And like I said, drawing a graph will help to visualize and understand what's going on. Okay, okay. Elevates rest and the velocity of the first 2m per second. It maintains the speed for 72s until it decelerates uniformly to station peak. So the journey sign is 112s, and the magnitudes of the accerary accary equal. Find the time it takes. Try and jusome from rest to 32m a second. Magnitare equal. Okay, so. That's 20. 20. Luis kind wrong. Thank you. I like the distance between the two nations. Distance is the area, this is 2032. A half times 20 times 32, 20 times 32, 640, 320m and then it's 72 times 32, I think. Two, three or 4m. And then it is again 20 times a half times 30 equals 326, 320, 640 plus two, three, o four equals two, nine, four, 4m. Two. Personal song to drop into agency ency. Okay, so this this acceleration. 0.6m per second squared. And then. 40s and then it's constant speed. Decelerates 0.4m per second. And its own distance is four closes. So it's 4000m. Okay. So okay. Did that be find the total time taken by the train to come from a to B2 two high? No speed and know acceleration time, acceleration and time. Another a equals U plus 80, v equals zero plus 9.6 40. 2420 4m second, so this is 24m second. Eight. A. Almost 78. Really over. Hey. Saleration point is going to 24 over 0.4, which is 60s. 4000m 4000 minus 60s. 18. Otimes. Oh, Yeah. 480-720 to eight zero meters, so 28o divide by 24. 160.67 point 7s 160.7 plus 60 plus 40 equals 216.7 seconds. How are you getting on? I think I finished the first one. Oh, great. Okay. Let's hear what you got. So the first one a is 20s. Yeah Yeah the gradient of that. Okay and okay. Yeah, Yeah, Yeah, Yeah, Yeah 20s, Yeah and then one b it's like pesium, I think so it's a to b is an upward slope, then constant, then to b is a downward slope, so it's like a trippeum. Yeah and then the x values or the time values would be at 20 and 92 and then 112 at the end with the height of the trapezium would be 32 on the velocity axis perfect. And question c is 2944m. Yeah, very good. Very good. Okay. And then question two. And then for question two, two a is also like a trapezium type, the same thing. And then two b is. Why the total sign to b is where is it, 216.7s? Whereshe get 216Yeah. Yeah, that sounds about right for. Your sketch of the graph of motion for two a, the height should have been 24. And then for the for the horizontal sections, like the first triangle on the left is like zero to 40. Then for the rectangular section, it's 40 to to capital or lower or whatever you've modeled it is an unknown time plus 40. And then the last section is 60s. Yeah no, but I think you've Yeah I think you've got that one correct. Lovely. Let me. Load the next ones, okay? Okay, power moving in a straight line with speed five mm meter per second, accelerates uniform before 6s, which reduces its speed two eu meters per second. It maintains and maintains the speed for the 16s before the accelerating uniforminterrest in a further two. Okay, sketch a graph. And then after these questions, what we'll do after these questions is maybe we'll just touch on you know like if you've done dy by dx for a polynomial, I would yes, suggest sts that we do the reverse operation, which is integration. It's very simple. It's just the reverse of it so that for next time we can do the next topic in mechanics, which it basically just requires knowledge of you know the basic operation of the derivative of a polynomial and also the reverse operation of that which is the integral of a polynomial. Like it's really simple. The integral ittake us like a few minutes, five, ten minutes to to kind. And then I can I can send you some exercises to do over the weekend just to sort of solidify it. And then we'll move on to a variable acceleration next week. Okay. Velcity 5m accelerates uniformly for six. Will you maintains for further 16. Oh very good. You meet us for second and then maintained. To rest in a further 2s. So then it's constant speed and then decelerates. So here is 16 sds and decelerates zero meters per second in a further 2s. Okay, combine an expression for each of the decelerations in terms of view, I do that. Write an expression for each of the deceleration. Deceleration deceleration. Two. Between you. A equals v minus. D D one d 2D one equals. T U minus five over six minus U over two t two equals this zero minus two U over was minus here. Total distance is cuand 20m. Find the real value of you. Four. I view this is to you. Five years, you. Five U plus two times. Six times a half. 21 year. Then it's two times 16 to 32, then it's a half times, two times two U goes to U so the total is 21 plus 32. 53, 55U 55U 220 divided by 55U 55 equals four. 风神有一共是福。嗯。Mmm. All thoughts from rest and accelerate at a constant rate ight to a speed of V V meters second and 6s, the car maintains the speed for 50s before the acceleration to rest magnitude of the deceleration, 1.5 times this magnitude of the acceleration. Okay, so. So thoughts. From rest accelerates. V need 2s. Second. 