 This is a practice video for the kinematic unit and it's on how to use and solve problems with those equations that were in the lecture video. The method I'm going to teach you is kind of a fun method and it's called the paved method. A colleague of mine and I, we came up with this and it truly was very helpful for our students. It paves your way to success with solving problems in physics. And so P stands for picture. Wherever possible, just draw out a little picture down. I'm not a very good artist, but you know, work stick figures and whatever works for you. A stands for axes, very simple arrows up, down, left, right. What is the direction of the object is? V lists your variables. So all the variables, what VF is or VI or A. E stands for equation. Any of the equations that you think would be beneficial to solve the problem. And then of course S substitute and then solve. And I'm going to show you how to use these in just a bit. Now let's just review the equations that we are going to use for kinematics. Well, we have our very simple equation. Average velocity is displacement over time. And we'll use this one anytime where you have a velocity that is not changing and there's no acceleration. There is acceleration then we get to our kinematics equations that we solved for at the end of our lecture. And VF equals VI plus AT, final velocity, initial velocity, acceleration times time. We have D equals VI T plus one half AT squared. And lastly, we have our VF squared, final velocity squared equals VI squared plus 2AD. One final thing is what's great to know if we are in free fall, if we're falling freely, we know already the value for A because that equals G. And in the metric system that is negative 9.8 meters per second squared. Sorry, I squished all that in. Now let's look at our first problem. A car traveling at 6 meters per second is uniformly accelerating at a rate of 3 meters per second squared for 15 seconds. What is its final velocity? Well, we're going to use our Paves method. So we have our picture. Well, I have a nice picture there. A lot better than I can draw. Axies, it's going in this direction. Variables, well, what do we have? We have our 6 meters per second. That would be our initial speed. That's what it's initially going. Uniform accelerating at a rate of, that's the acceleration. Also check the unit. That's a unit of acceleration, 3 meters per second squared. And then we also have 15 seconds, which is a time. Okay, so that's our variables. Now let's look at third. Oh, one more thing. I forgot. We are looking for a final velocity. I kind of like to do that. That's what we're looking for. Spell it now. What equation has all these in it? If you look, vf plus equals, excuse me, vi plus at. And now we're ready to substitute. Or as a good friend of mine used to say, let's plug and chug. Okay, so we have vf equals my vi, which was 6 meters per second plus a, which is 3 meters per second squared. That's a 2 times 15 seconds. Now, I really don't have a problem with you leaving out units as long as you don't forget at the end. However, at the beginning, it's kind of nice to have them just to make sure that this, you know, second cancels out here. So you have meters per second and meters per second and all this hunky-dory here. Okay, so we have 6 meters per second plus 15 times 3. Okay, so that would be 45 meters per second. All right, and then our final velocity then is 51 meters per second. Okay, and oftentimes we would have a direction for velocity, but because it's going in the same direction and we don't really, they didn't really specify what that is, it's okay not to list it. Okay, our next problem. How many seconds would it take a boat to accelerate from 13 meters per second to 26 meters per second over a distance of 1.25 kilometers? Well, using our Paves method, draw a picture of our boat. It's nice that I put one nice one in there, isn't it? And we drew in the direction axes. You'll see where the importance of the axes come in a little bit. Now variables. So we're looking for how many seconds we can start actually with t equals question mark. Would it take a boat to accelerate from 13 meters per second? That would be my initial velocity 13 meters per second to 26 meters per second would be my final velocity over a distance of 1.25 kilometers. Now units should all be in the same kind. So since there's meters up here in the meters per second, we really should put the distance in meters as well. So instead of 1.25, I'm going to put 1,250 meters because there is a thousand meters and one kilometer. Okay. All right. Well, do we have some kind of equation we can help out for this? And that might take some doing. So let's look and see what we got. Now there is another way to do this with one equation, but I really want to kind of show you how to do something where you might use two different equations to solve it. So I'm going to write down this first equation vf squared equals vi squared plus 2ad and you're saying, well, why are you using that? So I'm looking for time and this doesn't have time in it. But I thought it would be nice if we found the acceleration here. Okay. So acceleration would be found by plugging in, whoops, I'm sorry, 26 meters per second squared equals 13 meters per second squared plus 2 times a times that 1250 meters. Okay. All right. So when we go through and do all your math and do everything out comes out to be 0.2 meters per second squared. So just