hey friends my name is c and you're watching here mr eazy and welcome to a new video for igcc edmonton today we have the rules and examples for kinematics or basically mechanics and we'll start off with some basis but before you get into it don't forget to leave a like subscribe and ring the notification box and let me know in any future videos and you'll start off with differentiation and integration and one key thing to know is that for kinder matches or basically um like any thing that's related to trade we should use a radiant for trigonometry trigonometry or light trait so here's the relationship between distance or displacement and velocity and acceleration basically from distance to velocity we basically differentiate it and for velocity through acceleration you differentiate it so conversing the other way around acceleration to velocity will be integrate and from velocity to a distance you basically integrate and here are the two equations or like basically notations to use since distance is s or basically displacement we use this space more often for like kinematics and mechanics since distance is s we have to find the velocity right so we have to differentiate a distance with respect to time so it'll be d s on the change in distance over oops over dt which is basically the change with respect to time and that will equal to v which is velocity and conversely acceleration will be dv over dt but if you use the if you still stick with the distance notation it will be d square x over dt squared or basically the second derivative of distance and integration is the same right here where v equals the integral of a and s equals the integral of v and they are both with respect to time and here's an equation for the average speed basically average speed equals the total distance over the total time so what you need to know is that it's specifically the total and total then we have graphs with some basics so number one we have displacement time graph or basically distance time graph distance so we have the different ingredients which corresponds to different things so basically what um what the gradient in the displacement time graph tells us is the gradient the gradient is the velocity of the velocity so when there's a curve right here it means that it's accelerating if it's a straight line like a straight line going up it's constant speed if it's a curve that's only like this way it's slowing down and if it's a horizontal line it's sorry if it's a horizontal line it'll be stationary because think of it this way if we have a distance with a time and if let's say two second pass and your distance is still the same it means you're still stationary and basically the maximum point is the point furthest away from the starting point and the velocity velocity time graph the um the is the gradient is equal to the acceleration acceleration which is a bit messy so uh this line right here basically means positive gradient sorry positive acceleration and positive gradient zero acceleration and negative acceleration and the maximum point is the fastest speed at that point or at that journey and for acceleration time graph the gradient is we don't really need this because the gradient is not in uh spec but it's basically called jerk so jerk is basically the derivative of acceleration so one thing to note is that the area under these graphs right here is specifically what this is ie so the area under acceleration time graph will be velocity the area and the velocity velocity time graph will be equal to displacement and so on and then we have suvat or basically some physics equation so this one thing to note is that silver is only for equate like equations like they are equations of motions with constant acceleration this is very important but you won't really come across through any question with non-constant acceleration anyway so we have four basic equations v was u plus a t s equals u plus v over two times t v squared equals zero square plus two a s and s equals u t plus half a t square where x is a displacement u is initial velocity v is final velocity a is acceleration and t is time and we'll now move on to some examples number one a particular p moves in a straight line so that its displacement s meter is from a fixed point on the line is given by s equal three t squared minus 12 where t is the time in seconds after passing the point a on the line find the distance o aim so we can basically just just draw like a diagram a simple diagram so we have this straight line here this is more a straight line so we know that t is the time in seconds after passing point a on the line so let's just say it is a and this is a particular p and this will be let's say o okay so find the distance o a so the distance o a is when the time is zero and the p will be at the particle p will be in a and then you can just find the distance from o to a right so let all be like let's see in the origin um so when t equals zero or basically three t squared minus 12 when t is zero the this the displacement will be equal to minus 12 meter right but since distance is a scalar which has no direction therefore the distance sorry number seven the distance equals 12 well so this is this new color the distance equals 12 meters right so in question two so b find the velocity of p when it passes o so we have to find what the the time is at point o right like what time it passes by or like what time it is so at point o we know that the displacement is zero right because if this is specifically 12 but when p is at zero the displacement is zero so when the p is at o the displacement is zero so just set it as zero equals 3t squared minus 12 and solving for t squared equals 4 and since time common negative it will just be t equals 2. and remember how if we integrate sorry if we differentiate this basement you get type so you get velocity so basically the velocity is basically the inte the different the derivative of