 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue to talking about certain very simple kinds of motion where we can derive the equations of motion relatively easily. Now the previous lecture was about the simplest possible motion. It's the uniform motion when we are moving along the straight line trajectory with the same speed along it. The same velocity vector, so to speak. Now today we will consider the motion with constant acceleration. Constant acceleration. Now this lecture is part of the course of Physics 14's presented on Unizor.com website. I suggest you to use this website as the source where you are watching this lecture because the website contains lots of explanatory notes, very detail, and there are exams for those who want to take it for every section or will be. Not everything is ready yet. And decided by the way free, no advertisement, so use the website. Okay, so let's go to constant acceleration. What is constant acceleration? Well, obviously that's acceleration is constant. Now what is acceleration? You know this is the second derivative of coordinate functions of time. Now it's constant, it means that these are this way, right? Or if you wish, and I actually insist on using it in wherever I can, in a vector form when your position vector is three dimensional vector with three coordinates to the point where actual object is located. So this would be my second derivative of this vector is equal to, and ABC is another vector, which I call a vector of acceleration. So this is the vector in interpretation of the same, instead of three equations, I will have only one. The second derivative of the vector of position is equal to constant vector of acceleration. So acceleration is given, and our purpose is to derive equations of motion. And I will do it exactly the same way as I did in the previous lecture for uniform motion. If I know my second derivatives, I can guess what my first derivatives are, right? So if my second derivative is a constant, my first derivative is a linear function. So my first derivative is a linear function of t. Now why did I use this? Well that's very easy. It's some constant, unknown to me constant, and I have to add it because adding the constant to a function doesn't change the derivative. So this basically describes all the possible functions derivative of which is equal to constant a, right? Now if I will put the t is equal to 0 in this particular formula, I will have that this particular constant, this constant which I symbolize as vx0, is nothing but this. Now what is the first derivative of my coordinate function? That's the velocity, right? So if it's x, it's x component, if it's y, it's y component, if it's z, it's z component. So these are components of velocity. That's why I put v with an index x, which means it's an x component of velocity at moment t is equal to 0. So that's why this embolics this way. And similarly, this would be y components at point 0, and z is equal to plus z component of velocity at time 0. So as we see, we already have the velocity. So velocity at any moment t is equal to this particular formula where a, b and c are given components of my acceleration vector, and vx of 0, y of 0, and vz of 0 are components of my velocity at time is equal to 0, which means it's an initial velocity of the object. So there is something which is initial velocity. And for any case, whenever the initial velocity is, the formulas actually would be correct, and they have to really take the initial velocity into consideration. Because if there is a bigger velocity, and then I accelerate, it would be definitely even further my object will move during the same time, right? So it definitely depends on initial velocity and acceleration. But not only that. Now this is only my first derivative from coordinate functions, right? Now let's express this again in a vector format. This is my velocity at point t. That's what x prime, y prime, and z prime are. And it's equal to t times vector of a, b, c, which is t times a, plus. And these components are velocity of point zero. So this is the same, this is the same as these three equations, right? Velocity at point zero, this is initial velocity. These are components. A means components a, b, and c. And these three are components of velocity at moment t. Okay, fine. Now I have to go one more step and guess what are my coordinate functions if I know the derivative of these functions are these. Well, for those who understand the calculus, it's basically an integration. Alright, so what's the integration of this thing? It's again very simple. X of t is equal to, to get a t, I have to have a t square over 2, right? Derivative from t square would be 2t. 2 and t would cancel out, and I will have a t. Now to get this constant, I have to have this, times t. Derivative of this is the x of zero. And again, I have to add another constant, which I can call x zero, which obviously is the initial x coordinate, because if time is equal to t, this is zero, if time is equal to zero, this is zero, and this is zero, so x of zero. So x zero is definitely x of component, component x times zero. Alright. Now, obviously the same thing comes with y. That would be bt square over 2 plus vy of zero plus y zero. And z of t is equal to ct square over 2 plus vz of zero plus t, t plus z zero. Let's be more accurate. Okay. Now, let's talk about vector format of this. Now these three components are components of my position vector, right? At time t. Now these three components are my acceleration. So I have t equals, not plus equals, t square over 2 times acceleration vector a plus t times my initial vector at moment zero plus my initial position at moment t zero. This is x zero, y zero, and z zero. So this is the same as these three equations, but in the vector format. Now, you know that trajectories of uniform motion is always straight line, which goes exactly into a direction of the velocity. Now, when velocity is constant for uniform motion, now in case of acceleration constant, velocity is no longer constant because velocity is increasing. Since we have the same acceleration, we have velocity equally increasing as the time goes by. So what is the trajectory? Will it be a straight line? It depends. Let's imagine a physical experiment. You are on the top of the tall tower and you drop the ball, but you are not dropping it just down. You are throwing this way. So what happens with the ball? On one hand, the ball has the horizontal component because you throw the ball this direction. On another, from another standpoint, the ball should go down because the gravity. So the ball actually goes this way. Now, you will learn a little later that the gravity actually is causing the acceleration. The ball would go faster and faster down. Now, horizontally, if we will forget about the resistance of the air, the horizontal speed would be the same, right? So the speed would be this way and acceleration this way and trajectory would be obviously not a straight line, right? However, if your initial speed is vertical down and the gravity goes down, then the trajectory will be down. So what I am saying is that if my initial speed and the constant acceleration in this initial speed and constant acceleration are proportional to each other, then what happens? So let's say my v of zero, my initial velocity is equal to some constant a times a times vector of acceleration. So they are collinear velocity and initial velocity and constant acceleration are collinear. That's what collinearity actually means. Now, what happens with this? You will have p of t is equal to some kind of a function of time times a, right? So I replace v of zero with k a. Now, a as a vector goes outside of parentheses and in the parentheses, I have t square over two and tk plus p of zero. Now, this is the constant vector and this is the constant vector, right? So as a result, no matter what time parameter is, my p of t would be collinear to this. Now, this is basically everything I wanted to talk about constant acceleration as a movement. Again, acceleration is caused by forces which we have not yet touched. That's the dynamics part. Now we're talking about kinematics, which is only like analysis of the movement itself, considering you know the velocities, accelerations, initial position, etc. Now, the only thing which I can probably add is that if you are considering some case when there is no initial velocity, then again, you will have a simpler case and the formula would be simpler. And if you choose your system of coordinate with such a way that your initial position is at origin, now this also disappears. And the only thing you have is this one. So basically, if your acceleration is going along the x-axis, so this is acceleration along the this, and your position initial is here, and there is no initial velocity, then obviously you will move here and your x-coordinate would be t squared over 2 times a, where a is the acceleration along this. y and z would be equal to 0 in this particular case. And this is the simplest kind of a way of analyzing your motion with constant acceleration. But you have to really have this case when there is no initial velocity. That's something which is... All right. Basically, that's it for today. Thank you very much. I do recommend you to read the notes for this lecture. They are very detailed. Basically, everything whatever I'm saying is written in the notes. So it would be like an additional repetition of the same material. Okay, that's it. Thanks very much and good luck.