A Level Maths Tuition Lesson 22 Mechanics: SUVAT Exam Practice!

Okay, hello and welcome back to another one of these A Level Maths lessons. So today we're moving on to AS Mechanics which is very exciting but also pretty dang tricky. So mechanics can get a bit wonky so I'm just going to take you through some of the best ways to actually understand it and get it going. So first of all we're going to start with SUVAT because I think it's the simplest thing to start with. SUVAT just stands for the Constant Acceleration Equations. So you can only use them, shockingly, when you have a constant acceleration, which is the assumption we're going to use for pretty much all of A Level Mechanics. So these are the five equations that you need to be able to use. I wouldn't bother memorizing them personally but instead I would actually just write them down and have them on a piece of paper or a post-it note next to me whenever I do mechanics and then over time you're just going to remember all of them. So what do these letters stand for? So when we have SUVAT, what does it stand for? Well the S is for displacement. So it should be noted that displacement is not distance, it's distance in a direction. So it is a vector. If you like, it's just the vector for distance. So we need to actually make note of which direction is positive, which direction are we taking to be positive. U is the initial velocity. So again this is a vector for speed. So it's the initial speed but as a vector. V is for final velocity, exact same as above. A is for the acceleration, which again is a vector. So it can be positive or negative. And lastly we have T for time, which is of course a scalar. So almost everything in these equations is a vector, which means we have to take into account direction. The key part about mechanics, and this is true for AS and AT mechanics, for all of mechanics and even if you do physics as well, physics mechanics, you need to draw diagrams and take directions as positive. There are plenty of students every single year that come up to me and they say, well how do I do this? I don't understand. I'll draw a picture and they'll understand straight away. So drawing a diagram will save you time and it'll actually make the questions possible. I personally would never attempt a question without doing a diagram. And just to point that out, you tend to get quite a few marks for mechanics questions, but you don't need to do much work for them. Sometimes you may only need to do one or two lines and you get four marks. Why? Because you need to draw diagrams. So we're going to start off with a pretty tricky question. It says a ball is projected vertically upwards with a speed of 20 meters per second from a point h meters above the ground. The ball hits the ground five seconds later. Find the value of h and the speed of the ball as it hits the ground. So here's what we have. We have a ball and we have the ground somewhere down here. This distance is h and it goes upwards with a speed of 20 meters per second. And then what's going to happen is it's going to go up and then it's going to come straight back down and hit the ground. And that happens at t is equal to five seconds. So what else do we have? Well in questions like this it can be useful to take downwards as positive. So whenever we throw things up in the air or drop things, I personally take downwards as positive. However, you don't have to if you don't want to. You can pick whichever one you want and you'll get the same answer. So the other vector that we have is we have obviously gravity which you're going to take as 9.81. Now in your exam it will say what you should take gravity to be. Usually they'll say just assume it's 9.81 but they might say use 10 or use 9.8. So for part a we need to find the value of h. So what would be a good idea for this? Well there are two things that we could do. First thing is we could actually take the entire situation as our example. So what I mean by that is we're going to take it from the initial point here all the way to the end here. And if you notice the overall displacement is h meters downwards. So if I'm taking downwards as positive my s is positive and it's just equal to h. My initial velocity is going upwards so it's minus 20 because again I'm taking downwards as positive. So if you were to take upwards as positive u would be positive and h would be negative. My final velocity when it hits the ground or just before it hits the ground is you could take zero but I'm going to put as a question mark for now. My acceleration is positive 9.81 and my t is five seconds. So the two things I don't know are s and v but for part a I'm only interested in s. So now I'd like to write out what I have. So I have s, u, a and t. So what equation can I use? Well if you look through here can I use this equation? No because I don't have v. Can I use this equation? Well I have s, u, a and