 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about uniform motion. This lecture is part of the Physics 14 course of physics and it's presented on website Unisor.com as well as very detailed notes for each lecture and there are or there will be exams for each major section of this course. So I do recommend you to watch the lecture from this website rather than from let's say YouTube or anywhere else because basically you are looking at the video on the YouTube only as a video on Unisor.com it as well has notes and exams etc. And the site by the way is completely free there are no advertisements so you're okay. Now before talking about the laws which guide the uniform motion I would like to remind again I did it before in one of the previous lectures in the introduction actually that you really have to be familiar with certain basic mathematical concepts primarily of vectors and calculus especially derivatives you should be really comfortable with this and if you are not just forget this course go to the mathematics mass 14 there is another course on the same website and refresh your knowledge from that course it's also quite comprehensive course with problems with exams etc. So I do recommend you to get comfortable with vectors and with calculus especially differentiation before you attempt to listen to any lecture of this course. Now considering this is done and you are familiar with differentiation with vectors etc. Now we will talk about uniform motion. So what is uniform motion? Well in a simple language uniform motion is the motion along a straight line so trajectory is supposed to be a straight line in three-dimensional space and also you should cover the same distances during the same time. So if you are covering let's say one kilometer in the first five seconds you have to cover one kilometer in any interval of five seconds along this trajectory or meter or whatever it is the distance which implies that the first derivative of coordinates should be constant. This is velocity vector. So whenever talking about covering equal distance in equal time it means that x of t2 minus x of t1 should be proportional to time and x of t3 minus x of t2 should be proportional. Now this is t2 minus t1 should be proportional between them etc. So any intervals should be proportional to the same coefficient which is actually a speed along this particular axis in this case x axis. So we are talking about the motion with constant vector of velocity. This is vector of velocity. Its components are a b and c and this vector of velocity is supposed to be constant and independent of time and these are components of this vector of velocity the derivative from x, y and z coordinates by time. So this is the definition. Okay now from this definition I would like to derive the law of motion which means I would like to derive position of the object as a function of time. Now position are x of t, y of t and z of t. Well actually you can consider them as the components of the vector. Vector into the position. So if you have three-dimensional space and you have a point somewhere it has certain coordinates. So this is my vector of position. This is x, this is y and this is z. So this piece is x, this is y and this is z. Okay so the vector to the point where object is located is obviously the function of time because the object is moving and we have this particular vector which is the function of time. It has three components each one of them obviously is the function of time and now my question is how are these related to these. So I know that my velocity is the constant and I would like to find out what is my position. Well look at it this way. The only function whose derivative is equal to constant a is linear function of argument t of the time and I have to add some kind of a constant because the derivative of this regardless of the x o of x0 would be obviously a and I don't know what x0 is. Similarly I can have y of t is equal to bt plus y0 and z of t is equal to ct plus z0. So x0, x, y0 and z0 are unknown constants. Now what are these constants? We kind of found the laws of motion the equations which guide our motion however we have some unknown constants in them. So what are these? Well it's very easy substitute t is equal to 0. Now t is equal to 0 is the beginning of motion right and what do we see? The x0 is equal to x of 0 right this is 0. So x0 is equal to x of 0, y0 is equal to y of 0 and z0 is equal to 0, 0. So these are components of the vector which points to the beginning point of the motion. So if you would like to recreate completely the trajectory you not only have to know the speed or rather velocity I would say where the motion is actually occurring where we are moving but you also have to know the beginning point because again if you have a system of coordinates and let's say this would be a vector of velocity. Now if your position is here trajectory would be here if your position is here trajectory would be here. So it depends on the beginning position starting position of the motion and then if the starting position is defined then we can actually using a b and c construct equations of motion. So x0, y0 and z0 constitute basically the starting position of the motion we must have it if we would like to recreate the equations of motion based on the constant velocity. Constant velocity by itself does not define completely motion it defines basically only the direction and the speed of moving along this direction but it does not define the beginning point which is necessary to recreate the trajectory. Now I will allow myself to express this in a slightly different fashion. Now consider this vector p of t consider vector v constant I will put it as a vertical thing and consider the vector p0 which is the starting point vector to a point where we start. How can I express this using these three components? Well very easily I will do it here. Vector to the position at point in time t is equal to time t times vector of velocity plus beginning position. Now whenever I'm talking about this obviously this is an equation for a vectors for vectors in three-dimensional space so I have to know how to multiply vector by constant so what's the multiplication of vector v by constant t? Well whenever we are multiplying