A LEVEL MATHS: SUVAT Exam Practice LESSON 22

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 uh very exciting uh 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 Sue that because I think it's the simplest thing to start with uh suvat just stands for the constant acceleration equations so you can only use them shockingly when you have a constant acceleration okay 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 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 okay 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 four distance okay so we need to actually make note of which Direction's positive which direction are we taking to be positive U is the initial velocity okay 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 and the key part about mechanics this is true for as and A2 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 so uh 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 you know one uh one or two lines and you get four marks why because you need to draw diagrams okay so we're going to start off with a pretty tricky question okay so it says a ball is projected vertically upwards with a speed of 20 meters per second from a 0.8 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 okay and it goes upwards 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 5 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 down which is positive however you don't have to if you don't want to right 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 is 9.81 now in your exam it will say what you should take gravity to be um 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 to for this well there are two things that we could do first thing is we could actually take the entire situation as our example okay so what I'm 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 okay 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 down which is positive so if you were to take outputs as positive you would be positive and H would be negative my final velocity uh when it hits the ground or just before it hits the ground is you could say it's zero but I'm going to put it as a question mark for now my acceleration it's positive 9.81 and my T is five seconds okay so the two things I don't know are s and V but for part A I'm only interested in s okay so now I 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 it can I use this equation no because I I don't have V can I use this equation we'll have S U A and T I can okay 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 and sorry I have u t and a so that's the one that I'm going to use okay so we have H is equal to minus 20 times 5. plus half 9.81 times 5 squared okay 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 two five 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 okay so imagine this if I took upwards as positive this 20 becomes positive but this 5 becomes negative this 9.81 becomes negative okay so you're going to get Negative 22.6 which is accepted okay 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 marks it's 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 right so which one would I recommend well I personally would recommend trying to use one that doesn't have the letter that we worked at 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 a t 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 it's 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 of minutes to be honest so how long was that so we've been recording for about eight minutes yeah not bad next a car pass is 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 okay so 20 kilometers per hour if I want it in meters per second I'm going to change kilometers to meters by times in 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 let's do values so let's work out this so that is oh God we'll leave it as 50 over 9 just to get a really accurate answer same thing here [Music] 175 over nine yeah it's gonna be annoying okay so I'm going to 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 is positive so you could take to the left as negative if you're a bit Machiavellian um 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. V 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 B 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 is 1.39 per second squared and that again that is to 3 S F for Part B it then says work out the distance a b 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 the 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 UT plus half a which is 1.39 t 10 squared we get 125 meters and again that's to three 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 these questions this is called free fall and it's actually pretty uh blessed okay 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 down which is 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 u t plus half a t 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 full question it's a really common question and look how quickly 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 g plus a t is a very fast one to use so we have U which is zero so again we just getting rid of that uh 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 for 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 just going to call that a because we don't know what it is once again I'm going to take to the right is 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 of X so what we can call this is we can say that U is equal to just U right and on this one we're going to write 2u because it's twice the velocity yeah so the speed of the particle at X so again let's just write X and okay so let me just double check we've got everything we need to do 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 eight so the equation that we want has to have s u v and T in it which if you look you can use S V equals so everything except a basically so we can use S's V plus u oh God this is V plus u over two t um yes so V plus u over two t cool so we have 120 he calls B plus u is going to give me three U over two times eight so we can cancel this with this that gives me four so we have 120 equals three times four twelve U 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 uh 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 four 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 I'm not saying they can't be difficult there are difficult parts to this there are six smart questions that can come up with the 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 ten over eight which if you want to work out you are more than welcome to sometimes they accept just fractions though um 1.25 I don't know why I used to calculate for that but yeah perfect again not too bad I think so a car passes point over the speed of 5 meters per second accelerates at a constant rate and eight seconds later it passes point B speed of 20 meters per second find the acceleration the distance and the time it takes to reach the midpoint okay so um concentrate 80 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 five over eight 1.875 so you can say 1.88 meters 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 five V is 20. a is 1.88 and T is eight okay so we can use literally any question we like any equation we like so we use u t plus half a t squared classic so we have u t let's just do this perfect and we have a hundred meters again we do that to three significant figures perfect it's not too shabby at all so I'm getting through these pretty quickly we do have some more like uh recent exam questions as well but these are all exam questions two train movement with constant acceleration passes through three points a b and c where a b is four C and B C is 60. the train passes through the speed of 10 meters 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 mechanics paper Okay drawing good diagrams is definitely what is required so a b is 40 meters so and 60 meters 10 meters per second here and to get from here to here it's sick no sorry to get for the whole thing it's six seconds okay is there 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 uh what I'm going to see that for actually I should say is the whole distance the reason for that is the only 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 Sue that 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 meters U equals ten 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 u t plus half a t squared would be the best one to use yeah we have everything we need so 100 equals 10 6 plus half a six squared 100 equals 60 plus 18a so 18 a equals 40 which means a is equal to 40 over 18. which is ah 2.2 recurrings let's just do 2.22 to three significant figures again not too bad once you actually break it down right for a 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 V is equal to what we're trying to find out a is equal to 2.22 and T we don't know so okay what we can do here is we have s u a and we want to work out V so V squared equals U squared to a s 10 squared to effect cool so plus two point two 40. and that is 16.7 meters per second for that one so 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 meters 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 V uh through b B don't care about a again is still going to be 2.22 and T is what we're working out okay so if we have s u a and we want T then a good equation to use is s equals u t plus a half 80. squared savvy yeah 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 16.70 Plus so a half of a so 1.11 t squared so if I move the 60 to the other side we're going to so it is common to actually get a kind of 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 so minus B which is 16 point whoops okay call 0.7 plus or minus I'm gonna start with the plus because I feel like it 16.7 squared B squared minus 4 a which is 1.11 C which is minus 60 over 2 times 1.11 whoops that's one bracket 2.99 6. uh 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 I shouldn't say it's obvious we're going to take T is three 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 hopefully this has given you a pretty good foundation into like how Sue that works on 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