t. I can. So that's what you need to be doing as you go through these equations. You need to just have them written out in front of you and think about what ones you can use. Which ones can you use? Well it depends. You want your unknown in the equation obviously and then you want the other letters that you already have. So I have I want s, I need u, sorry I have u, t and a. So that's one that I'm going to use. So we have h is equal to minus 20 times 5 plus half 9.81 times 5 squared. And we should get, so let's go, we're going to use a calculator. So we have minus 20 times 5 plus 0.5, 9.81, that's 2, 5 squared and we get 22.625. So the key thing here is I need to check does this make sense. Well my h is positive because it's downwards. Now if you took upwards as positive you are going to get a negative value for h. Which is completely fine. So imagine this, if I took upwards as positive this 20 becomes positive but this 5 becomes negative. This 9.81 becomes negative. So you're going to get negative 22.6 which is accepted. So what they're looking for in the exam is that you've taken a direction as positive and you've stayed consistent with it. So if you get negative 22.6 they will give you the completely fine. Perfect. For part b it says then find the speed of the ball as it hits the ground. So we're working out what v is. Now if you look I have s, u, a and t. So I can use whatever equation I want. So which one would I recommend? Well I personally would recommend trying to use one that doesn't have the letter that we worked out in the previous question, just in case we get it wrong. There are error carried forward marks but I would personally just use v equals u plus at in this question. So we get minus 20 plus 9.81 times 5. So again can you see these are three marks each but realistically after you've drawn your diagram I'm subbing into an equation. That's one line. This is one line as well. I'm typing out a bit in a bit more detail just because I'm explaining it and we get 29.05. So we can say 29.1 meters per second to three significant figures. So we just got six marks and it took me less than six minutes to explain it. So in an exam you'll be able to do this in a couple minutes to be honest. So how long was that? So we've been recording for about eight minutes here. Not bad. Next. A car passes a point A with a speed of 20 kilometers per hour. The car accelerates at a constant rate and 10 seconds later it passes point B with a speed of 70 kilometers per hour. Find the acceleration of the car in meters per second. So the very first thing I want to do is I actually want to convert these two units. So 20 kilometers per hour. If I want it in meters per second I'm going to change kilometers to meters by timesing by a thousand and then hours to seconds you need to do divided by 60 because there's 60 minutes in an hour times 60 because there's 60 seconds in a minute. I'm going to do the same thing here. So 70 times a thousand over 60 times 60 is just 3600 though so you can just use that as your values. So let's work out this. So that is we'll leave it as 50 over 9 just to get a really accurate answer. Same thing here 175 over 9. Yeah it's gonna be annoying. Okay so I'm gonna leave it as a fraction just to get a super accurate answer but again you can avoid that. So let's actually draw this out now. So we have a point A and a point B. So here he's traveling at 50 over 9 meters per second and over here we're traveling at 175 over 9 meters per second and there's a constant acceleration and 10 seconds later. So t is equal to 10 seconds and we have this constant acceleration which we don't know we're going to call it A and obviously we're going to take to the right as positive. So you could take to the left as negative if you're a bit Machiavellian and everything would just be negative so I'm not sure why you'd want to do that. We don't know what the distance is so we can even write that down and we can just call it S and that's all the information we need. So for part A find the acceleration of the car. We have S is don't know but we also don't care. U is 50 over 9. B is 175 over 9. A is don't know and we most certainly do care and t equals 10. So we want A and we have U, V and T so what we can use is we can use the equation V is equal to U plus A T which rearranged for A would give me V minus U over T. So 175 over 9 minus 50 over 9 over 10 is my acceleration. Okay so let's type this in. It is best to keep your values as accurate as humanly possible. 