welcome to m1 today we're looking at chapter 2 constant acceleration now before we get into the formally for constant acceleration we need to look at the graphs that you need to know and understand so there's three types the first one is a displacement time graph now displacement is the letter s and obviously t for time if you're unsure what displacement is well it's essentially distance but as a vector okay so it has direction so think of it as distance from a point so imagine you know the school you go to is five kilometers away if you go from your home to the school and back you would have traveled a distance of 10 kilometers five kilometers there five kilometers back but your displacement would have been zero because your start point and your end point are the same okay displacement is a vector so it measures the distance from the start point to the end point so in that respect that's what's the real difference between the two now in terms of the graph the velocity of the graph is the gradient of the graph so for these the gradient of this graph is the velocity so if i have a horizontal graph then it means that this displacement this distance hasn't changed the distance has stayed the same value while time is changing so it means that it's stationary whatever we're looking at is stationary or not moving if we have like a straight diagonal line like this second one then it means that the distance is increasing at a steady rate so this means we've got a constant velocity and the velocity as i said is the gradient of the line if we've got more of a curve like an exponential this gradient is getting steeper and steeper therefore the velocity is getting steeper so the velocity let's say the velocity is increasing okay and if the velocity is increasing it obviously means it is accelerating okay the objects are accelerated so that's the displacement time graph okay so just to go with that a couple of obvious things that i have said so velocity equals the gradient as the velocity is the rate of change of displacement hence the gradient like ds by dt um and then you've got your average velocity is just going to be the total displacement from your starting point divided by time whereas the average speed will be a total distance traveled divided by time so the two things that this are different you know so like velocity sorry average velocity would be this total distance traveled divided by time and in this case the average speed would be the same but if this line did something like went up and say went down to a point here then the average speed and the average velocity would now be different okay because the distance let's say the displacement is only this much whereas in terms of the distance we traveled up to here and then we travel back to here so that distance is a lot more okay so there is some differences there um i will pop them up but that's the the essential stuff that you need to know second type of graph is a velocity time graph so this measures obviously velocity against time okay so time is in these ones is always going to be the horizontal uh axes now looking at a few scenarios for this one so first let's talk about the gradient um velocity and time so if i was to differentiate this and look at the rate of change of velocity which is my dv by dt that would be acceleration okay so in this case the acceleration is equal to the gradient okay that rate of change of velocity now you'll need to do this first with a different color but on the graph if i've got a line along that x axis then it clearly means that there is no velocity okay so zero velocity would be stationary so you can see you know there is a bit of a difference you know if you're looking at the displacement time graph stationary is a horizontal line but it can be anywhere whereas for the velocity time graph it obviously has to be at zero now if i draw a horizontal line kind of anywhere that's now when i've got my constant velocity so you can see here now that the graphs are different so this was the displacement time graph it was just a linear graph kind of in the form of y equals m x plus c whereas now it is a horizontal graph okay this means that wherever this value is here says 10 it means that the velocity is 10 all the way along even when time is changing so that is my constant velocity and then finally kind of matching up with what we've done before if now i have a first not a vertical line uh just a diagonal line in the form of y equals mx plus c now this is now when my velocity is increasing or i've got a constant acceleration so velocity is increasing okay i haven't written down about the acceleration as i think that's kind of obvious if the velocity is increasing it's obviously accelerating um but you can see now the difference for the same thing so stationary constant velocity and when it's accelerating okay now you need to be able to recognize these graphs okay when they come up in questions one additional thing to look at is area under the graph okay so area under my velocity time graph is going to be equal to the distance or in this case the displacement and i should clarify this a little bit so the area under the graph of like a speed time graph is the distance traveled now in terms of a velocity time graph as long as the velocity remains positive then the area under it is displacement if it isn't positive so if you have a negative velocity um then the area of the graph is distance