 Hi, I'm Zor. Welcome to a new Zor education. We continue physics 14's course. We are talking about certain aspects of kinematics and today I would like to present graphical representation of functions related to kinematics. Now, what functions? Well, position function, coordinate function. That's number one. The velocity function, number two, and acceleration. Now, these are all functions of time, right? So we have, for instance, three position function. Then we have their derivatives, which are components of the velocity vector. So these are all vectors. These are three components of the vector of the position, which is vector from origin to the point where the object is located. Now, these are components of every speed in every coordinate, which basically composes together becomes a vector called velocity. And the second derivative, which is acceleration along each of the axes. Again, together it becomes a vector of acceleration. So we have three vectors. Each vector is a function of time. Now, I would like to graphically represent something. Well, let's first think about something like this. This is the function of one argument. t time is an argument. x is basically x-coordinate of the position in this particular case. That's the value of the function. Well, I can represent it in graphical mode, t, and the value of the function, right? So starting from moment, let's say, 0, at moment t is equal to 1, I have this position, I have t equals to I have this position, and that actually becomes a graph of the x-coordinate as it depends on the time. Okay, now that's just one function. I have three. It's not easy to, well, if I had two, I can probably add another axis y. And in the three-dimensional space, I can represent a two-dimensional function. Two-dimensional, because it has two values. Now, to represent a three-dimensional function, which is vector, well, I need fourth dimension space to represent it. I mean, it's kind of difficult. So in all these graphical representation of the motion, its position, its velocity, and its acceleration, I will do much simpler. I will concentrate only on the movement within the x-axis. So we assume that out of this three-dimensional world where we live, we are talking about motion, which is only occurring which is only occurring within this x-axis. Which means y of t and z of t are equal to 0 for all moments of t, and that's why their derivatives, the first and the second derivatives, are equal to... So in the three-dimensional space, this is x, this is y, and this is z, motion occurs only here. It can be forward, it can be backward, it can be forward and backward, whatever it is, but motion is only here. Now, since I don't have these two other axes, I have only the x-axis, and now I can stretch it in time and show for each moment of time where exactly. So if my, for instance, if my object is moving from this point, x is equal to 0, all the way along the increasing x, so that's how my object is moving. Now if I stretch it in time, I will have its position at any moment of time. So at this position, at this moment of time, I have this position. At this moment of time, I have this position. So I have something like a graph, which represents my movement along the x-axis as the time goes by. All right, so I will concentrate only on the movement within the x-axis. Well, just because I can actually represent it graphically, I cannot represent how it moves in three-dimensional space as the time goes by. I can only watch it. I mean, if I can imagine my three-dimensional space and I have a point which represents an object, I can watch how it moves here and there, but I cannot graphically represent the same way as represent the graph of the function, right? All right, so we are concentrating only on movements within the x-axis, within one direction and we choose the x-axis along this straight line. And I will consider four different kinds of movement. My first is the body at rest. So at all moments of time, the body does not move. It stands still at the position x is equal to 5, at all moments of time. So what is the graphical representation of this type of, well, quote-unquote motion? It's not really motion because it's standing still, but it's still kind of a motion with a zero speed, etc. So how can we do this? Very simply, if this is my time and this is my x-coordinate, now this is my 5, so at any moment of time my function is equal to 5. At this moment, at this moment, at this moment, which means my graph is horizontal line which is intersecting the x-axis at point x is equal to 5 and in the future and in the past and at moment zero, it's always equal to 5. Now, on the same graph, I can actually represent how the speed actually looks like. Well, since the first derivative is equal to zero because it's a constant, right? First derivative of the constant is zero. My graphic of the velocity is always here, zero. For all moments of time, my velocity is equal to zero. Now, velocity is a vector as well as position is a vector. It's a vector from the my position is the vector from the origin to the point where it is located. My velocity is, as you remember, it's its increment of the position divided by increment of time. So it's like an average speed in the moment t. It's an average speed during the interval delta t when I take the limit of delta as delta t goes to zero, which is actually a definition of the derivative, right? So my vector of velocity is equal to zero in this particular case. Now, and so is my vector of acceleration, obviously, because, again, it's derivative from the constant zero. This is the simplest kind of graphical representation of the simplest kind of motion. Motion, which is basically standing still relative to this coordinate system, nothing's changed. And since nothing's changed, basically, we have a constant graph for position for velocity and for acceleration. Okay, next. Now, let's consider that I was standing at 0.5 only at moment t is equal to zero in the very beginning. And then I start moving proportionally to the time. As time is increasing, my position is also increasing. So what is the movement? If I'm watching my point, this is