plug in the numbers in your calculator and you're able to get that. So 26 squared minus, oh, that was an equal sign, minus 13 squared divided by 2 times 1250. Now we can use that to plug in Chaga, as I say, in our very easy equation, the vf equals vi plus at. So we have our too great at writing today. Our vf is 26, our vi is 13, our acceleration is 0.2, and we are looking for our time. Now, I did leave off the units there, but I can know that 13 from 26 is also 13. So that's meters per second equals 0.2 meters per second squared times t. So I just have to divide by 0.2 on both sides. But more importantly, I also wanted you to see that you are going to end up in seconds because this meter is going to cancel out with this one. And then one of the seconds cancels out, but you're still left with one. So when you go through and do 13 divided by 0.2, you get 65 seconds. So it's 65 seconds is the answer for the time of that both 65 seconds. Okay, now on to our third problem. It says a racing car traveling initially at eight meters per second accelerates uniformly at 10 meters per second squared for five seconds. How far does it travel in this time interval? All right, so let's go to our Paves method. So we draw a little picture of a car and we show the direction that it's traveling. So we have our picture in our axes. Let's write down our variables initially. So that's the vi is eight meters per second. I'm going to try to be a little neater here on this one accelerates uniformly at so that's our acceleration rate 10 meters per second squared for five seconds. So the way we know the time is five seconds. How far does it travel in this time? So we're looking for D. Well, let's see, do we have an equation that has D in it vi at and we do it's D. So now we're going to write that down equals vi t plus one half at squared. Okay, so now let's just like I say plug and jug equals so you have eight meters per second times your time which is five seconds plus one half 10 meters per second squared times your time which is five well that's a parentheses five a second squared. Okay, so we have eight times five is 40 the seconds cancel out I'm left with meters plus half a 10 is five and then we have five squared is 25 so 25 times five is 125 and let's make sure the units work this is second squared that's second squared they cancel out you're left with meters you can add meters with meters very nicely and that's very important units really can tell the story if you're on the right track so we get a total of 165 meters for my race car now we have this problem of a baseball hit straight in the air with an initial velocity of 38 meters per second how long does it stay in the air back to the original height and how high does it go so the great example for your axes are very important so we have our picture of our baseball and we know that the baseball is hitting hits hit straight up into the air right like that okay so we have our picture in axes now let's look at our variables initial velocity well it spells it out right there for you so we're going to have vi is 38 meters per second how long does it stay in the air back to its original um height so we want to know t and that's the total time t so let's look at that and then we want to know how high did this go that would be our d okay so those are our things well one thing is when you hit some any odd anything or throw anything straight up in the air okay there is a point up here it's highest point where then after that point it'll start going back down and so what you can think about is has initial velocity here at the highest point the final velocity is zero so the final velocity is zero there and and then on the way back down the initial velocity on the way back down would be zero and then the final velocity would be it's original one other thing we know about in this problem is we know the acceleration of gravity is negative 9.8 meters per second squared okay all right so what are we going to do here let's look for the time okay and so we have vf equals vi plus at our final velocity at the highest point here the highest point is zero initial velocity 38 plus acceleration times time which we're looking for so when i bring that over negative 38 divided by negative 9.8 so we get a time of 3.9 seconds that's not the total answer takes 3.9 seconds to go from here all the way up to the highest point and then 3.9 seconds to go back down again so the total time is 3.9 times 2 or 7.8 seconds now to find b how high does it go we are going to use our distance equation so we're going to use our for part b i made the screen a little too small on this one sorry about that d equals vi t plus one half at squared so since i ran our room i'm you're going to be on your own here to plug in but let me just show you so you would put your vi your initial velocity 38 times t now you don't use the 7.8 okay because how high does it go it only took half the time to go to its highest point or the 3.9 so you would put 38 times 3.9 plus one half a please don't forget the negative sign this is very important negative 9.8 times 3.9 squared if you don't have the negative sign you're going to get a whole eight different answer um and the final answer when you do plug in in that equation you'll get 73 meters okay okay here we go this will be a little better and the screen's a little better and a dull kangaroo can jump as high as 1.8 meters with what initial velocity must the kangaroo leave