welcome to m1 today's video we're looking at vectors in mechanics uh chapter three so a vector just very briefly is essentially something that has size and direction okay and we tend to assume in this mechanics we're looking in two dimensions so we're looking in that kind of x direction and y direction and that's how we're going to describe a vector now what we often use is if we think of one unit in the x direction this we use as i finishing at a coordinate of zero one so it's just one unit in the x direction and then in the y direction we have j and that would go as far as the coordinate zero one okay so one unit in the x direction is i one unit in the y direction is j so if i had say a vector that was 3 a plus 2 j what that means is that we go three units in the i direction okay or x direction then two in the y direction which is oj and this would be our vector there vector a okay and that's what it's all about obviously negative values would then move either left if it's a negative i or down if it was a negative j okay very straightforward and what that also then allows us to use is pythagoras and trigonometry to find the length of a line or an angle so if i take this one as an example if i want to find this length of a okay we write it like this modulus of a and that means it's distance its size is magnitude and that would be the square root of three squared plus two squared okay just pythagoras a squared is three squared plus two squared so a is the square root of three squared plus two squared okay which in this case would just be the square root of 13. now the other thing we often find is the angle so this will often say the angle it makes with the positive x-axis or the angular matrix of the positive y-axis so theta here would be the angle this would make with a positive x-axis and if i put an additional angle in here let's call this one um alpha so if i want to find theta it's going to be tan so tan theta is opposite which is two over adjacent which is three and that would be 33.7 degrees now if i want to find my alpha well if i've already found theta i could find theta and then i could do 90 minus theta couldn't i and that would give me my 56 point degrees alternatively i can think about right well let's think about this triangle here because this is going to be 3 this is still going to be 2 isn't it so opposite would be 3 this time adjacent would be 2 so i can do tan alpha is 3 over 2 and that will of course give me that 56.3 degrees so i have that option there the only times i really need to be a little bit more careful is if my vector is going in a slightly different direction so take this vector for example minus 2 i plus j so this is going to go 2 to the left and one up so this would be my vector b here it's two i two to the left and it's one j now if i wanted to take this one to the angle with the positive x-axis or the positive y-axis or what it often will say is to the vector i or the vector j because that is then you know not restricting where this vector is okay but it is just talking about it's what it's compared to so if i'm going to do this to the positive vector i it's going to be this angle here that i'm trying to find this theta okay so you know i would start by finding my alpha it's one option here you know opposite over adjacent so we're looking at you know tan alpha is say you do the opposite over adjacent that's what you're looking at it's going to give you a minus angle minus 26.8 what i like to do is just think about the size so you know i talk about just the size of this angle so i'm looking at the positive values and that will give me my 26.6 and then i think about what which angle i actually want which is this one here so therefore my theta is 180 minus my alpha which is 153.4 degrees okay so when i'm doing this myself what i tend to do is just use positive values for my distances my i and j or my x distance and my y distance and then i think about the actual angle that i need okay obviously if this was going to be find the angle with it compared to the positive j then it would be 90 minus alpha there okay i think you know doing this maybe even a little sketch to help you out is a good solid way to kind of approach these types of questions so quick example based on that find the magnitude of 6i minus 7j and the angle it makes with the positive y axis okay so this one you know the magnitude of this first is going to be quite straightforward so if i'm finding the magnitude i put it in these signs so it's called the modulus okay modulus sines for my magnitude and this is going to be 6 squared plus and it's 7 squared i'll put the minus in but you know it doesn't matter because you're squaring them you know you can just use your positive values there you know certainly when you're looking in your calculator and that's going to give me the square root of 85. okay sometimes it will ask you to put this you know to one or two decimal places or three significant figures or something like that so you know both answers are perfectly acceptable and depends on the question would depend on how you would leave it in terms of the angle okay if we're gonna do it with the positive y axis then we can assume that this is coming from the origin so i want to go six to the right seven down so i'm just doing a very rough sketch it makes no no real difference here but we've got six i and that's the minus seven j okay and we want to find this full angle here which i'll call theta and i'm gonna start by finding this little alpha here so to find my alpha tan alpha now i'm just going to use the positive values opposite is seven adjacent is six and that gives me 49.5 degrees therefore now my theta is going to be 90 plus my alpha this time so it's 139.4 degrees and that really comes from you know a quick little sketch just helps you to visualize what's going on okay it doesn't have to be super accurate as you can see mine here but it's just to visualize what's happening to make sure you don't make mistakes and just makes your life a little bit easier so i'm going to give you three very quick questions like this to have a go at then we'll look at the next bit so the next bit is combining two vectors together so you can see here we've got vector a and vector b and vector a plus vector b is very simple it's just about adding them together so a plus b is my three i plus two 2j plus my 5i minus 4j and for this i will just add my eyes and add my js so i've got 3 plus 5. is going to be 8 i i've got 2 plus a negative 4 it's going to be -2 j if i want