WEBVTT physics/high-school-physics/selhorst-jones
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Hi, welcome back to educator.com. Today we are going to be talking about one-dimensional kinematics.
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Kinematics in a single dimension. What does kinematics mean first of all?
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Kinematics just means study of motion.
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Something that is going to talk about how does something move.
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That is what we are going to be learning about today.
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How do we talk about motion! How do we talk about things moving around! That is clearly an important part of Physics.
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First idea, position. Position is simply the location of an object at a given moment in time.
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For this one, we could have something like this, and BOOM!, it would be a 3. Whatever 3 means. That is a really important point to bring up.
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We are the ones who have to impose the coordinate system.
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Nature does not come with an inherent coordinate system already told to us.
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We have to say, okay, this place is zero and then we are going to measure metres this way.
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So, we get 3 metres, or 3 whatever it is that we have measured.
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But it is up to us to decide where we are going to put that zero and how we are going to measure off it.
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We impose the coordinate system. It is us who are assigning position value ultimately.
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It is coming from nature, but where our starting place is, that is on us, it is on us to figure out how we are going to orient things.
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That is a really important thing in Physics sometimes.
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Next idea, distance. Distance is just a measurement of length.
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For example, let us say you start at your house.
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And then you walk a 100 m to the North.
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And then after that, you feel like walking a bit more, so you walk another 50m to the North. How far would you have traveled?
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Well, you just put the two together, and BOOM!, you have got a 150 m of travel to the North.
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What if instead, we want to talk about displacement.
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Displacement is change in position. Displacement is very similar to distance, but sometimes they can be very different.
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Consider the following idea: Once again, you start at your house. And you walk a 100 m to the North.
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But then, you keep feeling like walking, but you do not feel like walking to the North anymore, you decided to walk to the South.
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Now, you walk 50 m to the South. What is your displacement?
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What is your change in position from where you originally were.
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Well, your change in position from where you originally were is, that much.
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You are now 50 m to the North of your house. So, your change in displacement is 50m North.
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But what is the distance that you traveled!
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The whole distance that you traveled, you wound up going up here, and then you turned back and then you walked another.
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So, you walked a 100, and then you walked a 150 total.
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You walked a 100 up, and then you walked a 50 down, so the total amount of distance you traveled was a 150 m. Very different numbers here.
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Important to think about this. Displacement is your change in position.
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How did we get from where we are now to where we end up, versus distance, which is what was the total length we took.
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Just to compare them once again, distance: how far we traveled, how far an object travels, where displacement is the change between the start and the end points.
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Notation. When we are talking about these ideas, we want some kind of shorthand variable to denote them.
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We are going to solve for this stuff eventually.
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It might seem confusing at first, but we are going to wind up denoting them all with just simply 'd', 'd' would be the letter we use to denote it.
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Location, distance and displacement, they are all going to get denoted with 'd'.
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And it is going to wind up getting obvious that, over time, we know what we talking about.
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We know we are talking about distance, sometimes we know we are talking about displacement. We are going to get it contextually.
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We are going to get it based on how we are working with things.
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We will know what we mean by 'd', do not sweat over the fact that it currently means three different things.
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It will make sense when we are actually working on problems.
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Speed. Speed is an idea I am sure you are all already familiar with. It is just how fast something moves.
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If we want to know how fast something moves, it is going to be the distance that it traveled, divided by the time it took to travel it.
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You can travel a 100m, but there is a big difference in it if you travel it in 1 s, or if it took you a 1000 hours to travel it.
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That is where speed comes in.
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If a car travels 100 m in 5 s, What would its speed be?
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Well, speed, simply the distance traveled, 100 m divided by 5s, and that is going to get us 20 m/s.
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It travels 20 m/s, because every second it goes 20m.
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This is an important point. m/s is the standard S.I. metric unit for motion, the 'm' and 's' being the units, which brings this into begin with.
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Velocity. Velocity, is very much like speed, but instead we are going to be talking about displacement.
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Velocity is equal to the displacement divided by the time.
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So, it's how much the position or the location has changed from beginning to end, not the total distance that it travels over.
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A race car can travel around a circular track 200 times in a couple of hours, but what was its velocity on the average!
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Well, it did not leave the track, it just eventually parked in the same spot it left from, so it had no average velocity.
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But it would have had a very great average speed because it whipped around that track very quickly.
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We denote velocity with a 'v'.
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New idea. Delta, Δ.
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It is the Greek letter Δ. Δ is a capital Greek letter pronounced Delta.
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It is used to indicate 'change in'.
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If we want to indicate a change in location, we can say Δ d, which would be the starting location subtracted from the ending location.