five. At moment t is equal to zero, my object is here. Now, what happens as the time t is increasing? Well, at moment t is equal to one. It's eight. At moment t equals to two. It's eleven, etc. So the body, my object, my point is moving upwards. So if I'm watching, that's where it's moving. Now, if I want to represent graphically each position as a function of time, obviously, I have to have at zero, I have five. At one, I have eight. At two, I have eleven. So it goes, it goes something like this. So that's my graph of the position expressed as this function. Oh, obviously, it's a linear function. Let's just look at this from the algebraic standpoint. It's just a plain linear function. Obviously, I have a straight line as a graph. So whenever I want to determine a position at any moment of time graphically, well, I have to set the time. Let's say the time is equal to three, for instance. Then I go the way up, take this point and take the coordinate x-coordinate, and I will see that at point three my position would be fourteen. Right? So this is a graph of the position. This is x of t. All right. Now, let's talk about the velocity. Well, velocity is the first derivative, right? So velocity is the first derivative. Velocity of the first derivative of this function is three of this, right? So this is this is three. This is my x prime of t. This is my first derivative. Now, speed is constant and equals to three. Now, why actually this way? Well, just think about it. No matter what kind of two different points of time I choose, let's say I choose t1 and t2. My position at t1 would be this. My position at t2 would be this. And if I will subtract one from another, we're talking about movement along the straight line, right? So if I want to know the distance between these two points, I have to just take the difference between them. And I have to divide by the difference in time. That's my average speed and the interval on the interval of time from t1 to t2, right? Well, and obviously this is t2 minus t1 would cancel out, five would cancel out, three would be outside, so it would be three. So no matter what t1 and t2 are, now average speed would be three. And that's exactly what my first derivative shows. So at any moment of time, speed of movement along this straight line would always be equal to three. How about acceleration? Well, acceleration is, as we know, the rate of change of the speed, right? Rate of change of velocity to be more exact. But in this case, we're talking about straight line, so we have only one component, same speed and velocity, almost the same thing, except direction, positive or negative. But in any case, so what's the acceleration? Well, again, rate of change of the velocity, velocity is always constant, so the second derivative of the first derivative from the constant would be zero. And obviously if it's a constant, the rate of change of the constant is zero. Constant doesn't change. So my second derivative acceleration is equal to zero, so the graph of acceleration would be along this t-line. So I have position graph, I have the velocity graph, and I have acceleration graph. So this is all about this particular uniform movement along the x-axis. Now, why is it uniform? Because my velocity is constant. So we are always covering the same distance in the same amount of time, no matter where we are. From any t1 to t2, the speed would be constant, which is equal to 3. Okay. That's my second problem. Now, the third problem would be just a bit more complicated. 3 minus t-square over 2. Now, it doesn't really matter how I got this formula. I got it from somewhere. Now, in theory, this formula actually reflects the falling down object from certain distance above the ground, let's say 3, and this is also not exactly correct. I should put some kind of a constant here, but I just simplified the whole thing. So this is the law of motion. This is how our point is moving. Okay. So first of all, let's just do this graph. Now, this is my tree, and at t is equal to 0, I am, well, let's use the physical concept. I'm at the height of 3 of something, meters, let's say, right? And then as t is increasing, my height is decreasing, right? So somewhere at t is equal to what? When this is equal to 0, so it's 3, 6, square root of 6, right? Square root of 6. My x would be equal to 0, right? Square root of 6, square would be 6 divided by 2 is 3, 3 minus 3. Okay. So in between, my curve would be something like this. This is a parabola, right? Now, I didn't really put the whole parabola, but on the time interval from 0 to square root of 6, it will look like this. Okay? This parabola. Now, physically, my body, if I'm watching this body, this is x axis, and this is t. So if I'm watching from this point, it just goes down. And as it goes down, it goes down faster and faster because I'm losing the same, you see, during the first interval, I'm losing only this height. Then during the second interval, I'm losing greater height. And this is the height I have lost during the third interval. And this is the height I've lost during the fourth interval. So my height which I'm losing is increasing as t is increasing. Well, that's because it's a parabola, right? Now, from here, we can calculate my speed. Now, the derivative of this is, well, this is the constant, so it will be minus 1 half times 2t, so it's minus t. Yeah, I specifically put 2 in the denominator to have it cancel out the 2 from the derivative. So this is my speed. Now, first of all, why is it negative? Well, obviously, because my object is going down, it's against the growing of the x. This is the direction where x is decreasing, and that's why my velocity is negative. So that actually, I'm sorry, that actually goes well with the fact that this is not just a speed, so to speak. It's a velocity. It's a vector. This negative sign means that the vector goes down. Speed is still t. That's the absolute value of the vector. But the velocity is actually minus t because velocity is a vector, right? Now, my acceleration minus 1, derivative from minus t is minus 1. So acceleration is constant. It's also negative because my movement is against the growing