not displacement okay so you do have to be a little bit careful about that but for that to happen obviously the velocity would need to be below the x-axis so it should be obvious now my final type of graph is the acceleration time graph now for the acceleration time graph we don't have to worry about diagonal lines as all sloping lines you know as these are way beyond the scope of this syllabus all we need to know is a kind of oh three main scenarios so the first one under this in black is if we have it go along the x axis okay that is obviously when acceleration equals zero so that is a constant velocity so it's kind of like linked with these ones down here um the second type of graph we've got is when i've got one kind of appear in the positive so that is an increasing velocity as we have a positive acceleration so increasing velocity and then my last one as you might have guessed it is a negative acceleration so it's decreasing velocity okay and that's essentially what you need to know about our three types of graphs now in this third graph the area would be the velocity okay so if i was if i was to work out the area it would be equal to the velocity but obviously you have to be careful as some of it might be underneath the axis some might be above so you'd need to do it in sections okay so you know you would do it just for specific sections of your graph where you get that horizontal line we like to change it and while i can't quite remember to be honest i've told my head whether this is in this m1 syllabus but it's certainly worth mentioning if the velocity if i look at the displacement time graph if the velocity is the gradient of that then the velocity is equal to ds by dt looking at my second graph velocity time graph then the acceleration is the gradient there so that's equal to dv by dt okay and this is quite nice because what it essentially means is that you if you integrate acceleration you actually get velocity and if you integrate velocity you should get displacement and here's another kind of way of looking at it differentiate displacement you get velocity differentiate velocity you get acceleration because we're looking at the gradient of the displacement time graph for velocity gradient of this for acceleration so differentiate differentiate and then going backwards is integrating isn't it so integrating acceleration i would get velocity therefore area under the acceleration graph is velocity integrating velocity i get displacement therefore area under the velocity type graph is displacement okay so that's a bit more simplistic and hopefully just helps you out there now i'm not going to do any examples on these graphs i think they're very straightforward and you should be able to do it from this information obviously if you do have problems or struggles with this please you know let me know in the comments and you know i'll expand this video to include then some examples and work through questions on the graphs also just before we get into our constant acceleration formally if you haven't subscribed to my channel so far please just hit that subscribe button now help me out okay and maybe help yourself out in the future also you know if you're in school and anyone else is and your friends are finding it difficult or need some help you know feel free to point them in the direction of my channel um it would certainly help me out as well next we want to talk about the constant acceleration formally more commonly known as suvad and this is just because of the letters we use so what i'm going to do first is let's go through each letter make sure everyone knows what each one means so s is my displacement u and v are both velocities u is my initial velocity whereas v is my final velocity a is acceleration and t is time and just going with this i'm going to pop it up there sometimes we do look at acceleration in terms of gravity things acting under gravity and for m1 we generally use gravity as being 9.8 meters per second squared i know in physics you often use 9.81 to go an extra decimal point um luckily most of the questions that you will see where gravity is involved it will often say use or take g to be 9.8 meters per second squared okay so that's kind of everything in general we'll be looking at displacement in meters our two velocities in meters per second our acceleration meters per second per second means per second squared and time in seconds so that is what we'll be using it can't say this is a hundred percent for every question um but generally even if it isn't you'd often change it into these units in terms of kind of a little diagram to kind of help you out too and these are the kind of things you know i do often draw diagrams of mechanics i find them useful and some questions certainly necessary but here you've got your initial velocity and then the final velocity that your particle or object is moving at and during all of this you know you'll have some sort of constant acceleration now the distance between your start point and your end point is going to be your displacement okay this is because we're going to be moving in a straight line so that is our displacement and then finally this will be our position [Music] after t seconds okay so that's just a little diagram to kind of visualize you know what we're talking about next going with this as i was talking i'm gonna give you all the formally in one go here and then we'll actually go through