50s. Celeating to rest. I was total time taken for the journey a 60. U is zero so a equals v over T A. Equals v. Over six. Magnitude of acceleration is 1.5 times this magnitude of acceleration. So 1.5a equals 1.5v. Over six equals v over four hearts for decelbration. I'm not sure how to do four a. Okay, shall we go through question three first? Yeah so three a so it's like the graph is like it starts from 50m per five U meters per second, then so it's a straight line, then there's a deceleration line like sloping down to like 2m per second and then it it goes constant for 16s. So between 16, between six and 22, Yeah Yeah there 22s, it decelerates down to zero at 24. Yeah good. Yeah and three b is so the first like from from. Like the deceleration from five U meters per second to two U meters per second is minus U over two. Well, second, well, it's a deceleration that it's asking for. So you would leave it as a positive sort of be U over two. If it was asking for the acceleration, then it would be negative U over two, okay. But deceleration, Yeah because deceleration by definition is the negative of acceleration. So it's just a subtle but important point. Okay? Okay. So then so the first deceleration, then it's going to be U over two and then the second one from 22 to 24 seconds, so it's gonna to be you. Yeah okay. And three c is four U equals 4m per second. Yeah Yeah U equals four. Lovely. Okay, so then for question four, so you've got it starting from rest to accelerate at a constant rate to a velostv in 6s. Car maintains the speed for 50s before decelerating to rest. And you need to show that the that the total time is 60, right? So for that first section, you know that the acceleration is v divided by six Yeah. Yeah and then it says the magnitude of the deceleration is 1.5 times the magnitude of the acceleration. So for that section you can say 1.5a is equal to v over t where t is are unknown amount of time. So you need to solve for that amount of time and hopefully you should get around four because six plus 50 plus something should be equaling to 60. So it's an exercise of showing that from those two equations that you've got you know a equals v over 61.5a equals v over t hopefully you can show that t is equal to four. Does that make sense? Yeah. Okay, the equthree of the six. Be other thing. Five a it's 1.5b. 56. The time equals v over n. Zero. So that's zero. But me of acceleration at v over v four which equals to four. Seconds that pases 4s. 60Yeah 56 plus four 60s okay that's fair. Total distance is 1320m. And the value of v. So the acceleration triangle is. Six times v, times a half, which is three v and then this part is 50 times v equals 50v and then this part is four times v, times a heart equals two v so 50 plus five, 55V1 32o, divide by 55v. 传奇凤凰传。与你如。Okay, sir, I'm done. Good. What have you run? Okay, so for a so I got because I did like because decelerations magnitude is 1.5 times the accelerations magnitude and the acceleration is v over six. So the deceleration should be 1.5 equals 1.5v over six, which is v over four, which is the deceleration of the Yeah so the deceleration equals v over four. So then I did because time equals v over acceleration. So I did v over v minus v over v over four. So I got 4s and then 56 plus 4s is 60s. Yeah, Yeah, that sounds sounds good. And four b so zero to six is acceleration and then it accelerated to v meters per second and then a constant speed of v meters per second from six to 56 seconand then a constant acceleration from 56 to 60s. Yeah Yeah you got trapesian good. And then four c is 20v equals 24m per second. Yeah because you just the. I guess the way to do that given the distance would be would be you would just be using the trapezium. Yeah they Yeah Yeah Yeah perfect. Okay. So so I think that one was actually a relatively shortish booklet. Let me just double check that there wasn't any more on that one. Yeah. So I think what we'll do, we'll have a quick little talk about the reverse of differentiation because it will be useful when looking at when looking at variable acceleration, because we already looked at we already looked at constant acceleration here, which was the case of you know when you're using sv equations, Yeah there is another sort of area topic where you can solve problems where the acceleration is not constant. Yeah but usually this will end up being as a function of time. And so between the dislacsement and velocity and acceleration functions, which are functions of time, you can flick between them, right? You know how the acceleration is the gradient of a velocity time graph. Well, it doesn't have to be a straight line, it could be a curved line. And in that sense, you would effectively have to draw a tangent at any point. But if you know how the function of v as a function of t is prescribed, you can apply calculus, you can differentiate, okay? That function is a function of time. And if you know the exact form, and in sort of as level maps, that would be limited to a polynomial function. Yes, maybe t squared or t cubed or t to the half, you can differentiate it and then just evaluate it at a particular value of t and then you could find the acceleration at that point. Yeah and then it's the reverse operation for finding the distance of the displacement. Yeah. So rather than differentiating a velocity time graph to get the acceleration, you would do apply an operation which is known as integrating, okay, which is essentially just the complete reverse. And for as levels 's sake, it will just be, to start off with, restricted to let me just try and get an image to and into great. How are your function? What would be good? I just want to get like a good image and upload it. Suppose that one will do just so you can see the. The sort of mathematical notation of how it looks like that. So have you seen both of these before or just the top one? Just the top one. Okay. So is that I mean, the bottom one? You