! I hope you all aced the quiz. It s important for you to understand what velocity and acceleration mean and how they differ. To test your understanding of these terms, how about a physics challenge? (Physics Challenge on screen) You are looking at sketches of three hills. If a ball is allowed to roll down each hill, what will happen to its velocity? Will it increase, decrease, or remain constant? And what will happen to the ball s acceleration? Will it increase, decrease, or remain constant? If you can answer these questions, you really understand the difference between velocity and acceleration. Sketch the three hills and then your teacher will pause the tape to give you time to make your decisions. Remember that your choices for each blank are increase, decrease, or remain constant. Let s see how you did, starting with the first hill. (graphic I on screen) Most of you probably figured that the ball would pick up speed as it rolls down the hill, and you re right. Velocity increases. Acceleration is a little harder. Because the hill gets steeper as the ball rolls down, it will pick up speed faster and faster. So acceleration also increases. 2

(table on screen) To help you understand, let s look at some sample data. You can see that velocity increases, but look at this. During the first second, the ball picked up one meter per second. During second two, it picked up two meters per second, and during the third second it picked up three meters per second. So acceleration is increasing. (graphic II on screen) On hill number two, the ball is still rolling downhill, so common sense tells us that it will pick up speed, as in hill one. Velocity increases. But this hill becomes less steep as the ball rolls down. So the rate of increase in velocity will decrease. So acceleration decreases. (table on screen) How can acceleration decrease while velocity increases? Look at sample data again. You can see that velocity increases as time passes. But look at the rate in increase. During the first second, velocity increases from zero to five meters per second, for an acceleration of five meters per second per second.. During second two, it increases from five to nine meters per second, for an acceleration of four meters per second squared. You can see that the rate of increase keeps going down as time passes. That s why acceleration decreases as velocity increases. (graphic III on screen) Finally, look at the third hill. If you said velocity increases and acceleration remains constant, you are correct. What do you think the sample data will look like? (table on screen) Let s see. As time passes the velocity increases, but notice that during each second, the ball s velocity increases by two meters per second. That makes the acceleration a constant two meters per second squared. So how did you do on the challenge question? Were you a physics surfer? Or did you dive deeper and try to understand the difference between velocity and acceleration? Our goal this year is to turn all of you into divers. And speaking of divers, it s time to dive into some linear motion math. Solving motion problems will require some basic algebra. Since the prerequisite for taking this course is two years of algebra, this review may insult some of you. But we re going to do it anyway, because it s so important. So here it goes. (cartoon on screen) In physics, we use algebraic equations. But instead of x s and y s, we use symbols which represent specific quantities or variables like distance, velocity, acceleration, and time. It is very important that you learn to rearrange equations to solve for one variable. We do this before any numbers are inserted in the equation so that when you re ready to calculate, you punch one set of buttons on the calculator and have the answer. This eliminates errors and lets you round only once. 3

Let s start by having you rearrange some equations. Your teacher will pause the tape and give you a set of equations, with instructions to solve for one variable. Rearrange the equation.. Don t worry about subscripts. They are part of a symbol. When you ve finished, your teacher will check your answers. If you need some work on this skill, we ve put some examples and practice problems on the Peachstar website. Your teacher will show you how to find it. Local Teachers: Turn off tape and give students problem set number one from facilitator's guide. Number one was easy, so let s start with number two. It looks easy, but tricks many students who try to rearrange in their heads. When the unknown variable is on the bottom of a fraction, you must get it to the top. That means you start this problem by multiplying both sides by t. That gives you the equation, v times t = d. Now you can get rid of the v by dividing both sides by v. That leaves the unknown variable, t isolated. t = d divided by v. Now let s try number three. We need to isolate the a. Remember that you want to work your way into the unknown. That means getting rid of outside obstacles, like the term, v delta t before you work on the ½ and the t 2. If you started on this term before the v delta t, that would be like trying to remove the handcuffs from the princess before you got rid of the guards and dogs. It just won t work. So let s look at the v delta t. Since it is added to the term containing our unknown, we remove it by subtracting it from both sides. Now our equation reads, d v delta t = ½ a delta t 2. How do we get rid of the ½?. Since ½ times a is the same as a divided by two, all we have to do is multiply both sides by two. Now all that is left is to divide both sides by t squared, and we have our answer. Number four is the most complex. Don t worry about these subscripts. m1 is just the mass of object one. Since there is not a term added to or subtracted from the term containing our unknown, the first thing we have to do is to get the term containing the d off the bottom of the fraction. That term is d squared. Can we get rid of the square sign yet? No, that s the last thing we ll do. It s like pulling the evil twin away from the princess. First, we have to get both twins out of the basement. We do that by multiplying both sides of the equation by d squared. Now we have to get rid of the F. Since d squared is multiplied by F, we ll divide both sides by F. Now how do we get rid of the square sign? What is the opposite of squaring a number? You ve got it taking the square root. So we take the square root of both sides of the equation and we have our answer. (New set of equations on screen) If you had any trouble with this basic algebra skill, you ll need to practice. Local teachers, this is problem set 1A. (screen fades to black) 4