finding it so we've got five in total now i'm pretty sure those of you who do physics will say that there are four formerly which is kind of true it was definitely just the four that we used when i was in school and the fifth one really we just used by it would come up so rarely you know if you had a question you just used to the formerly to get it to get the answer okay so here are my five formally i've kind of just rearranged them in the order we will be working them out and you know i'm just gonna put what we're missing so you'd use a formula or formulae based on what information you had and you know essentially what you didn't have so v equals u plus 80 you can see there's no s in there my second one you can see contains no acceleration v squared equals u squared plus 2 a s is no time s equals u t plus a half a t squared is no final velocity and then very similarly s equals vt minus a half 80 squared there's no initial velocity so these four formally you need to know them and you're also going to need to be able to derive them okay so the first two my v equals u plus a t and s equals u plus v over two times t these can be driven from a velocity time graph the remaining three come from these formulae okay so as i said you do need to be able to do this you could get asked this in an exam is in the syllabus for you to know it so i'll go through it now before we do uh look at some questions using these formally so as i initially said this first two formally come from a velocity time graph so if i start by just getting an idea of a velocity time graph something like this line okay where it's representing the motion of a particle because we've got a constant acceleration okay it needs to be a straight line constant acceleration and let's put some things lines on the graph to make it easier so at the end here that will be a time t this start point is going to be when my velocity is zero so that's quite important on this top point is my final velocity there so this is u and this point here is v so that is essentially all i need to be able to get my first two formerly so the first thing we want to do is think about the gradient now this gradient is going to be my y distance divided by my x distance that's what we're looking at yeah so gradient of line now the y distance will be this distance here which is v minus u and that is going to be divided by the horizontal distance which is from 0 to t so that's going to be t now we already know previously as we were talking about the gradient of a velocity time graph is acceleration it's dv by dt so acceleration equals v minus u over t and what i can now do is rearrange this so a t equals v minus u take my u to the other side and we get v equals u plus a t and this is my first equation or formally i shouldn't say the second one i can get from the area under the graph so that's the shaded area so area under the graph but you can see here is this purple shaded area well it's a trapezium isn't it you know trapezium has the formula a b and h and the area is a plus b over two times h so in this case my a is this distance here which is u my b would be this distance here which is v divided by 2 times my height which is this distance here which is t now area under the graph of a velocity time graph is the displacement as long as the velocity is positive in this case so it is the displacement so s equals u plus v over 2 multiplied by t and now we have our first two formally so you need to know how to derive these two formally from this graph so exactly essentially exactly the steps i've just taken here now that we have these let's go ahead and show you how to get the remaining three formally so first we're gonna work out v squared equals u squared plus two a s this has no t so we wanna eliminate t so from the one we can see we can get v minus u divided by a and that will be t so i've just rearranged it and sub into two so s equals u plus v over two multiplied by v minus u over a so what i want to do is first multiply by 2a so we get 2 a s is u plus v multiplied by v minus u so 2 a s equals now i'll write this out in full but you know not always completely necessary with these types of ones difference of two squares but uh uv minus u squared plus v squared minus u v these will obviously cancel take my u squared to the other side and we get v squared equals u squared plus 2 a s and this becomes my third formula now if we look at s equals u t plus a half a t squared there's no v so that means we want to substitute equation one straight into equation two s equals u plus u plus a t times my t divided by two so you can see now if i just expect add these two together i get two u i've got eighty times t i just just put it like this first of all shall we so now i've got 2u times my t divided by 2 so it's going to be u t plus and this one now is a t squared okay um you could have taken an extra step in there you know if you want to you could have then traps then the two u t plus a t squared over two u t plus a half eighty squared or anything kind of along those lines but that gives us equation or formula i should say number four and finally finding my final equation is very much like we did with the last one so first we want to rearrange one to make it u equals so u equals v minus a t and then sub it into equation two so that means i get s equals now instead of u i've got v minus a t plus my v times t over two so now you can see that we've got