the ground to reach this maximum height so there we have our picture see i i'm really trying to get around my really lousy picture drawings but so there we have a kangaroo and it's jumping up as high as 1.8 meters so we know that it can go a total distance of 1.8 meters so that's one of my variables there's my axes with what initial velocity that i am looking for that okay do we leave the ground to reach this maximum height there's something else we know we know the acceleration of gravity it's still negative i know even if he's jumping up the acceleration of gravity is downward it's slowing you down on the way back up as you're going up all right there's one other thing we do know we know if it says the adult kangaroo can jump as high as if the adult kangaroo can jump jump jump jump jump jump jump jump and then the maximites 1.8 why because it's going to start coming down that means at this point at the highest point the final velocity is zero well we have one equation that has a vf and a d and a vi in it and has no time and that is our vf squared equals vi squared plus 2ad and we're just going to plug in chug so we have zero equals vi squared plus two times negative 9.8 meters per second squared oh i'll keep writing the units times distance 1.8 meters and notice for the units for a second you'll have meter square per second squared and then you're going to be taking the square root because when i bring the vi over it's negatives in the negative signs cancel out don't worry you won't be taking the square root of negative numbers ever in physics um and then we take the square root of meter square per second squared you get meters per second so to just plug out in that math in your calculator the square root of two times 9.8 times 1.8 is let's see i have to do that out of my little calculator here two times 9.8 times 1.8 and we are going to take the square root of that and you get 5.9 meters per second and it's 5.9 meters per second and if the direction was asked it would be up all right here's our final one and of course it has to do with graphing my love of graphing okay a car starts from rest and accelerates at a rate of 4.5 meters per second squared i think for this problem let's just make it as easy as we can let's make that two meters per second squared we're going to make a distance time table and graph for the first five seconds and then a velocity time data table and graph for the first five seconds so let's see what we have here um we're going to have time in distance uh zero one two three four five there's my data table uh before we do that let me follow my paves method we have our car it's accelerating from rest that means vi is zero there's my acceleration of two um and let's see what we got we're going to use an equation vi t plus one half at squared the great thing about this equation is when you're starting from rest this just goes by by so to find our distances in this table we're going to take one half of two and put in whatever time squared so that would be zero one squared is one half of two is one so we have one then we have two squared is four four times one is four okay three squared is nine nine times one nine that's why i use the two it's pretty easy right because half of two is is one and so we have four squared is sixteen and then five squared is 25 okay so that is um what we're going to have and now if we just graph that you would get if equals zero you know one two three four five for the time and for the distance it would look like a nice parabolic curve because it's showing the acceleration that's the what the acceleration is shown in a distance time graph a changing slope so parabolic okay if we make a velocity time data table and graph for the first five seconds so we're going to do that now what the you might want to do here is say oh well distance over time is velocity so one over one is one and four over two is two nine over three and sixteen over four and that would make sense but it would be wrong okay so if you are looking for the velocity and you are accelerating you have to account for that so to find the velocity you have to use an a velocity acceleration equation so here again this is zero so we have vf your final velocity which really is what this is final velocity i guess i better get better and put my units up here in the tables you know because i'm the one always it's man people don't do that and we're going to go out let's just do one two three four five okay so acceleration times time so we're going to give two meters per second squared times the time in each one so one times two we have two and then we have two times uh two is four two times three is six two times four is eight five times two is ten okay and we got that simply by our equation here now if we graph that if we graph the time and the velocity starting at what we would have for this is a slope an upward slope and it says what is the slope of this graph if i did rise over run okay so well i guess i never did put a zero zero in i apologize but there would have been a zero zero in i could do any two point any points at all and that rise over run at any any area here the slope of that equals the acceleration and the slope would also yield two meters per second squared because that is what the slope of a velocity time graph gives you so we really intermixed our kinematics equations with graphing and we'll see a lot more of that so now i'll have some practice ones for you to do as well