All right, we re finally ready to do some motion math. We ll start with the simplest kind of motion, which is uniform motion. There is only one basic equation that is used for uniform motion. And that s the definition of velocity. You already know that velocity equals displacement over time. Now some books use the delta signs for displacement and time, but some don t. So we re not going to be picky about it. Using the equation is the important thing. (green chalkboard on screen) Since most motion problems in physics are word problems, you ll need to know some commonly used phrases. Look at these questions. For which variable do you think each is asking you to solve? I ll wait while you write them down. OK. Here are the answers. When a problem asks how far, you are solving for the variable, delta d. How fast means that you want to calculate velocity, v. Do you know why we didn t use the delta sign here? It s because in uniform motion problems, velocity does not change. Got it? And finally, how long means time, not distance. For example, an appropriate answer to the question, How long have you known her? would be six years, not six miles. Now let s try two uniform motion problems together. Let s go over the first problem. Here it is. Anita Break and Earl E. Byrd drive 48 km east. Anita drives at a constant 88 km/h while Earl drives at a constant 92 km/h. How long will Earl have to wait on Anita at their destination? (interrupts) Before we begin, I must point out that the writer of these problems often uses strange names like I need a break and Early Bird. If you think these names are weird, just wait. I just want everyone to know that I had nothing to do with naming the characters in our problems, although I must admit that I sometimes enjoy trying to figure them out. In this problem we want to solve for time for both Anita and Earl. The displacement for each is 48 kilometers. Anita s velocity is 88 kilometers per hour and Earl s is 92 kilometers per hour. Remember that how long means length of time. We want to solve for "delta t" for each. The equation we use for uniform motion is v equals d over t. Since we re solving for t, we need to rearrange the equation. We ve already talked about this. We multiply both sides by t to get our unknown off the bottom, then divide both sides by v. So t equals d over v. Now all we have to do is plug in the numbers and chug out the answers. 5

Anita s time is 0.55 hours and Earl s is 0.52 hours. So Earl E. Byrd is early has to wait on Anita for 0.03 hours, which, by the way, is less than two minutes. Now for significant digits here, you ll have to stop at the hundredths because you re subtracting. Now you need to practice. When you finish, your teacher will go over them with you. Local teachers, turn off the tape and give students problem set number 2 from the facilitator s guide. Now we re ready to tackle problems involving accelerated motion. The only type of accelerated motion we will deal with is uniformly accelerated, because the value of a is a constant number. There are three basic acceleration equations that you will use. We re going to show you how these are derived from formulas you already know. Put your pencils down and watch. This is for your information only, not for your notes. (equation on screen) To derive acceleration equation number one, we start with the definition of acceleration, which is change in velocity divided by change in time. Let me give you an example before we introduce the next equation. If I asked you to calculate your change in weight for the month, how would you do it? Did you say take your weight at the end of the month and subtract your weight at the beginning? Well, in physics we use the terms, final and initial, and we use the subscripts f and i after the v s to get this equation for change in velocity. You can either read this equation in the words, change in velocity equals final velocity minus initial velocity or in symbols, delta v equals v sub-f minus v sub-i. Now we do a little substituting. We replace the delta v in the first equation with its definition from the second equation. Now all we have to do is rearrange this equation, obeying basic rules of algebra, to get our final equation. The equation we have derived is acceleration equation number one, which you will be using to solve accelerated motion problems. Deriving the second and third acceleration equations is a more complex process. But if you want to see how we get these equations, look in your physics book. (green chalkboard on screen) Here are the three acceleration equations you will use to solve problems involving uniformly accelerated motion. Put them in your notes, keeping them to the left of the paper. We ll add something on the right later. 6