two v minus a t times t over two or two v t minus a t squared over two which is v t minus a half a t squared or equation four and there are all five of my equations just popping them all back up for you in one go before we head into some examples now in this example the stone is sliding across a frozen lake in a straight line so i want to do a little quick diagram just to put this information on it's not necessary in all questions but it's often can be useful so we've got a stone sliding across a straight oh a frozen lane in a straight line so u is equal to 12 meters per second the acceleration is going in the opposite direction so what we want to do is we always want to kind of give one direction as the positive direction we're going to use so i'm going to use left to right as positive therefore my acceleration will be a negative 0.8 meters per second squared and if i'm thinking about the total distance the stone traveled well it's slowing down so the total distance will be when the velocity equals zero and at this point i don't know the time and i don't know obviously the distance which is what we want to find so what i'm looking at here is we know u well let's put them down u is 12 v is zero a is minus 0.8 we actually want to find s and we could find t we don't need to so when i look at this i go right we've got u v a and we want to find s so no t no t is v equals u squared plus two a s so v is zero so zero squared zero that's equal to twelve squared plus two times minus zero point eight s so i'm gonna multiply out my bracket and take it to the right so 0.8 times 2 is 1.6 s is equal to 144 therefore s is 90 meters now we want to find the speed of the stone at the instant air traveled half of this distance so in this case u is still 12 v is what we actually want to find the acceleration is still the same as is constant and this time s would be equal to 45 half of that distance so now i'm looking at uv a and s so it's the same formula isn't it so v squared equals u squared plus 2 a s so v squared equals 12 squared plus 2 times minus 0.8 times 45. so v squared equals 72 that means that v would be equal to six root two or eight point five meters per second this is the two significant figures here or eight point four nine if you did it to three significant figures okay and that's an example of what you need to do you just put a sketch down where appropriate sometimes it's a bit too obvious this one's quite an obvious one i don't really need a sketch but a lot of them when there's more steps can be really useful and also put down your suvat values that you know and what you don't know okay because you need well basically you need three values aren't you plus one defined to use any of the formally now this second example here is one under gravity so first we want to start off with a little diagram and we've got a ball that is thrown so ball a is thrown 46 meters above the ground so thrown from a tower directly down so u a equals five meters per second pull b is thrown directly up so u b equals 18 meters per second and obviously somewhere they collide we need to find that distance and we need to find that distance from here where a was thrown so let's look at ball a we'll take downwards as the positive so we know that u is five acceleration is gravity going in the same direction um v we don't know t we obviously don't know an a sorry s we don't know now we're not worried about v okay the velocity they hit each other we obviously want to eventually find s but when i'm thinking of the two balls their s values are going to be different aren't they because one is going to travel a distance or displacement from the top the others going from the bottom t is my key thing here that i want to first initially work out okay but let's put that s back in there because we need something in terms of s as well don't we so s equals u t plus a half a t squared so s is equal to five t plus a half g t squared and this is the distance that a is traveled let's look at ball b now ball b i'm going to take positive as going up so u is going to be 18 where the acceleration is now negative g time t we don't know but same as the other one and s let's put that as s b which will mean that i'll change this one to s a whereas in terms of the times both my times are going to be equal to each other so i can put them both as t they're the same t so s equals u t plus a half a t squared remember as always write the formulas down every time you use it every time you're using it and you're writing it down it's just giving your brain more chance of remembering it which is a good thing so here we have 18 t plus a half times minus g t squared so we've got 18 t minus a half g t squared for s b so we've got two pieces of information here we've got at the moment three unknowns though t is the same but the sb and the sa are different however we know one's thrown from the top one thrown from the bottom they're gonna meet somewhere along this lines and when they do wherever they meet they would have traveled the total distance of 46 meters between them because even if this one had traveled 40 meters up then this one would have traveled six meters down okay because wherever they meet that distance plus that distance is going to equal 48. so i can say that the two distances say 46 is equal to 46 meters that now allows me to put these two equations together and have just the one unknown so now we've got five t plus a half