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Final location, dfinal - dinitial, which is what displacement would be, right?
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To figure out what your displacement would be, it is going to be where you ended minus where you started.
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So, Δ, the idea of 'change in', lots of time, we are going to have to consider not what the absolute value is, what its whole value is, but what was the change from the beginning to the end.
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Really important idea.
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Get this in just a simple, this nice Δ symbol.
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For velocity, v = the Δ of the distance.
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Δ (Displacement) divided by time, Δ (Displacement) divided by time for velocity.
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If you want to talk about speed, it would be Δ (Distance) divided by time.
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Acceleration is the rate at which velocity changes.
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It is a big difference to be in a truck that accelerates really slowly, it gets from 10m/s to 30 m/s over a course of 10s, versus an incredibly fast sports car, that gets from 10m/s to 30 m/s in 2 s.
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So, acceleration is how fast we manage to change velocity, or in our new letter, our new formation, that we can use Greek letters, new way of talking, we have got acceleration = Δ v / t .
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Change in velocity divided by time.
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If we had a car moving with a velocity of 20m/s, and accelerated at a constant rate of 40 m/s over 10s, what would be the car's acceleration?
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Well, acceleration, is equal to Δ v, so Δ v/t,
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Δ v is vfinal - vinitial, divided by time, we plug in the numbers we have, vf = 40, ended at 40 m/s, it started at 20 m/s, and it is divided by 10s.
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So, that is going to get us, 20/10 or 2.
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And the units we get out of this is, we had m/s on top, and we divided by seconds on the bottom, so it is going to be m/s/s.
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Gravity is a constant acceleration that always points to the centre of the Earth.
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It is always -9.8 m/s/s.
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So, if you are anywhere on Earth, you are going to experience an acceleration of -9.8 m/s/s.
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The reason that you are not currently falling through the floor to the centre of the Earth is, you got something resisting that.
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If you jumped to the air, for that brief moment, before you eventually landed on the ground again, you would be accelerated down towards the centre of the Earth at -9.8 m/s/s.
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That negative sign denotes that we are always pointing down, that we are always going down with it, because generally we talk about up as the positive direction.
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So, -9.8, that negative is there to denote we are going to go down.
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As a special note, some people denote acceleration as m/s/s, others denote it as m/s^2, as we are dividing by second twice.
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I prefer m/s/s, because it gives you a little bit more of an idea of what is going on with the acceleration, it is how many are m/s changing for every second.
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A 5 m/s/s means that 1s later you are going 5 m/s more.
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m/s^2, same meaning mathematically, but we lose a little bit of the inherent sense of what we are getting at.
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Formulae. So, the most important formula for all kinematics is what shows us all of these different ideas interact.
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The distance, your location, location at some time, t is equal to 1/2 times the acceleration the object is experiencing times t squared plus the initial velocity of the object times t plus the initial location di. ( dt = 1/2 at^2 + vit + di)
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We cannot actually explain where this formula is coming from, because it comes out of calculus.
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The way we get this formula, it does not take much difficulty, if we had a little bit of calculus under our belt, we will be able to pick it up fast, but at the moment we do not have calculus.
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So, We cannot understand where this is coming from quite yet, but we can just go ahead and accept this on a silver platter, so this is a really important formula, we will be taking it for right now, and we will be running with it.
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We will be using it a lot.
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Next idea is one that allows us to relate velocity, acceleration and distance without having to use time.
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And this one we can actually derive.
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First of, remember, acceleration is equal to, change in 'v' over time.
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Well, that is the same thing as vf - vi over time.
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We could solve this for time if we felt like it.
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Time is equal to vf - vi divided by acceleration.
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We can take this idea right here, and we could plug it into this.
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Now, first thing is m let us state first this is actually dfinal because that is where it is at the time t that we are looking at, the final location for our purposes.
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1/2 at^2 + vit + di, because that is the starting location where the object was.
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We plugged that into there, and we are going to get, just box this out, so we have little bit of room to work.
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df = 1/2 a (vf - vi/a)^2 + vi(vf - vi)/a + di.
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That di, we are not going to need it, we are going to move it over to the other side for now, so we get df - di.
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And then, while we are at it, let us multiply in that 'a', on that one, so we got, 1/2 times, well, it is going to cancel out, the squared of the a^2 on bottom, so we are going to get, vf, let us change this up, so we have a fraction that split now.
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It is going to be just 'a' on the bottom, because it used to be a^2, but we hit it with an 'a'.
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vf - vi, but this part is still squared, plus vi × (f - i)/a .
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At this point, we got an 'a' on both of our sides, we got this 2 showing up here.
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So let us multiply the whole thing by 2a, and let us also realise, df - di is another way of saying that it is just 'change in', Δ.