25 over 18 so we'll write our answer as 1.39 per second squared and that again that is 2, 3, F. For part B it then says work out the distance AB. So that is our S. So you can work this out in pretty much any way you'd like because we have all of the equations. So we could just use S equals U plus V times T or because it's one that I have on the front of my mind S equals U T plus half A T squared. Okay so we have U which is 50 over 9. T which is 10. U T plus half A which is 1.39. T is 10 squared. We get 125 meters. Again that's 2, 3 significant figures. Not too bad. So if we keep going to the next one now we have the stone is dropped from a point 120 meters from the ground. So we can assume it's above the ground. Find the time it takes for the stone to reach the ground. Okay so with these questions this is called free fall and it's actually pretty blessed. So when something is dropped the only thing you actually need to know is the height. If you know that you can work out literally everything else which is why it's so kind of impressive. So the acceleration is obviously A so 9.81. I'm going to take downwards positive because we're dropping something. If we're dropping it it means the initial speed is zero because we're just letting go. So if we begin to drop something it will start at zero and then it will accelerate and V is we don't know. Time we also don't know which is actually what we're working out for part A. So again if you're dropping something the only thing you need to know is the height or the distance that you're dropping it from. So we have S is 120. We have U which is most certainly zero. V which is don't know, don't care. A which is 9.81. T which is again don't know, don't care. Now the equation that you always want to use when you're dropping something is you're going to be using S equals UT plus half AT squared. The reason for this is U is zero. So we actually get rid of that instantly. Really really powerful equation. So then T is just going to be 2S over A square rooted which we can just shove into the calculator. And again you're going to use this pretty much every single time you have. Just a free fall question. It's a really common question and look how quickly we did it. Three marks again. 4.95 seconds to three significant figures. Really quick, really easy. B, the speed at which it hits the ground. You can use pretty much any equation you want because we have everything else. So V equals U plus AT is a very fast one to use. So we have U which is zero. So again we're just getting rid of that. A which is 9.81. So we can just do our answer times 9.81. So we even keep a perfectly accurate answer. 48.5 meters per second. Three marks for that. Not too bad I think. Okay. The particle moves along a straight line from point X to Y with constant acceleration. The distance XY is 120 meters. The particle takes eight seconds. The speed of the particle Y. Ah okay interesting. Cool. So we have this is 120 meters. Constant acceleration. So I'm going to call that A because we don't know what it is. Once again I'm going to take to the right as positive because I'm pretty sure everything's going to be here. It takes eight seconds as well. Now the speed at Y is double the speed at X. So what we can call this is we can say that U is equal to just U. And on this one we're going to write 2U because it's twice the velocity. So the speed of the particle at X. So again let's just write X and Y. Okay so let me just double check we've got everything we need. Yeah we have. Cool. So for part A, speed of the particle at X. So S is 120. U is U. B is 2U. A is don't know. T is 8. So the equation that we want has to have S, U, V and T in it. Which if you look you can V equals. So everything except A basically. So we can use S is V plus U. Oh god. S is V plus U over 2T. Yes. So V plus U over 2T. Cool. So we have 120 equals. V plus U is going to give me 3U over 2 times 8. So we can cancel this with this. That gives me 4. So we have 120 equals 3 times 4 12U. So U is just equal to 10 meters per second. Not bad. Which means here it's 20. So again like the question sounds complicated. Even kind of looks complicated. But if you draw a diagram and write down what the variables are then you're done. You're literally just kind of it's like you're doing like a crossword or something right. You're just checking which ones you have and which ones you need. So I knew that I didn't have A. So I can't use anyone except number 4. That's it. And then from there it's basic algebra that you guys have been doing for like a decade at this point. Really easy. Really quick. Not saying that it can't be difficult. There are difficult parts to this. There are six mark questions that can come up. With acceleration we have literally every single thing that we could possibly need. So you could just use for example A is V minus U over T. So that's just the definition of acceleration. So we have 20 minus 10 over 8. So that's just 10 over 8. Which if you want to work out you are more than welcome to. Sometimes they accept just fractions there. 