g t squared plus 18 t minus a half g t squared equals to 46. simplifying this 18t plus 5t is 23 t half gt squared minus a half gt squared would cancel out so that's going to equal 46 so the time equals 2 seconds now that i know the time i can effectively substitute it into either formula okay my b or my a but i want the distance from point a where a was thrown so it makes way more sense to work out s a if i worked out s b i then need to take that value away from 46. so we have t plus a half g t squared t is two so you got five times two plus a half g times two squared and that gives me 29.6 meters and that is my distance from a now a point on accuracy here if we use in [Music] g is 9.8 what we're basically saying is that this has corrected two significant figures which means that for accuracy i should really put my answer correct to two significant figures i shouldn't be marked down in an exam if i didn't on the left is 29.6 but just for accuracy that's what you should really do um not i wouldn't worry too much about it i just want to make a note of it so i'll give you a couple of questions to have a go at and as always i'll pop the answers up at the end of the video now in this example we can see it starts at a height of 2.5 meters above the ground and it's fired upwards and then comes down and catches it just before the hit the ground so you can basically assume that it's hit the ground there and what we need to do is a velocity time graph so if i think of my velocity time graph i'm going to be starting with like a velocity of 24 meters per second and as it rises in the air it's going to slow down at a constant rate until it hits zero at which point then it'll start to drop back down to the ground so we've got exactly the same these two should be symmetrical to this point and at this point it will be hitting that kind of 2.4 meters sorry 2.5 meters which is the height of the machine obviously if it drops a little bit further it's going to go a little bit faster so that graph will finish a little bit above this 24 here see just a bit above now part b we want to calculate the speed of the ball when ahmed captures it so it's quite straightforward it's quite simple if i look at this graph on my left here let's do this in red at the point of the machine so back at 2.5 meters it's going to have a velocity or a speed again of 24 isn't it so what i want to think of is u being 24 meters per second acceleration is g gravity because this is heading towards the ground isn't it and all we're trying to find is basically the point or the value here okay so we know that that distance is going to be 2.5 meters and we need to find v so we've got three unknowns and one oh sorry three values that we know one unknown so we can do this we've got v u a and s so v squared equals u squared plus two a s v squared equals 24 squared plus 2 g times my 2.5 that gives me 625 so v is going to be 25 meters per second sorry about that i just realized i've actually worked out c first read the wrong one so let's call that c and then let's work out b down here sorry about that so b is obviously the height so starting value again 24 meters per second so this is you know we think we've got a floor we're at 2.5 meters and then it's going to shoot up to a point somewhere up here which is you know what we're initially going to find at that point the velocity is going to be zero acceleration is going to be gravity however as it's acting in the opposite direction we go in and we want to think about it'll be minus g and s is what we're trying to find so u v a and s so v squared equals u squared plus 2 a s so it's 0 equals 24 squared plus 2 g s but g is a minus so that is then the same as taking my 2g s to the left hand side and then dividing by it so 24 squared over 2 g and that gives me 29 meters or 30 meters to two significant figures there now that is just the distance above the machine isn't it so height above ground is going to be my 29.4 plus my 2.5 and that gives us 31.9 meters now just double checking the question it is actually the nearest centimeter so it's actually 31.89 meters to nearest centimeter and see i've already done as i did the wrong order there so let's look at d calculate the length of time the ball is in the air now you could do this in sections find the time from here to the top double it and then that time at the bottom or you should be able to do it in one go you just need to think about how you're gonna set it out so if i want to think of say down being positive it's going to go up at 24 meters per second which will be minus 24 the final velocity we know is 25 meters per second as we found that already the acceleration is positive g this is going down and then t is what we're trying to find so v equals u plus a t 25 equals minus 24 plus gt so that gives me 49 equals gt and 49 divided by 9.8 is 5 seconds now again you know like i said you could do it in sections here doing it in sections which should pretty much mean you don't need to use some of your workday values but it's much quicker to do it in one go like so and even if i took up as positive i will still get the same value so we got a card driven along a straight track passes o at time zero let's call that o and it starts to decelerate so that is my direction of a car passes point l when t equals one and point m when t equals four and o l is 54 and l to m is 90. okay so we've got quite a bit of information here so looking