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So, we got 2a Δ = (vf-i)^2 + 2 vi(f- vi) .
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Let us take all this, move it over here, for ease of work.
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So, we have got 2a Δ D = (vf)^2 - 2vfvi + (vi)^2 + 2vi vf - 2 (vi)^2.
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So, at this point we see, this shows up here, and here, so we cancel them out, -2v (i)^2 is going to cancel out.
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So, we are going to get 2a Δ D = (vf)^2 - (vi)^2 .
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So, we can move this over, and we have got (vi)^2 + 2a Δ D = (vf)^2.
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That is exactly what we see up here.
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So, if you know the final velocity, and you know the initial velocity, and you know the acceleration, then you can find the change in distance.
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Or if you know three of these four elements, you can do it.
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We can do this without having to work around with 't'. Sometimes, that is a nice thing for us.
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Remember, acceleration will have to be constant for this, and also for the problem above this, but everything else works out great.
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Final couple of formulae to point out. Just definitional formula.
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So, we defined acceleration = change in velocity/time and velocity = change in distance/time.
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With this point, we are ready to hit the examples.
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If we have a rock, and we throw it directly down from a very tall cliff, the initial velocity is -7.0 m/s.
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We ignore air resistance. What would be the speed of the rock 4 s later?
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Well, gravity is equal to -9.8 m/s/s.
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What is acceleration? Acceleration = change in velocity / time .
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Since there is nothing else acting on it, it is just a rock falling, we do not even have to worry about air resistance, acceleration is just equal to gravity, so we got -9.8 = change in velocity / time.
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What is the time? 4 s .
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We multiply both sides by 4, we are going to get -39.2 m/s = change in velocity. That is not quite enough, we started at something.
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This is our change in velocity. We have to take -7, our initial velocity, and add it to our change in velocity.
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So, that is going to be equal to -7 + (-39.2) = -46.2 m/s .
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Next problem. Second example:
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We got that same cliff, but now we are going to say, it is precisely 200 m tall.
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Once again, we chuck a rock down it, at a speed -7 m/s .
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Ignoring air resistance, how long will it take that rock to hit the ground below?
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Gravity = -9.8 m/s/s, and what formula we are going to use, we are going to use location, based on time, the final location at some time, 't', is equal to 1/2 at^2 + vit + the initial location.
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For this one, what is our initial location, 200 m above the ground.
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Let us make the ground, zero.
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That is our base location.
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So, our initial location is, positive 200 m (+200 m) above that ground.
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Our initial velocity is -7 m/s towards the ground.
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And our initial acceleration, the only acceleration we have throughout, the constant acceleration we have throughout is, -9.8 m/s/s, the acceleration given by gravity.
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We plug all those things in, what are we going to want to solve for, we are going to want to solve for the time one were at zero.
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So, d(t) = in general, for any time it is going to be 1/2 t^2 × (-9.8) = -4.9 t^2 + (-7t) + 200 .
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So, we want to know our location at any time 't', we just chuck in that time, and we will find out what our location is.
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At least until we hit the ground, or if we go back before zero.
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Because this equation is just Math, it is supposed to tell something, assuming we are using it right.
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But if we go past the time it hits the ground, it does not know where the ground is, it is just a Math equation.
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So it starts giving us negative numbers.
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It is going to operate like a problem, it is up to us to be careful of how we use some of these equations.
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But, we know what we are doing here.
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So, what final location do we want to look for? We want to look for the zero.
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We want to find out when is that rock at the ground.
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So that is the case, -4.9 t^2 -7t + 200 .
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We want to solve this equation for what time, it will give us the location, the final location of zero.
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How do we solve something like this?
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We got a couple options. One, we could look to factor, but that does not look very easy to factor to me.
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Two, we could put it into a numerical solver or some sort of good calculator, and we could figure it out.
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We could graph it with a graphing calculator, and look for the zeroes, or we could put it into the good old quadratic formula: (-b +/- (b^2 - 4ac)^1/2)/2a.
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In this case, what is our b?
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It's -7. So, we got (7 +/- ((-7)^2 - 4 × (-4.9) × (200))^1/2) / 2 × (-4.9) .
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It does not look very friendly, that is not super friendly and easy, but we can definitely punch that out if we work through.
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So, equals (7 +/- (49+3920)^1/2)/(-9.8) = (7 +/- (3969) ^1/2)/(-9.8),
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We are going to wind up getting two different answers out of this, the two possible answers for our time, this quadratic formula would give us, are -7.1 s and 5.7 s, because of that plus/minus.
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We are going to get two optional answers depending on if we work with plus, or if we work with the minus.