the displacement with respect to time ds over dt will be equal to 6 t and the minus 12 just disappear and t equals 2 in this condition therefore 6 times 2 is 12 so it will be 12 meter per second right so all right so and number three or c find the average speed of p during the first three seconds so as you saw just know the average speed every speed is equal to the total displacement or the total distance of a total time the total time we know is three seconds as given in the question so three seconds so therefore we can basically find um the displacement where oyp is from o after three seconds which i've mentioned just now so that displacement equals 3 times 5 sorry 3 times 3 squared minus 12 equals 15 meters that means that after 3 seconds p will be 15 meters away from the initial point right so therefore the total distance travel is 15 which is after three seconds plus 12 which is the initial um speed is the initial distance from 0.0 therefore if we were to simplify it you would get an average speed of 9 meter per second like so and last question we have question number 2 a stone is projected vertically upwards from a height of 13 meters above ground with velocity eight meter per second calculate the greatest height above uh the greater height the greatest height above the ground so let's say this person throws this stone up was just like um the top it'll be like an arc right like so and it'll go down so this pawn right here will be when the greatest height above the ground and this part here will be whenever that's with the velocity is zero right because it's not moving up or it's not and no it is not moving down so we can use the equation the v squared equals u squared plus 2as and we can resolve the force going upwards and it's important to resolve a force in one direction so that you keep all your signs the same for example if we have acceleration since we know the earth is right here and the person standing on it the acceleration to the ground is equal to g which is that gravitational force and if we were to observe the force downwards g will be equal to positive but if we were to resolve the forces upwards g will be negative because it's going against the direction of motion right so v squared the v will be zero plus you define the fact this point is equal to zero the initial velocity is given as eight eight meter per second so would be eight squared plus two acceleration will be minus g or minus 10 because we use 10 for acceleration and remember that we use -10 it's because we're basically resolving the forces upwards and it'll be s so we can basically just rearrange the equation it'll be minus 20 s bring it around to get 20s equals 64. therefore s equals 64 over 20 or if we were to simplify it 64 over 20 will get us 3.2 meter so 3.2 meter but wait that's not the greatest height because we are basically finding um it it's projected vertically upwards from a height of 13 meters right and the ground is basically 13 meter below so the the greatest height above the ground will be the initial height which is 13 meters plus the height after the height when the velocity is equal to zero so it will be three plus at the 3.2 meter plus 13 meters equals 16.2 meter and that's the final answer and b it's velocity when it hits the ground so we have to find you know to use the free square equals u squared if v weight was used to pass through s as well use square plus 2as the velocity when it hits the ground or like right before it hits the ground it'll be v right and we know u is the initial velocity that we toss upwards with so the initial initiates right here eight meters per second eight squared plus two a s will be the a will be basically will be equal to positive ten because in this situation is a bit more different because this is through an accurate diagram so we have this 13 meter above the ground 30 meter we're finding the speed of the stone when it goes upwards and then downwards like here we have this point it's the speed here similar can model this whole journey as one whole journey in this case right here so we know that the v is here the v the u squared is the whole this that the journey which is useful right here 2 a s a you might think that we might actually use a as negative because it's going upwards but we're more focused on the downwards motion and it is um the most like sensible option to use as the ball is accelerating downwards when the final velocity is that occurs so it'd be plus 10 times the displacement which is the displacement from the initial position which is 13 meters right here so 13. therefore v squared equals three to four meters and therefore v equals if we were to square up both sides it'll be equal to 18 meter per second and in this case it's important to note that i'm resolving the forces downwards and i should have stated it at the start so therefore the initial velocities to actually be minus eight but it doesn't matter anyway because you have to square the number and therefore v equals 80 meters per second and that's the final answer and this is for this questions video for igcse match but today we look into the kinematics the rules and examples for kinematics and i hope you find it useful and helpful and use it please leave a like and subscribe and ring the notification button let me know in any future videos and if you have any comments or questions feedback about my channel or my youtube or my instagram you can leave them in the comment section and reply to them and check my social media in the description for example youtube or linkedin or instagram and if you need any learning resources or any teaching resources you can check out my website in the description or you can check it out in your browser at www.emuez.com and i hope you'll find it useful and helpful and i'll see you all in the next video which will be the questions for kinematics which would be interesting but until then stay safe and [Music] happy learning [Music]