 Hi, I'm Zor. Welcome to a new Zor education. Today, we will solve a few problems on kinematics. Basically, we have talked probably about everything we wanted to, as far as the kinematics is concerned. So it's time to solve some problems before taking an exam, right? Okay, so the problems on kinematics are part of the course called Physics for Teens. It's presented on Unizor.com. That's the website. The website is free, has no advertisement. So I suggest you to watch this lecture from this website, because, first of all, these are problem solvers, right? So you can actually take a look at the problems themselves first, and they are on the website as notes for the lecture. And try to solve these problems yourself. These are very, very easy problems, and there are answers provided, so you can check your solution. And, again, besides, I was talking about exams. The website has exams for every topic, or will have exams for every topic. So I do suggest you to go to the website for this and every other lecture. Now, the problems. Well, first of all, we are talking about physics, about the movement, etc. So we are dealing with certain units of measurements. And I will prefer, in most cases, I will actually do it, to use system international units, like meter for the length or distance, second for the time, and all the derived units. In time, we don't really have any discrepancy. Everybody is using seconds and minutes and hours. And as far as the distance is concerned, yes, you can use meters, or you can use kilometers, millimeters, whatever the derivative from the meter is. Most likely, I will not use non-standard units, like miles, or inches, or anything like that. All right, so my first problem is, let's assume we have a certain object, and it moves according to certain coordinate functions. X of t is equal to t square, and y of t is equal to zero, and z of t is equal to zero. Now, what does it mean? Well, it means that in this Cartesian system of coordinates, we are talking about movement only within this x-axis. So, it's a straight line movement. Now, this is the way how my distance from zero is changing with the time. Well, obviously, time is equal to zero. I'm here. And then I can actually put my position on this line at every moment in time, substituting the time into this formula. Now, here is what the question is. Let's assume we have a completely different system of coordinates. Now, its origin is at point five, and the axes are parallel and similarly directed to the old system. So, y one is parallel to y, z one is parallel to z, but the origin is at new point, which is five zero zero. Five along the x-axis, and zero and zero along y and z. So, my question is, in this new system of coordinates, how will my motion be described using the functions, basically? So, I need x one of t is equal, y one of t is equal, and z one of t, which is equal. But for y and z, it's obviously zero, because we are still moving within the same x-axis, or x one, actually, axis. So, there is no change in y and z. y one is equal to zero for all t's, and z one is equal to zero for all time. How is x one? Now, since my coordinate system is shifted by five to the right, every point which used to have coordinate x, now will have a coordinate which is shorter by five, right? So, this was x, now this is x first, right, if this is the position. So, my x one will always be five shorter than x, which means this is t square minus five. My old coordinate, which was t square minus five. By the way, at time t equals to zero, my point, my object positions here, right? And this is exactly minus five along the x one coordinate, right? Because it's to the left of the new origin of coordinates. So, this is the transformation, basically, of coordinates. If I shift by five, my system of coordinates, well, parallel to itself along the x axis, then my new coordinate x one in the new system will be shifted by five with a minus sign. That's my first problem. By the way, I do recommend you first, before you listen to all these solutions which I'm suggesting, do try to do it yourself. I mean, in the website, there are all these problems listed with answers, so you can check yourself. That's a very useful exercise. Next, next I have the movement within the z axis. I'm pulling down from point zero, zero, twenty-five, so this is twenty-five. The coordinates are zero and zero on x and y, which means I'm on the z axis itself, and the z coordinate is twenty-five. So I'm falling down from the height of twenty-five. Now this is the ground level. Now my equations are obviously x and y are always equal to zero, but my height is changing as twenty-five minus t squared divided by four. So t is equal to zero, I am at twenty-five. And then I'm falling down using this particular formula. Now my question is, what is the time until I reach the point zero, zero, zero? So what's the time of falling? Well, obviously this thing is diminishing from twenty-five at t is equal to zero to zero at some point. So to find out the time I have to solve this equation, whenever my height is equal to zero, my time t would be equal to exactly the time of falling. So the solution of this is twenty-five is equal to t square four, t square is one hundred, t is equal to square root of one hundred, which is ten. So my answer is the falling time is ten. Let's say seconds for instance, if everything is in seconds and meters, then this is seconds. Okay, now my next problem is very similar. Now I am asking what is the speed at the moment of falling to the ground. Because the speed is obviously increasing as we are moving down. And what is the speed? Well, the velocity vector obviously is the derivatives of this. Now derivative of x is equal to zero and this is exactly the same as y because we are not actually moving in these directions. Now derivative of z, so it's a z component of my velocity is equal to, this is the constant, this is the minus from t square is two t divided, okay, so it's minus t over two. So this is my velocity vector and we are interested actually only in the z component. So my velocity, first of all, it's with sine minus y because we are moving the opposite direction to the direction of the z axis obviously. And again at point zero we are at rest, so velocity is equal to zero and then as the time is increasing the velocity is also increasing by absolute value being negative. Now so what is the velocity at the time when I fall down? Well, all I need to know is t, but I have already calculated t in my previous problem. So obviously my velocity is z of ten is equal to minus five. In this case the unit of measurement will be meters per second if my original is meters and time is in seconds then this is meters per second. So at the moment of hitting the ground the object moves with the speed minus five meters per second. And again minus because I'm moving opposite to the direction of the z axis. Velocity is a vector so it's very important which direction we are moving. All right, fine, next. Now before we were talking about one-dimensional movement along some axis, along the x axis or down along z axis. Now in this problem I will have a two-dimensional movement. So let's consider this is my z, this is my x, this is my y and direction now y would always be equal to zero y of t. Now in this particular case what I'm doing is I'm shooting the