 Hi, I'm Zor. Welcome to Unisor Education. The previous lecture was about power, and right now it's an opportune time to do some problem solving. So I have three problems here, all about the power. Now, this lecture and all the preceding ones and all the following ones are part of the course called Physics for Teens. It's presented on Unisor.com. Now, this is a completely free website. There are no advertisements, no financial strings attached. And I do suggest you to watch this lecture and everything else from the website because it's presented in a logical course order. So if you found this lecture, for instance, on YouTube or anywhere else, it's much better if you will start from the Unisor.com because every lecture has a textual material and there are some exams for those who would like to take exams and you don't even have to sign in. I mean, it's optional. Okay, so problems about the power. Well, let me start from something which I did talk about in the previous theoretical lecture about the mechanical power. But it's a very important formula which I'm going to use in all these problems. Now, we all know that in a very simple case, the work is basically a product of the force times the distance this force is acting. But this is basically a very kind of simplified definition because obviously the force can do variable and the distance during which the force is applied. Obviously also it's not a fixed, basically distance where the fixed force is applied. Everything is a function of time. So let's just concentrate on this time-based thing. So if you have a motion and the motion is described by the function of distance from the beginning as a function of the time t, then what we can say is the following that the increment, infinitesimal increment of the work at time t is basically from t to t plus dt. This is a time frame we're talking about. During this time frame, our force can also be, by the way, a function of time. Force can be variable. And during this infinitesimal time period, the distance the object moved under the influence of this force is infinitesimal differential basically of the distance. So this starts from this. It's a much better definition. It's a differential definition of the work. Now from which we can definitely derive, if we will, basically divide it by dt by the differential of time. And this would be a definition of power. Power is the rate of the change of the work which is done. Instead of f of t using the second Newton's law, we can substitute m times acceleration, mass-time acceleration. And the differential of the distance is the speed. So this is the function which is function of dependence between the power which is exhorted by the force. The mass, acceleration and the speed, this particular object is moving. So this is the formula which is very important and I am going to use it in these problems. So let me put it here. Power is equal to mass times speed times acceleration. Okay, now problem number one. We have a car. I mean all problems about the cars. Now it has mass m and I know that during the time t it accelerates from 0 to speed v. This is the maximum speed with constant acceleration. So I have two problems here. What's the power as a function of time? Now the A is constant. Power might or might not be constant in this case. We will find out actually what is the power to basically assure the constant acceleration. Power will not be constant. So what's the power as a function of time? And I would also like to know, to relate it to the power, usually the speed of the burning of the fuel in a car is kind of proportional to the power. So if I will know how the power behaves, I will have to basically make a judgment about what will be my fuel consumption as the time goes by. This is the, not the quantitative, it's the qualitative. Now, so how can I approach this? So first of all, if I know that the A is a constant and I know that during the time t my car accelerates from 0 to v, I can definitely come up with what is the acceleration. Sorry, it's capital T. And this is constant, as I'm saying. Now knowing the acceleration, and I know that this is a constant in this case, I know the force, also constant, right? Now, again, knowing acceleration and acceleration is constant, I know the formula for the distance covered. It's basically a t squared over 2 as a function of time. So now, since I know the acceleration, I know the distance as a function of time. This is a simple formula from kinematics. And so the speed as a function of time, I should probably put max here, that would be better. Now, speed as a function of time, if the constant acceleration is basically a times t, right? So it's v max over t times t, right? So I know my force, I know my acceleration, I know my speed from which I can derive the power using this formula. Mass times speed, which is v max over t and time, and times acceleration times v max over t. Alright, now, since I know this particular function, I can actually make a judgment about the fuel consumption. So first of all, well, let me just write it square, square, and get rid of this. So this is the formula for the power. And as you see, the power is supposed to be increasing as the time goes by. So to assure the constant acceleration, if you have a very low speed in the very beginning of your motion, it's easier, so to speak, for the engine to give you certain acceleration than if you have already gained certain speed and you would like to continue the same acceleration. The same acceleration from a larger value of the speed would take more power than the same acceleration from the zero speed or from the lower level of the speed. By the way, the same thing with economy. If economy is in a very poor shape, in the very beginning, so to speak, of development, it's much easier to accelerate it than if it's already developed economy and you would like to continue acceleration of the same level, like 4% a year, for instance. 