We derived number one, final velocity equals initial velocity plus acceleration times time. The second equation is displacement equals initial velocity times time plus one half acceleration times time squared. The third is final velocity squared equals initial velocity squared plus two times acceleration times displacement. Now for the big question. Student 1 Do we have to know these? Student 2 Are we supposed to memorize all three equations? Student 3 I was afraid it would come to this. Student 4 Surely you don t expect us to know this stuff. The answer is No, you don t have to memorize these equations. They will be given to you on every quiz and test. But you do have to use them correctly. And remember that the units must cancel out, so be consistent in the units you use. As long as the four of you are here, hang around. Local teachers are going to pause the tape and give our classroom students some more phrases from acceleration word problems that you ll need to recognize. After everyone has copied these phrases and thought about what specific information that gives us, we ll come back and the four of you can help me go over the answers. Local Teachers: Turn off tape and give students problem set number three from facilitator's guide. We already know, how far, how long, and how fast. Let s go over the other phrases. Jot down the symbols each phrase involves. Student 1 How fast was it going? asks for initial velocity and How fast will it go?, asks for final velocity. Student 2 Sometimes the object will start at rest. This tells us that initial velocity is zero. 7

Student 3 And sometimes the object slows down. This means that acceleration is a negative number. Student 4 Finally, when an object stops, we know that final velocity is zero and acceleration is negative. You ve got to slow down in order to stop. Good. We re about ready to try some problems. But first, let s go back to the three acceleration equations and add something. Many problems will involve objects starting from rest, and you know that this means initial velocity is zero. (table on screen) The first term on the right of each acceleration equation involves initial velocity. When initial velocity is zero, what happens to this term? Anything times zero is zero, so it goes away. And look at the other two equations. The same thing happens to their first terms when v sub-i equals zero. The term disappears. Let s rewrite the equations for problems where the object starts at rest. We re left with three very simple equations that are easier to rearrange and work with. Now we re ready for an example problem. Starting from rest, a ball rolls down a hill, uniformly accelerating at 3.2 m/s 2. How long does it take the ball to roll 24 meters? The first thing we see is starting from rest. This means that initial velocity is zero. So we write it down. Acceleration is 3.2 meters per second squared. And displacement is 24 meters. We want to solve for time. Look at the three acceleration equations, but use the simple ones on the right. Which one will we use? If you said number two, you re right. It s the only one that includes d, t, and a. We need to rearrange to solve for time before we do any calculating. Multiply both sides by two and divide by a. The take the square root of both sides. That gives us t equals the square root of 2 times a times d. When we plug and chug, we get the answer. It s 9.8 seconds. Don t forget to round. Skid marks at the scene of an accident show that Justin Time s car moved 64 m before it stopped. If the car decelerated at a rate of 8.0 m/s 2, how fast was Justin driving before he applied the brakes? In this problem, we know that displacement is 64 meters and that the car comes to a stop, so final velocity is zero. We also know that the car decelerated. That means that acceleration is a negative 8.0 meters per second squared. We don t know the car s initial velocity. Now, which equation will we use? We need one that includes initial and final velocities, acceleration and displacement. That s equation number three. 8

Now you can t use the short form, because initial velocity is not zero. Let s rearrange to solve for initial velocity. To isolate it, we need to subtract this last term from both sides. Then we ll take the square root of both sides. That gives us initial velocity equals the square root of final velocity squared minus two times acceleration times displacement. Now watch your signs here. The two negative signs will multiply out to be positive. If you re wondering if you can make two mistakes and get full credit for the right answer, the answer is no. In physics, two wrongs do not make a right. The answer is 32 meters per second. What do you think? Was Justin Time speeding? Police computers can actually use skid marks to find out, as long as the car didn t hit anything to stop it. You can calculate Justin s speed by converting from meters per second to miles per hour, using the fact that one kilometer equals 0.62 (zero point six two) miles. We ll talk about it at the beginning of our next program. Now it s time for you to try some motion problems. Your teacher will give you the problems and time to work them. Remember to watch your units, watch your signs, and pay attention to the type of motion you are dealing with. We might just slip in some uniform motion among the acceleration problems, so don t be tricked. We ll go over some of these problems in the next program, too. And we ll learn how to tell how tall a bridge is by dropping a rock from it and counting one Mississippi, two Mississippi, three Mississippi 9