a vector by a constant it means we are multiplying each component so the components of this vector will be what? tA, tB and tC right? So this is tA, tB and tC this is the vector. Now this vector is x0, y0, z0 and if you add them together again whenever you are adding vectors you are adding component by component separately you will get exactly this tA plus x0, tB plus y0 and tC plus z0 which is exactly this. At and tA same thing. So basically my equation of motion is much shorter to express in this vector format which in many cases I will do. Not all of them in that sometimes I would like actually to go component by component but in any case this is a vector interpretation of the equation of motion or again if you want you can do it x of t is equal y of t is equal z of t is equal. Now basically right now I am stating that this is representing a trajectory which is a straight line in the three dimensional space. Now why is it straight line? Well it's actually very easy. Take two points here this is trajectory. Let's take the beginning and point p1. So this is the beginning of motion and this is motion at some moment p1 at some moment t1 so p1 is actually p of t1 and then I will take some other position p2 which is p of t2. All vectors. Now this is a straight line basically to prove that this is a straight line. Vectors from p0 to p1 and from p0 to p2 must be collinear right? Now probably I shouldn't really put p0 in the middle between p1 and p2 since this is the beginning of motion and we are moving in one direction probably it would be more appropriate if I will put p0 here so forget about this point because the time moves only forward. Well technically there is nothing wrong with moving the time backwards but in this particular case it's kind of more philosophically understandable that if time moves forward. So at moment t0 this is p of t of 0 just of 0. I'm here this is the beginning of motion then this is the moment of time t1 and this is the moment of time t2. So let's just compare this vector and this vector. There must be collinear right? Well let's just check it out. Now this vector p of 0 if I substitute 0 I will have p0 right? Now this vector p of t1 would be would be tv plus p0 and this point would be this is t1 and this is t2 v plus p0 again from this equation right? Now what is this vector and what is this vector? These are differences between vectors from the from the origin right? So vector from point p0 to point p1 is a difference between p1 and p0. Now the difference between p1 and p0 obviously is tv no t1v this is my p1 minus p0 which is equal to t1v. Now this vector from p0 to p2 is t2v plus p0 minus p0 which is equal to t2v. Are they collinear? Of course they are because this is the same vector v multiplied by either t1 or t2. Obviously they are collinear so the same vector of speed whatever the vector of speed is if we multiply it by one constant or another constant we will get the collinear results it will be within the same line right? So I can always say that they are all collinear and they are in particular collinear to the vector of speed vector of velocity v. So what I'm saying is the trajectory is a straight line and the direction of trajectory is exactly the direction of the velocity vector v. Okay so that's easier right? Now considering this is a movement within a straight line it's kind of obvious that you can choose the coordinate system if it's up to you to choose by the way. It's natural to choose the coordinate system with one of the axis let's say x axis going along this line. So if this is the trajectory I will change my coordinate system instead of this one I will use the coordinate system which has origin somewhere on the line and x axis along the line. Now my x axis is along the line obviously my y and z coordinates would be equal to 0 so x of t would be equal to still the same but my y of t would be 0 and my z of t would be equal to 0. So that's kind of easier and in many practical problems let's say the car is moving along the road well you probably choose a coordinate system somewhere on that road originating and the x axis along the road so the car is basically moving along the x axis and perpendicular to it y and z axis are basically all with 0 so that's kind of easier and your equation of motion becomes actually now I should not put the vector here I'm sorry I should put the component of the vector now my components I a b and c so this would be a so that's kind of easier in which case a is just a number basically which represents in this particular case we can say speed instead of velocity it's not a vector it's a component of the vector x component of the vector and it's actually we have agreed that we can talk about speed in this particular case that's the magnitude of the vector and obviously if my direction of my x axis is the same as direction of the movement a is positive and if its direction is opposite which doesn't make any sense actually to do then a would be negative so so that's kind of easier to deal with because you're not dealing with vectors you're dealing with scholars right one dimensional case and if you choose your coordinate system in such a way that the origin is exactly where movement starts then x 0 would be equal to 0 right so you will have even simpler equation of motion this is the simplest equation of motion possible when you choose the coordinate system and origin exactly where the motion starts you choose one particular axis along the trajectory of the motion along the vector of speed in which case these two components of the vector of speed obviously are equal to 0 and the movement within those other two components also non-existent it will always be 0 and this is the simplest equation which represents the motion along the straight line with constant speed which we call uniform motion well basically that's all I wanted to say about this particular simplest of all possible motions I will also spend some time to talk about the motion with constant acceleration and and also the rotation that would be another two types of motions which we will analyze well again as a reminder make sure that you are comfortable with vectors and calculus that's a must for this course of physics for teens that's it thank you very much and good luck