4% a year is easier if the economy is a very small one. But if economy is a very developed one, it's much more difficult to achieve the same 4% a year, because 4% of economy growth is exactly the same as acceleration in the car. And now, speaking about fuel consumption, since the power is increasing as the time goes by, fuel consumption also should increase. So the rate of the fuel consumption per unit of time, let's say, is greater and greater if you would like to maintain the same acceleration with increased speed. So that's my first problem. The second one is basically much similar to this one. I'm just giving you different initial data. Now what is in this case is I do have again the car and I have its mass. Now my power is constant. So the previous problem, acceleration was constant and the power was supposed to be increasing. Now my power is constant. And I have to find out what are my speed and my acceleration. And in particular is acceleration constant. And which is the same thing is my speed linearly increasing, which means acceleration is constant. Well, let's see. The answer to both questions is no. But let's just derive it the way how we are supposed to derive it. Again, I'm using this formula. Now my P is constant. So I don't have to put P of t as a function. I'm just putting P. And this is equal to m v of t. And instead of a of t, I will put d v of t by dt. Since acceleration is the first derivative of the speed, I can do that, right? Well, now what is this? This is a differential equation. And again, we did actually talk about these very, very simple differential equations in the course of mathematics, which precedes this course, Physics 14. On the same website there is a mass for teens. So differential equations and calculus are very, very important for physics. So this is one of those cases. But in this case, it's so simple that I can actually just do it right now. What I will do, I will multiply everything by this infinitesimal increment of the time. And I have a differential here. And I have a differential here. Now what is this? Well, if I will integrate this, this is actually a derivative of P times t, right? So if I will integrate it from 0 to t, I will get this. And if I will integrate this from 0 to t, integral from 0 to t, integral from 0 to t. What happens now? What is this? Well, m is just multiplier. This is differential of v square divided by 2. So if I will integrate it, I will get this. From which follows, v of t is equal to 2pt divided by m square root. This is my function, speed as the function of the time t. In the beginning, obviously, speed is 0. P is a constant. We apply the constant power. But the speed is not growing linearly. So it's not a linear speed. It's a square root of time. So the speed is the rate of speed, which is acceleration, is diminishing as the time goes by. If it was linear, the speed would be linear and it would be the same. But in this particular case, it's a square root of time. So the graph would be like this. Incidentally, a of t, which is a derivative from this, which is what? So square root of 2p times m is just a multiplier. So I have a derivative of square root of t, which is 1 over 2 square root of t, which is equal to square root of, this would be 4. So it's p over 2mt, right? Yes. And again, it's not a constant. So acceleration is not constant. If your power is constant, acceleration is diminishing as the time goes by. In the beginning, when time t is small, it's rather large. But as the time goes by, this is denominator, it's increasing. So my acceleration is decreasing. If you have a constant power, you started the car, for instance. You feel acceleration if you're just starting the movement, right? But as you move forward and forward, if you keep the gas at the same level, so the power of the engine will be at the same level, you will reach certain plateau and your increase of the speed will be really diminishing. And then, obviously, you will have a resistance of the air and the friction, etc., which would completely put a plateau on your speed. So that's my second problem. And now let's go to the third one. And again, it's very much similar and it's using exactly the same kind of formulas for power, speed, acceleration, etc. I'm just giving different parameters. So the car, again, mass m, power is constant. And what I'm interested is how soon I will achieve certain speed which I'm calling Vmax. So how much time it will take for me to reach, let's call it Vmax, how much time it will take under these conditions to reach this speed. Well, actually this is the problem which we basically solved because let me just remind the previous problem we had. We have come up with this equation, if you remember, square of T divided by 2, from which we have derived what is the speed as a function of time. In this case we have a completely reversed thing. We know the speed. The speed is supposed to be Vmax and we are asking about the time. So what I'm doing is right now I'm just resolving this equation for time. And now if my speed which I'm looking for, it's supposed to be equal to Vmax, so the Tmax, the time to achieve this speed would be VmVmax2 divided by 2p. So again qualitatively, obviously if your power is greater, this is denominator, your time to achieve this speed would be smaller, shorter, right? Another thing is the more massive the car is, the more difficult it is to accelerate it to this particular speed and that's why with increased mass the time to achieve this speed would also be increasing with the constant power. So that basically corresponds to all the qualitative analysis. Okay, so what I'm suggesting to you right now is to go to Unisor.com, open this lecture and just try to solve all these problems yourself. I think it's a very useful exercise. Check against the answers which are provided on the website and that basically it. I mean that would complete my introductory lectures about the power and that basically completes my part of the course which I called mechanics. Now the next part is related to, well partially it's related to mechanics but I call it energy and the energy is not only mechanical energy, obviously there are some other aspects of the energy but that would be the continuation, that would be the next part of the course. Mechanics is more or less finished. So thank you very much and good luck.