to think about this visually you know we did we would have done three across two up and then you would have gone five across four down okay and you know the resulting vector is this vector a plus b well that was a and that was b that's what we're kind of it's all about here so adding these vectors together is nice and straightforward you know thinking about it very logically there's total distance from the x here is this three plus this five the distance i've come down in terms of my y you know i've gone up two and down four so overall i'm just down two okay makes perfectly logical sense hopefully to you guys uh and nice and easy vectors that's one of one of the most enjoyable bits i think in mechanics now it's like a b3a minus b so we get three lots of vector a minus one lot of vector b so expanding my brackets i get 9a plus 6j and i'm going to expand this bracket as well minus 5i plus 4j because of the two negatives there so nine minus five is going to be four i and then we've got six plus four is going to be 10 j and then finally find the magnitude of two a plus five b so before i can find the magnitude i first need to write this as a single vector so we got two lots of a plus five lots of b expand your brackets first just methodical this is just taking the logical steps you know and then we've got 6i plus 25i so 31 i and then we've got minus 16 j so now i can say that the modulus of 2a plus 5b which is the magnitude is the square root of 31 squared plus my negative 16 squared remember you can just put in you know a positive 16 into your calculator as you know the sign doesn't really matter there does it that will give me the square root of one two one seven or as a decimal that would be thirty point to 3 significant figures and that is pretty much it nice and easy so again give you a couple of questions first question very similar to this one second question a little bit different get you thinking but obviously feel free to look at the answers afterwards if you're unsure about that one and then stay tuned for the next little bit on vectors now for this one what i've done initially just a plus lambda b and then put them in terms of just i and j hopefully nice and easy but i need to make it parallel to i plus j so if i look at this case i and j are both the same size therefore this i and this j must also be the same size so i can say that 5 plus 2 lambda equals one minus two lambda because they both have to be the same size so rearranging i get four lambda on the left minus four on the right so the lambda is gonna be minus one so that's me finding my value of lambda now at this point this one's a little bit harder because we've got to make it parallel to this 2i minus 3j don't we and what we want to do is we actually want to make these the same size so what i want to do is i want to multiply this one [Music] by three and i will multiply this one by minus two and that would make them essentially like a 6i and a 6j and that would make them equal in size so i want to take my i which is here and multiply it by 3. and then i want to take my j and multiply that by minus two okay once i've done that bit it's quite straightforward after that it's just that initial step to make sure they're equal in size so from here is just a simple you know expand and rearrange now this one now is bringing in velocity so if i compare and velocity and speed the big difference is you know velocity is a vector and speed is scalar you know speed just has size whereas velocity has size and direction okay so you know if i look at this kind of question here and we've got our velocity as three i plus j the speed of it is just the size of that so if i think of my velocity is going to be three across you know one up three i plus j the size of this vector is just its speed therefore all i'm doing is finding the modulus again the magnitude so writing it down you know that's what i want for speed so speed is going to be the magnitude of v so it's going to be 3 squared plus 1 squared square root it which is just the square root of 10 or you know in terms of decimals 3.16 meters per second so it's still meters per second whichever value we go for now for part b we're talking about the distance moved every four seconds and we kind of have two options to solve this so our first option is i can use speed is distance sorry we're trying to find distance army distance is speed times time as we've already worked out the speed which is our 3.16 and our time is four seconds now obviously i'd use the full value in the calculator for this one and i get 12.6 meters so that's my first option and that is an option i would probably take if i've already found the speed however if i haven't found the speed then i'd probably go with the second option which is looking at the displacement so displacement for those you don't know is essentially distance in terms of a vector so distance is just distance traveled you know you could be running around in a circle and that total distance that you traveled running around in a circle would have potentially a large value whereas in terms of its displacement that's the distance from a specific point so if you run right around in a circle your displacement would be zero if you started and finished at the same point okay but if you run in a straight line that distance on that displacement will be exactly the same because displacement is a vector you know from start point to finish point okay so displacement is just velocity times time so we can take our velocity which is three i plus j and we can times it by a time which is four so that will give us 12i plus four j and all i need to do then is find the magnitude of that so that's gonna be 12i plus four j and as we've done before 12 squared plus 4 squared and then square root it and that gives me 4 root 10 or 12.6 meters so i think ultimately it doesn't matter which way you do it and as i said personally if i've already found the speed i would just use speed this does equal speed times time if i haven't found the speed they didn't need to then i would just use displacement and having that kind of options open to you guys and remembering both of these options can save you time in an exam and time is often your biggest enemy again just a few questions for you to have a go at answers afterwards and then we'll move into the next part on vectors so now we're looking at more of the kind of problem solve inside of our vectors and in this example you can see now our start time t equals zero particle has position vector