of the x component. So it goes against the increase of the x. That's why it's negative. My velocity is negative. My acceleration is negative. Basically, what's important is it's constant. It's a constant acceleration, which means my rate of change of velocity is the same. I'm losing more and more of a position, but my velocity's absolute value is increasing, or since it's a vector it's decreasing. But in absolute value it's increasing by exactly the same factor. So that's what's important, but because it's t square obviously. Second derivative from quadratic polynomial is obviously constant. So in this particular case I just chose the polynomial with nice values of the velocity and acceleration. Okay, that's my number three program. Let's go to the fourth one, the last one. Now what we are doing here is, okay, now we are talking about the situation when my object, which is initially at this point at the origin of coordinate, is going first up and then down, up and down, so it's oscillating between minus one and one. So as the time goes by, starting from moment zero, I'm oscillating according to this function, sin of t. So as t is increasing from zero I will eventually reach point one when t is equal to pi over two, right? Then I go down again when t is equal to pi, sin of pi is equal to zero. Then when t is equal to three pi over two I go to minus one. Then at t is equal to two pi I return to zero, and then the cycle repeats. So how can I reflect it in the graph? How the graph would look in this particular case? So if I look at the object, object moves like this, but if I would like to represent it graphically as a function of time, I have to stretch it. So at zero it's zero, at pi over two it's one, at pi it's zero again, at three pi over two it's minus one, and at two pi it's zero again. So I have this, this, this. So my graph would be this, this, this, etc. So this is my graph of the position x of t as the function of time. If I want to know where exactly my object at any moment of time, I put the time here, then I draw the perpendicular and find the place where my object is located. So this is the graphical representation of the position. Okay, what's the velocity? Velocity would be cosine of t. Now when I'm just starting the motion from zero, or I'm continuing the motion when I'm crossing the point to pi, I'm basically moving like a straight line here, right? My velocity is at t is equal to zero, cosine of zero is equal to one. Okay, now as I'm reaching the top, my velocity is changing down, and at the very top velocity is supposed to be equal to zero, right? Because I'm changing direction. I'm moving up, basically I'm stopping at that particular moment of time, and then moving down. So my velocity is changing from being equal to one at this point to being equal to zero at this point. So velocity would be like this, the cosine, that's my velocity. So again, at moment zero, or two pi, or any other two pi multiples, I will have velocity at maximum, at one. And then as my position goes to its maximum, my velocity is going down to zero. And then my position going into opposite direction. So my velocity becomes negative. And again, it's increasing in speed, it crossing zero with maximum by absolute value, but negative in this particular case because I'm moving down, right? And again, it reaches the zero whenever I am reaching, my position is reaching the minimum point, etc. And finally, my acceleration minus sine of t, that's the derivative of cosine is minus sine. So how that would be arranged? Well, graphically, if this is a sine, then this would be minus sine, right? So acceleration would be basically zero at this particular point, because my speed is along the straight line, it would be basically constant. It does not increase or decrease at this particular point. Now, my speed is decreasing at this moment, right? Which means my acceleration should be negative since my speed is decreasing, my acceleration is negative. And at this particular moment when my speed is actually hit zero, because my position is equal to maximum, acceleration should also be negative, but maximum by absolute value. In this case, it's minus one. That probably is not as obvious from the physical sense of this, but mathematically it is obvious. I mean, you probably understand intuitively that acceleration should be greater whenever we are really changing direction from moving up to moving down, and the speed actually is reaching zero. At that moment, the change of speed actually has the maximum value, by absolute value. So that's the graph, all the three graphs of the motion of the object which oscillates along this sinusoidal law between minus one and one. So this is a graphical analysis of the movement when we stretch the movement along the timeline. Now, as I was saying before, it's kind of difficult to do it with two-dimensional movement. If you have x and y changing as functions of time t, we need three-dimensional like surfaces would represent this particular movement. And in the general three-dimensional case, we cannot really use any kind of a graphical representation because it needs fourth dimensional for the time, right? So we cannot imagine it graphically or draw. So basically for the movement within a straight line, which we basically take as an x-axis, it's possible, it can look nicely, etc. And that's the only thing which probably is interesting because in most cases, all these physical problems, they are related to movements along the straight line, at least these problems which we will consider in classical physics. All right, well, basically that's it for today. I think I will introduce a few problems for the next lecture and I will solve these problems. And then I will probably put certain problems as examination material, which I do encourage you to take. Everything is on unizord.com, including solutions to problems which I will represent in the next lecture. And the exams will also be there. Your score will be registered if you are a registered student or it will not be registered, it will be just given to you if you are anonymous. Okay, that's it for today. Thank you very much and good luck.