between o and l what do we know we know the time taken is one second and we know the distance is 54. and between l and m we know the time taken is three seconds and the distance is 90. what we also know is the speed at l is going to be the same so for this one our v and our u are going to be equal so let's call this v for o l and u for l m and we'll just make them both equal to v and then obviously not time acceleration is going to be the same for both of them and that's what we're trying to find so for the first one we got s t v and a so i'm going to use s equals v t minus a half a t squared so 54 equals v t minus a half a t squared and t is one isn't it so let's get rid of that so it's actually going to be v minus a half a so there is one equation now looking at the second one we know u a s and t so this is s equals u t plus a half a t squared so again s is 90 u is this v that we just called it and t this time is three so three v plus a half a times nine so here we've got 90 equals 3v plus 9 over 2a and this is our second equation now we've got two equations two unknowns but we actually want to combine the v's together just to find our a so we want to rearrange so if i call this one and call this one two look at from one we can say that v equals 54 plus a half a and then let's sub that in two so 90 equals three lots of 54 plus a half a plus 9 over 2 a so 90 equals 162 plus 3 over 2a plus 9 over 2a taking away 162 gives me minus 72 and 3 over 2 plus 9 over 2 is 12 over 2 or 6a so a is negative 12 meters per second squared and that's obviously because i'm taking this weight as my positive direction the direction i'm going in another way i could have done this is this distance o l and then instead of l m i could have just done o m i just noticed that afterwards and that way i could have used the same formula as in like s equals u t plus half 80 squared for both same value of u or an idiot that would have made a bit more sense okay maybe some of you did it that way but you should still get the same answer now part b tells us now we've got a new point n we don't know the time of that but we know that the velocity is zero at that point find this distance from m to n so let's do that total distance o n and i o and we don't know what u is do we but we know that the distance well that total distance is going to be my uh 144 plus that little s but we'll just work out that total distance first and then we'll take away the 144 from it so the distance i'm trying to find v i know is zero a i know is minus 12. so as it stands i don't have enough information so i do need to work something out now since we know the acceleration we can find at least u at the very start so let's look at ol again we know the time was one second we now know the acceleration is minus 12 we want to find u [Music] and we know the distance is 54. so s equals u t plus a half a t squared so 54 equals u plus a half times minus 12 times one squared so we got minus six added on so u equals 60 meters per second so now this can go up here so back to oh n we now got u is 60 we got s which we want to find v is 0 a is minus 12. so we've got v u a and s v squared equals u squared plus 2 a s v is 0 60 squared plus 2 times negative 12 times s so we get 24 s equals my 60 squared so s is 150 this is the total distance though distance of mm m n is going to be that 150 minus my 54 plus 90. and that's gonna give me six meters and then we're done so we've got the carnival to bike at rest adjacent to each other jason just means next to set traffic lights long straight stretcher road set of the same time t equals zero we've got the motorcycle accelerating at six meters per second until he hits 30 and the car accelerated uniformly for nine seconds until she hits 36. find the acceleration of the car as the first thing but before i jump into that let's get all this information down so this is basically what we know now find the acceleration of the car we know u v t and we want to find a so that is v equals u plus a t so 36 equals zero plus a times nine nine a equals 36 a equals four meters per second squared now for me to find the graph because we've got to draw a speed time graph it would be useful if i know how long it takes the bike to get up to speed so we know u v a and we want to find t so same formula we've got 30 equals zero plus six t so t equals five seconds okay so that'll make my life a little bit easier so the bike takes five seconds to get to 30. so we're gonna line here and then after that it goes at the same speed doesn't it constant velocity so you can keep that going now we want to take a dashed line across and we also want to take a dashed line down don't we to show our time and our velocity at that point so time is 5 and our velocity is 30 and this is the bike now our car takes longer doesn't it to get to 36 36 isn't much bigger than this but nine seconds is almost double the distance so we want to go somewhere around right here so again you know take my ruler and a pencil it's gonna be kind of almost double the distance in terms of the time and this will probably be good enough i mean it's just a sketch really again dash line across for my speed dash line down for my time and that is the car obviously i made this quite quite big it doesn't really matter you know how long i make this part of the graph but there we have the graph now for part b now for part c obviously the level at the start here the next time