four seven okay so four i plus seven j so position vector is if you think of it on a x y axis your position vector would just be essentially a coordinate so four cross seven up so that's my coordinate for seven and that is essentially position vector is that vector from the origin okay so particle has position vector four seven it means from the origin it's gone for the four units in the x direction and seven units in the y direction okay so that's its initial starting point and then you can see it's moving in the direction of three i minus four j so that's three across four down so that is the kind of direction that this particle is going to move you know afterwards so 3i minus 4j so it's essentially this is where it's starting and this is now the direction it's moving in okay so hopefully that makes some sense to you guys now we're told moving at a speed of 15 meters per second in this direction okay so we want it to move at 50 meters per second that means if i just take this vector to the side we're going in this direction of the 3 i minus 4j but the length of this line has got to be equal essentially to that 15 the 15 meters per second so it has to be the size of 15 15 units okay but how do i do that so the first thing i want to do is take my vector 3 i minus 4 j and i want to find out the size of that vector so the size of that vector is going to be its magnitude so 3 squared plus negative 4 squared and we all recognize this from a three four five triangle so the square root of this is going to be five so the length of this vector is five we want it to be essentially the length of 15. so you can think right i just need to make it multiply this by three and that would work okay but i'm not going to do that in this particular one i'm just going to show you just because you might not always have a nice straightforward number so if it's length five and we want it essentially length one okay well we wanted length 15 don't we really so we want one over five of three i minus four j and that would then take my vector here and make it length one because i've divided by this five and then we're gonna multiply that by 15. okay so you know if i had this if this came out there's like seven in length i'd have one seventh here and then i just multiply by y whatever my actual speed is for the size of the vector this one is nice numbers which makes our life a little bit easier so we get 9i minus 12 j there for the new vector in terms of its size so back to essentially diagram we'd already said you know we've started at essentially 4 7 from that vector 4i plus 7 j then we moved in the direction of 3 i minus 4 j but this time we're looking at how long so two seconds we're moving 15 meters per second so 15 meters every second so in two seconds that will be two lots of that so this distance this vector is going to be quite large you know it's going to be 2 times my 9 i minus 12 j and that will then give me you know my new vector here which will then give me the position because it's from the origin so looking at my position we've got our think about you know we start at the origin we move four across seven up four i plus seven j and then we went two lots of 9i minus 12 j because that distance that's where we go in but in one second but in two seconds it'll be twice as much so 4i plus 7j plus 18i minus 24j so in total that's gonna give me 22i now different type of question here we've got a velocity at timing t equals zero so initial velocity here constant acceleration and we need to find the speed after time t so if you think we've got initial speed or initial velocity there we've got our acceleration here and we've got our time here and we want to find our final speed okay our velocity we've got here v u a and t so v equals u plus a t nice and easy minus 3 i plus j obviously we can use this without constant acceleration plus a is two i plus three j and t is three so minus three i plus j plus six i plus nine j so here we can see we've got 3i plus 10 j in terms of our velocity so that's my velocity okay velocity is obviously a vector remember it actually said speed in the question so speed is when i find the magnitude of that velocity so 3 squared plus 10 squared and then square rooted so that's the square root of 109 which is 10.4 meters per second and this is to three significant figures or one decimal place final example here we've got constant force here and mass so we are probably thinking along the lines of f equals m a we've got time so let's put everything down so we've got f equals we don't know yet so we want to find m equals four kilograms t is gonna equal five seconds uh it was initially at rest so it was at t equals zero and five seconds later it was at a velocity here so if it's initially at rest u would be equal zero and v would be equal to four i minus three j so looking at what we've got here from our super equations we've got t u and v and from f equals m a we've got m we want to find f so it makes sense to find a we can find a from our suvat here so v equals u plus a t v is 4 i minus 3 j u is zero a we want to find and t is five so a is going to be one fifth four i minus three j and then substitute that into f equals m a we've got m is four times one fifth four i minus three j so this is four fifths of four i minus three j newtons now of course you can leave it as a vector you could simplify this vector a little bit as in you know our options i'll do this in a different color because it doesn't you know these aren't necessarily the way you have to do it so that's another option there as a single vector you know or even in terms of its size which you know kind of in this one does make a little bit more sense so we'd have you know four fifths multiplied or you know actually i can just use my 3.2 squared plus my 2.4 squared and then square rooted and i get my 4 newtons there you know which makes sense because if you look at this vector it's a 4 and a 3. so the size of this vector is going to be five from a three four five triangle five times a fifth is one times four is four so in this kind of question you know leaving it as four units does make sense but what i'd always say is do you want to do extra work in an exam if you don't need to and potentially make a mistake so most of the time i'll leave it as a set answer depending on the question you know the questions and vectors i'd probably leave in vectors unless i it was specified otherwise so finish with kind of four more questions along these lines and as always the answers will be at the end you