might also sort of see it written as the integral. It's actually probably more common like this. I didn't check this x to the end. Dx. Yeah you might more commonly see it like that. So you can literally see that it's just a complete reverse operation. Yeah when you take the derivative of a polynomial, you bring the power down Yeah and then subtract one from the power. So in words, the derivative. Of a polynomial function, you. Bring down the power. And then subtract one of the power here. Whereas for the integral. Of a polynomial function, we first of all, you know, and you do it in the verof order, rather than bringing down or dividing, we add one to the power and then divide. But new power, Yeah. So what I'll do, I'll get a very basic integration. Exercise boclip just so you can start practicing with some of those. Okay. Before we move on to applying this with variable variable acceleration. So you're just gonna to have a go, you know some sort of simple. Integration just to practice you know this this notation, but you can see it's not not too hard. And Yeah if you've got more than one term, you just do them term by term. And for question three, you'll have to expand out the brackets. There's just one extra thing that I forgot to mention. When you're integrating, if it's an indefinite integral, which these are, you need to add a plus c term where c stands for. Because in the reverse, if you differentiate a constant like five or ten or a million, Yeah it just goes to zero. So when you do it, when you the reverse, you have to assume that there could theoretically be some constant. There might not be, but you have to sue. Yeah and that can only be evaluated with boundary conditions. It's just like a little Yeah thing that you got to got to remember. Okay, let's start with some. I. Two x is five x squared minseven x ten. Three. So it's going to be X N plus one. Half x two powerful four dis. Square D X five X X. 25 over three execute and then minus seven x dx. So minus seven over two x squared and then ten dx ten x six going to be like. A half x to power I four plus five over three x cubed minus seven over two x squared plus ten x plus c. Px squared plus seven x minus two dx. Three x第x three X Q three execucubed seven x dx equals seven x squared over two equals seven over two x squared minus two dx equals minus two x so it's going to be。X cubed plus seven over two squared minus two x and the last one x plus four x minus three dx. X squared minus three x plus four x plus x minus twelve dx x square x equals x cubed over three. Goes one over a third x cubed x dx equals x squared over two half x squared minus twelve dex equals minus twelve x so x plus four x minus three dx equals two third x cubed plus a half x squid minus twelve x okay, sorry, I'm done. Okay, let's let's hear what you've got. Question one is a half x to the power of four plus five over three x cubed minus seven over two x squared plus ten x plus c plus see. Yep, sounds good. And then question two is x cubed plus seven over two x squared minus two x plus c Yeah hopefully vely. And then three is a third x cubed plus a half x squared minus twelve x plus c Yeah great. So let's do a couple more just to reinforce it. Itbe slightly harder, but still, I think quite manageable. So wait, why won't it let me out? Yeah, there we go. Question. Four just uses a slightly different notation. So instead of saying f of x and integral of f of x, they've given you f prime and you need to find f of x, which is just the integral of f prime. And then for question four also, you can evaluate the plus eta, and you do that by the fact that they give you a coordinate which lies on it, because you can say when x is one, y is equal to eight, or when x is one, f of x is equal to eight. And that enables you to evaluate the plus c ta. Question five, you're going to have to rewrite them in index notation. Yeah index notation. And for six, just stick before and five for now. And then we'll check those answers. And then I'll sort of mention about what's going on in six. But we probably won't go into the depths of evaluating it with limits just because you see that there's two numbers in the in the elongated s, the integrground symbol, that would just be evaluating it between two points. But I think let's stick with 45 for now. Okay. Let's add six square rored minthree x plus. Makes sense. Very good. Eight. 没事。And then say y eight equals 212. One squared plus 81 plus c. It equals two -1.5 plus eight plus c eight equals. 8.5 plus C C equals 1.5c equals one is a half. So then this is going to be to execute minus three over two x squared plus eight x minus a half. 一个多。Ten so y equals. Or x to alhalf plus x two of five minus g plus ten. And then. Just going to be. Squid. One sets x -11 equals minus x of minus one. Ten dx is going to be ten x so this four x two plus ten x equals 83x three over two minus x. No x plus ten x plus c. Okay, so I'm done. Okay, let's see what you got. So for the first two questions, question four is two x cubed minus three over two x squared plus eight x minus a half good. Good. And question five is eight over three x to the power of three over two minus one over x plus ten x plus c. Yeah, great, great. That's good. I think that will be Yeah that will be good for then starting the next topic next time on variable acceleration. Yes, the next topic in in mechanics. Okay, okay, okay, perfect. All right, then enjoy your weekend. I'll see you next time. All right. Thank you. Bye bye.