in this video you'll learn how to derive the four kinematic equations we'll start with this velocity versus time graph on this velocity versus time graph we have a straight line the slope of this line represents the acceleration and then using the slope equation we can write that the acceleration is equal to the change in velocity over the change in time so slope is just the change in y variable divided by the change in X variable from here we can expand that and write V final minus V initial divided by and delta T is T final minus t initial but if the T initial zero we're just going to go ahead and write t from here I'm going to move the T on the bottom right to the left so we get a t is equal to V final minus V initial if I add the initial on both sides I get that V final is equal to the initial plus a t oftentimes you'll see it written as it as it a bit flipped B final equals V initial plus a t so that is our first kinematic equation for the second kinematic equation we're going to look at the area under the curve so we're going to look at this area right here and the reason we're doing that is because the area under the curve on a velocity time graph represents the displacement so the area under the curve here we can see that there is a triangle and a rectangle so the rectangle the area of the rectangle is the base times height so we can take v i times the t v i times the T so that would be the height times the base and then here we have a triangle the area for triangle is one over two base times height the height is the final minus V initial and then the T times t from here you'll notice that there's a v final minus V initial and we saw that earlier we saw that over here so we can take this and substitute that over there so with that substitution we get V initial it's time plus one over two V final minus V initial is a t times T and if we multiply that out we get v i t plus one over two a t squared and that is our second kinematic equation for a third kinematic equation we remember that the average velocity is equal to change in position or the displacement divided by the change in time and we also know that if we were to solve for Delta X that would be equal to V the average velocity times the time and once again the change in time is the final time minus initial time is the initial time is zero then you're just going to have the final time so I'm just going to call that t and then next I'm going to um for the V average I'm going to substitute that with this equation V final plus v initial okay so if if it is any constant acceleration and which we are using these kinematic equation for then the sum of the initial and the final velocity divided by 2 kind of like if you want to find the average of two numbers the average two numbers add them together divided by 2 and that works for the initial and final velocity if it's going to constant acceleration if it's not then this is not going to work okay so we're making an important assumption that we're dealing with constant acceleration so and then we have t here the next step is that you'll notice that t is equal to if I go let me go over go back here over here okay let me come back here you'll notice that t if I were switch to T and A I get T is equal to V final minus V initial divided by a and what we're going to do is we're going to take this and we're going to substitute this over here to this T right there okay and what that gives us is Delta X is equal to V final plus v initial divided by 2 times and then this purple what's circled on the left on the purple here is V final minus V initial divided by a yeah all right that looks kind of like a mess here and then I'm going to keep going so on the top I've noticed the only difference is that one is VF plus VI the other is a minus v f minus VI so we're doing a little algebra or you can use foil you know that this will turn out to be V final squared minus the initial squared divided by 2A okay I'm going to go to my next page so I just copied what I had from the previous page here so this is where we left off we have Delta x equals VF squared minus VI squared divided by 2A I'm going to move the 2A over so I get 2A Delta x equals to V final squared minus V initial squared then I'm going to add VI squared on both sides VI squared on both sides and that gives me VI squared plus 2A Delta x equals V F squared and often you'll see this written as VF squared equals v i squared plus 2A Delta X that's how you'll usually see that written and that is your third kinematic equation actually this is your fourth kinematic kinematic equation so let's go back and just kind of recap the four kinematic equations I need to point out the third one I didn't Circle that one but I want to point that out all right to recap your four kinematic equations we have this one right here which is your your first one so let me go put a little one right there this is your second kinematic equation right here and I didn't Circle this one but this is actually your third kinematic equation let me go ahead and circle that one so this right there right there that's actually your third kinematic equation right there and then we have our fourth kinematic equation right there okay so let me write all of these out for you so you can have them nice and neat all four kinematic equations so here are the four kinematic equations and also on the right hand side I indicated what the variables stand for so Delta X is displacement V is velocity a is acceleration T is time and typically we'll be using these kinematic equations where problems are dealing with constant acceleration a common issue students have with these kinematic equations is trying to figure out which equation to use for a particular problem so in the next video I will show you a kinematics chart that would be really helpful to help you decide which equation to use to solve a particular problem