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Because it is two different answers that will work. Both of these answers are going to solve this equation right here.
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But which one is the right one? Well, this equation does not apply negative time.
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We start the time, we want to start the clock when we throw it, that is when we set our initial distance, that is when we set our initial velocity, that is the moment of throwing.
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So that is when we need to set our time as zero.
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So that is our zero time, just as we throw the rock.
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So, that is the case, we are going to have to go with the positive answer because it is the only one that make sense.
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-7.1 s, that works because what we have described here is, we have described a parabola.
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We are looking for when does that parabola cross the zero.
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When does it hit the x-axis. What are the zeroes, the roots to this parabola.
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Of course, there is two answers to this.
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But, it is up to us to pay attention and to go, 'Oh yea, that would not make sense for that to be a..'.
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Math, it is a really useful tool when we are solving Physics, but it is up to us to keep it rained in.
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It is just going to give us the answers, it is just going to do its own thing, because we are using it to model real world phenomena.
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It is up to us to pay attention to how we are applying it.
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So, in this case, we will get two answers out of it, but only one of them makes any reasonable sense. So, that is the answer we have to choose.
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Third example: We have got a person on a bike, traveling 10 m/s.
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The person begins braking, and then comes to a stop 10 m later. What was its acceleration then?
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In this, did we say what the time involved was? NO, We did not!. We could probably work it out.
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We could solve for it. But, it would be easier if we did not have to.
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So instead, we have got that formula (vf)^2 = (vi)^2 + 2a × (change in distance) .
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Let us say, what he started out was just at zero, because we are setting it arbitrarily.
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It does not mean to us where we start, and our change in distance is that 10 m.
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Final can be 10, but the important thing is, wherever he started, he stops 10 m later, so the change in distance is 10.
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The acceleration, we do not know the acceleration, that is what we wanted to know.
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We know what he started at, he started at 10 m/s .
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Do we know what he stopped at? Yes, we know that he came to a full stop, so final velocity must be zero.
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We plug these in, we get 0^2 = 10^2 + 2a ×10 , 0 = 100 + 20a .
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So, we have got -100 = 20a, we have got -5 m/s/s , because that is what acceleration comes in, equals our acceleration, which makes a lot of sense, because if this person is moving forward, for them to come to a stop, they are going to have to have a negative acceleration, they are going to have to be slowing themselves down.
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They are going to have to be opposing the movement they already had.
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Final example: We have got a UFO that is currently 500 m away from the surface of the earth.
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Here is the Earth, and here is our little UFO, hovering in space, and it is currently 500 m above the Earth.
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And at this moment, it has velocity of 50 m/s, so it is not really sitting there.
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We take a snapshot and in that first snapshot, it is 500 m away from the surface of the Earth.
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And in that snapshot, it currently has a velocity of 50 m/s.
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It also has a constant acceleration, of a 100 m/s/s.
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So, what would be the velocity of the UFO in 10 s?
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If we want to know the velocity of the UFO, the acceleration = change in velocity / time.
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We know that the acceleration is 100, change in velocity divided by, we are looking 10s later, so we have got 1000 m/s = change in velocity, so our final velocity, is going to be, (our initial velocity, let us denote it as vi).
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vf is going to be those two added together, so we have got 1050 m/s.
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It's moving pretty fast at the end there.
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If we want to know what its location is, how far is the UFO from the surface of the Earth in 10 s, we need to set up that same equation we used before, distance (t) = 1/2 at^2 + vit + initial starting location.
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We plug those in, we are looking for its distance at 10 s.
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Is equal to 1/2 times, what is the acceleration?, interesting point to look at, is gravity affecting this, yes, gravity is affecting this, but we do not need to subtract for it, because we are told in the problem that it has a constant acceleration of 100 m/s/s, away from the Earth.
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So, whoever gave us the problem, however we got this information, we know the acceleration.
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There is some forces involved, gravity is pulling, and it probably has got thrusters or something, causing it to move away from the Earth.
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But, we do not have to worry about that, because we are told what its acceleration is precisely.
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So, we are good with that. So, 100, (and its positive because it is moving up), times the time squared, we do know the time, plug that in, 10^2 plus, what is initial velocity, 50 times the time, 10, plus its initial location, 500.
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After plugging things in, 50 × 100, 1/2 times 100 becomes 50, 10^2 becomes a hundred, plus 500, plus 500, we get 50 × 100 becomes 5000, plus 500 + 500 becomes 1000.
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So, we get in the end, it is 6000 m above the Earth.
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Hope you enjoyed that. Hope it made sense.
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Next time we will look at how multi-dimensional kinematics works, when we are looking at more than one dimension.
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See you at educator.com later.