the level is here obviously the graph is no good to me in terms of you know finding the value i need to use my formulas but this is what we're trying to find the same time when they're level okay so this will be when they essentially have the same distance isn't it but obviously we're trying to find the time but this is the same distance so first let's just assume that they're both gonna meet somewhere down here at the time t we don't know where along here they're gonna meet okay don't make the mistake of thinking they meet here that's just when they meet in by a time it's not when they're actually meeting okay so they're working down there at time t so let's look at the distance the bike has gone so the distance of the bike is going to be the area of this triangle so 5 times 30 over 2 plus the area of this rectangle which is 30 multiplied by t minus five as obviously that length is t minus five that height so the height to the by q is 30. and simplifying this we get 75 plus 30 t minus 150 or 30 t minus 75 now let's look at the distance the car's gone so the car's done this triangle here which is 9 and 36 so 9 times 36 over 2 plus then it's done this rectangle here so that's 36 multiplied by t minus 9 again let's simplify and we get 162 plus 36 t minus 3 2 4 which is 36 t minus 1 6 2. now these two distances are equal aren't they so that means that 30 t let's do this this way s b equals s c so 30 t minus 75 is equal to 36 t minus 1 6 2 so 6 t is going to be equal to 87. so divided by 6 and we get that the time is 14.5 seconds okay hopefully not too difficult to follow that one final question here question four so we've got a guy jim it's just up on a cliff cliff is 122.5 meters growth stone and it's just dropping down so this initial velocity is zero and we can ignore air resistance here so we also know that our acceleration is gravity and if we take down as positive it'll be a positive g now since we also know the height which is 122.5 we can find v we can find t so we need to take the time taken first so remember we've got u equals a a equals g and s equals 1 22.5 and we want to find t initially and then part b will end up finding v so part a we've got u a s and t u a s and t so that is no v so s equals u t plus a half a t squared s is 122.5 u is zero half g t squared there so t squared is going to be 122.5 divided by a half g which will give me 25 so t is gonna be five there because obviously we're not going to have negative time part b we're going to take the same things and we're obviously finding v in this case so we've got v u a and s so v squared equals u squared plus 2 a s if you're wondering how quickly or why i can i'm quickly saying the formulas and spotting them it's because i write them down all the time makes your life so much easier if you just keep writing everything down so don't take short skips or shortcuts when you're revising so this is 2401 so v is the square root of that which is 49 we just put second well they give us nice values for a change now as we go into the question further it tells us that after two seconds stone up two seconds after the storm against the fall jim throws a tennis ball the tennis ball's initial speed is u and hits the stone just before they both hit the sequel it doesn't say just before but before they both hit the c find the minimum value of u so the minimum value will be the slowest he can throw it which means that we have to assume it hits the stone pretty much at the instant it hits the c so if i'm thinking of now for part c we know the information of the stone so i'll put that down here and that's based on it just about to hit the c isn't it and then if we think of the tennis ball well u is u that's what we told a is g it starts two seconds after the stone so for this one it must get there in three seconds if it's gonna hit it just before it enters obviously we don't know v um oh and we know the height don't we 122.5 because that's got to be the the same distance there now since we don't know v and we don't know u and u we want to find we're looking at u a t and s so s equals u t plus half a t squared so 122.5 and that is equal to u times three plus a half g times three squared so we got three u on this side so three u equals 78.4 dividing this by three and we get that u is 26 and a third no 26.13 and the three is currently recurring so here it goes 26.1 meters per second hopefully that wasn't too difficult there now part d if you had taken air resistance into account what effect would this have had on your answer to part c so the tennis ball we can pretty much assume is bigger than the stone at least i would have thought so if you just kicked it off the edge um that is probably the assumption i would make um otherwise you'll be talking about rock or a boulder if you're talking about bigger things so we would say the tennis ball is bigger so it would experience more air resistance as it's bigger and it will be moving faster because it covers the same distance in three seconds so that's what we need to say that means that the real value of u then would be greater than this value u because the air resistance would be slowing it down so we'd need to start off faster and there we have it hopefully you found this video useful if you did don't forget give us a like thumbs up uh comment down below anything you want to ask or you know anything you want to say and subscribe to the channel if you haven't done so already