1.25. I don't know why you need to calculate for that. But yeah. Perfect. Again not too bad I think. So a car passes point A with a speed of 5 metres per second. It accelerates at a constant rate and 8 seconds later it passes point B with a speed of 20 metres per second. Find the acceleration, the distance and the time it takes to reach the midpoint. Okay. So constant rate and 8 seconds later. Okay. So again I'm thinking here that we can just use the definition of acceleration. Acceleration is V minus U over T. So we have 20 minus 5 over 8. 8 is 1.875. So you can say 1.88 metres per second squared. B, the distance between A and B now. Well if we actually write down all the variables we have. S is what we don't know. U is 5. V is 20. A is 1.88 and T is 8. Okay. So we can use literally any equation we like. So we've used U T plus half A T squared. Classic. So we have U T. Let's just do this. Perfect. And we have 100 metres. Again we do that to three significant figures. Perfect. Not too shabby at all. So getting through these pretty quickly. We do have some more like recent exam questions as well. But these are all exam questions too. So train moving with constant acceleration passes through three points. A, B and C. Where A, B is 40 and B, C is 60. The train passes through the speed of 10 metres per second and six seconds later passes through C. Find the acceleration of the train. The speed at which the train passes through B and the time it takes for the train to move between B and C. So again we're going to use a diagram because again this is the only proper way to get all of the marks in a mechanics paper. Drawing good diagrams is definitely what is required. So A, B is 40 metres and 60 metres. 10 metres per second here. And to get from here to here it's six. No, sorry. To get the whole thing it's six seconds. Okay. Anything else? And it has a constant acceleration between A, B and C. Okay. Cool. So part A we're going to use SUVAT. So S equals. What I'm going to use SUVAT for actually I should say is the whole distance. The reason for that is if I just do between A and B for example I only have two pieces of information which is the initial speed and the distance which isn't enough information to do SUVAT with. You generally need three pieces of information. So I'm going to use the full distance which gives me the full distance. One piece of information, two pieces of information, three pieces of information there. That's how I'm deciding. So S is 100 metres, U equals 10, B equals don't know, A equals don't know, T equals six. So we have S, U, A and T that we want. So S is UT plus half AT squared would be the half A6 squared. 100 equals 60 plus 18A. So 18A is 40 which means A is equal to 40 over 18 which is 2.2 recurring. So let's just do 2.22 to three significant figures. Again, not too bad once you actually break it down. For part B, the speed at which it passes through B. So we know what the acceleration is now. So again if we just look from A to B we have S is equal to 40, U is equal to 10, B is equal to what we're trying to find out, A is equal to 2.22 and T we don't know. So what we can do here is we have S, U, A and we want to work out 40. And that is 16.7 metres per second for that one. Again, if you want me to go through that one more time, I'm more than happy to. Hopefully that makes sense. Lastly, for part C, it just says calculate the speed or the time it takes for the train to move between B and C. So now we have S is equal to 60 metres because that's the distance between B and C. U is going to equal 16.7 because that was the speed when it passes through B. V we don't care about. A again is still going to be 2.22 and T is what we're working out. So if we have S, U, A and we want T, then a good equation to use is S equals UT plus a half AT squared. So if we want to rearrange for T, it's actually going to give us a quadratic, which makes it ever so slightly harder. But let's do it. 16.7 T plus a half of A, so 1.11 T squared. So if I move the 60 to the other side, so it is common to actually get a quadratic like this. And then we're going to get two values for T. We're going to use the good old fashioned quadratic formula. So which is minus B plus minus blah, blah, blah. So minus B, which is 16.7 plus or minus, I'm gonna start with the plus because I feel like it. 16.7 squared. B squared minus 4A, which is 1.11. C, which is minus 60 over 2 times 1.11. Whoops, that was the wrong bracket. 2.996. Okay, I'll make it 3.00. Or it's T equals negative 18. So which one are we going to take? Well, we're obviously going to take, the question says obvious, we're going to take T is 3, because it's positive. Okay, you can't have a negative time. What this actually means in terms of our model is that it was before the model started. But again, we can't have negative time, so we're not going to deal with that. Okay, so I think this has given you a pretty good foundation into like how SUVAT works and everything. It's not too tricky. But if you do have any kind of questions about specific questions, then do let me know. But for now, I'll let you get on with your evening. Thank you.