projectile within this plane x z plane. So that's why y is always equal to zero. So I'm shooting the projectile and obviously it moves like this. So the gravity pulls down the object but I'm shooting at certain angle to horizon so it moves a little bit up first and then the gravity actually pulls it down. Now what I see is that horizontally my object always moves with uniform motion. And uniform in this case is 2t. Now that's uniform because the velocity along my x axis which is the first derivative is equal to 2, it's a constant that's why it's a uniform motion. At every interval, in two equal intervals I will cover two equal distances, right, with this constant speed along the x axis. Now as far as the z is concerned, z is my movement which is basically caused by the gravity. So first it moves a little bit up and then the gravity pulls it down. So my z function is given as 10t minus t square. So my question is what is this distance, how far my object will fly in the air, right? So I'm not talking about forces of gravity or inertia or anything else. No forces are involved. The kinematics is studying only the movement. So movement is given. That's it. I'm not talking about y and half. Movement is given. So all I need to know is how far it will go. Alright, how can I address that particular problem? Well let's just think about it this way. What is the characteristic of this particular point? Well at this point my z-coordinate is equal to zero, right? It's a ground level as well as in this particular point. So this is the parabola because it's t square, right? Minus and that's why it's directed with its horns down, right? So all I have to know actually is what are the moments of time when this particular z function is equal to zero. The one I know already is t equals to zero. What is the second one? Well again, the same problem was before. All I have to do is solve this equation. Now this is obviously ten minus t times ten times t is equal to zero. This is one solution t is equal to zero and t is equal to ten is another solution. So this is t is equal to ten. So I have the time moment when I reach. Now what happens during this time as far as the x-coordinate is? Well it moves with this particular function, with this particular formula, right? So I know the time. So what is my location, my position along the x-axis at this particular time? It's equal to two times ten, which is twenty. So it goes for twenty meters in this particular case. If everything else is meters and seconds and whatever, all right? So first we get the time of the motion until it hits the ground using z of t is equal to zero equation. And the t substituted into x and that's my distance, okay, next. Now this is a horizontal movement along the x-axis so I will only put an x-axis. Now I have an AB, let's say this is zero. Now and this is 720 kilometers, kilometer is one thousand meters, right? So what I know is that half of this distance I was traveling 60 kilometers per hour. And another half of this distance I was going with 90 kilometers per hour. So my question is what's the average speed during my trip? Well let's just recall what is the average speed. Average speed is total distance divided by total time. So all we need to do is to find the total time because the distance we know, 720. So how can we calculate the total time? Well, the total time contains actually two halves, so to speak, not halves in time but halves in distance. So half of the distance, which is 360, half of this kilometers, I was driving with the speed 60 kilometers per hour, right? So my distance, as you remember, if this is a uniform motion, now the distance is related to speed with this formula, right? So speed is constant and this is the only function which gives me the constant speed as a derivative, right? The derivative of this, well, probably I should say x of t is equal to v times t. So the derivative of my position is the speed v. Now if I have the distance and I have the speed, how can I calculate the time? So 360 is equal to 60 times t. So t is equal to 6 hours, right? Kilometers, this is kilometers divided by hour per hour, so this is hours. So we've got the first half. Now the second half, analogously, I have 360 again because this is another half of this distance. Now it's equal to 90 times t. So what the t is? Now let's say this is the t1 and this is t2. So t1 and t2 is equal to 360 divided by 90, which is 4 hours. So my total time is 10 hours. So I have the total time and I have a total distance, which means if I divide one by another I will have an average speed, right? So it's 720 divided by 10, 72. Now what's interesting is this is 72 is not a arithmetic average between these two speeds. That's very important. The average is, sum is 150 divided by 2 is 75. So the average between these two numbers is 75, but that's different. So you have to really think about it. If you divide distance in half and then you have two different speeds, the average speed is not the average between these two numbers, these two speeds. Average speed is the total distance divided by total time. So you have to calculate separately time and time and they are obviously different because there is a different speed with the same distance 360 here, 360 here. But speed is different, so the time is different. And that's why you don't have exact equality between the arithmetic average between these two numbers and the real average speed on the entire trip. Now my next problem is kind of similar, but in this case instead of dividing the distance in two halves. I'm dividing the time. So my trip from A to B took 10 hours. Now my first hour I'm accelerating and my average speed is 60 kilometers per hour. And there at the very end I'm decelerating, I'm slowing down and my average speed is also 60 kilometers per hour. Now this is one hour and this is one hour. So I'm accelerating and one hour decelerating at the end. Now in between eight hours, total is 10, right? So one here, one here and eight in between. In between my average speed is 90 kilometers per hour. Now my question again is, what is my average speed? Well, let's think about here again. Average speed for an entire trip is the entire distance divided by entire time. Time we know, 10 hours. So what's my total distance? Well, I don't know the total distance, I have to calculate it. It contains three independent pieces. Now one piece, I know that within one hour, 60 kilometers per hour, I was driving during one hour. So what's the distance which I have covered? Now this is average speed which means it's distance divided by time. So basically this is 60 kilometers. Now the next piece is 90 kilometers per hour times eight hours, that gives me 720 kilometers. And decelerating, again, average speed was 60 for one hour, so I have covered 60 kilometers. So what's my total distance? 12840 kilometers in 10 hours. So my average speed is 84 kilometers per hour. Well, that's it for today. These are six different very, very easy problems which do not require anything except the concept of distance, time, average speed, which is distance divided by time. And in some cases I was using derivative to find instantaneous speed. But that's only in those cases where it was given the formula for the movement, like x of t is equal to t squared divided by whatever. So that's it for today. Thank you very much and good luck.