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 multi-dimensional kinematics.
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What happens when we are moving in more than just one dimension!
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So, previously we dealt with everything as if we were only one-dimensional, always just one dimension.
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Otherwise, everything is a scalar, a single number on its own.
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But, now we are going to see some of the concepts we were dealing with before, actually are vectors in hiding.
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All the vectors, all the ones we were talking about, are getting from one place to another place, because a place can be many dimensions.
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We are used to living in a three-dimensional world.
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You do not just go to the store on a direct path, not everything is always going to be a straight line, if you look at two paths or more paths.
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So, we are going to have to talk about things using multiple dimensions, either in x-y axes or in x-y-z coordinate system, some kind of coordinate system that has more than one dimension, if we going to be talking about the real world.
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All of our vectors are going to be the ideas that we have a change, displacement, when we are moving from one location to another location.
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Velocity, moving at a certain speed.
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But, more than just speed, it is going to talk about the way we are moving.
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Acceleration, when we are changing the velocity.
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Scalars, on the other hand were the things that were just raw length, if we changed it, it is really one-dimensional.
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The distance between two points is, what it would be if you take a tape measure between those two points.
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It does not care what the angle is, it just cares where, how far it is from 'a' to 'b'.
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The displacement on the other hand, would care how did you get there, it is just not the length, there is an entire circle you could go.
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If you were talking about some length, that length could be pointing in any direction on a circle.
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You need more than that if you are going to really talk about displacement. Speed is similar to distance.
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It is just distance/time, if you are looking for the average speed. That is how fast you are traveling, but once again, it is not going to say anything about where you are actually heading.
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Finally, time.
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We are going to treat time as a scalar, however I do want to say that if you are to get in to more heavier Physics, you would wind up seeing that time can actually be treated as another part of the space vector, where we are located.
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That is going to have to do with the idea of space-time, that is beyond what we are talking about right now.
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So, we are going to be able to treat time as just a single dimensional quantity all the time.
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First, before we get started, a quick note on vectors.
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When we are indicating a vector, I like to use a little arrow, like this guy here.
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Other people, they prefer to use a bold font. If you are looking in a text book do not be surprised if you are seeing bold fonts everywhere, I am writing little arrows.
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Or, if you look in one text book, and it has bold fonts, and another has little arrows, now you know why!
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They are both ways of indicating vectors.
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Sometimes, when we already know we are definitely talking about vectors, it is just assumed that we write it like that.
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Finally , when we are writing it with handwriting, I tend to write it like this, because frankly, I am a little bit lazy.
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I could write it like that, but that takes little bit extra effort, so, instead I make this little harpooned hat that lands on top of it.
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So, little harpoon on top makes that little arrow, and that is how I write it when we are dealing with a vector, when we are writing it up.
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But when it is written, it has that actual arrow above it. Alright! So, position.
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Like said before, position is just the location of an object at a given moment in time.
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But instead of having position be just along a single string, it is no longer just a single dimensional coordinate axis, we now would have to have some sort of grid, may be two dimensions, may be three dimensions.
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We are going to have to talk about more than just one dimension though.
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Here, we would be able to talk about the point (3,2), just the x axis distance, and the y axis distance.
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We go over 3, and then we go up 2. Simple as that.
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Distance and displacement. So, like before, distance is a measure of length.
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We got two points 'a' and 'b', the distance is just that line, that straight line distance from one point to the other.
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It is just the length that you would measure if you were using a tape measure, or walking it out with your feet.
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The displacement on the other hand is a vector that indicates the change occurring between those two points.
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So, 'a' to 'b' would be very different vector than 'b' to 'a', and even if we are dealing with the same length over here, but pretend it is the same length, I think it is about the same length, If we had 'c' over here, 'a' to 'c' would be a completely different vector than 'a' to 'b', even though these are the same length.
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So, the length 'ab' = the length 'ac' = the length 'ca' = the length 'ba'.
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But, each one of those would be a totally different vector.
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We are talking about taking a different path, We have to go there in a different way.
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We 6ake the same number of steps, so to speak, to get there, but we are taking a totally different path to get there.
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Because we are going to a different final location.
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For example, say we walk 3 km North of the house, and then 4 km East.
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We start off at home, we go up 3 km, and then we go East another 4 km.
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So, what would our displacement be? Our displacement will be from where we started, to where we ended.
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So, we go like that, and that would be our displacement vector.
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Our displacement vector would be (4,3).
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And also notice, we are not saying this explicitly, but we almost always assume 'up' as positive, and 'right' is positive.
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Because, that is what they are on the x-y axis on the normal Cartesian coordinate system.
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That is what we are used to in Algebra, we tends to translate over.
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Once in a while we might want to change which we consider to be positive, and which we consider to be negative, but for the most part we are going to treat going North as positive, going South as negative, going East, to the right as positive, going West, as negative.
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And if we are dealing with on a flat ground, may be on a table, and had a box, if the box moved to the right, that would be positive, if it moves to the left, that would be negative.
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If the box moved up, that would be positive, if the box moved down, that would be negative.
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It is up to us to impose a coordinate system.
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It is always an important thing to keep in mind.
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It is us humans who impose a coordinate system on the world and make sense where things are, by giving them assigned values.
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The assigned values can vary, but we have to choose how we are going to start with to make the window frame to look at the world with.
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So, our displacement would be this vector right here, (4,3) km.
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We traveled 4 to the East, we traveled 3 North.
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It does not matter which direction we have done it.
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We could have alternatively done it, it does not matter which order we do it in. We could have alternatively done like this, we would have landed at the same spot.
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But what is the distance between where you started and where you ended? We have got a right triangle.
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By Pythagorean theorem, we know that 3^2 + 4^2 has to be equal to, here is the symbol that we use to show distance, when we want to show how long that vector is, we use the magnitude or the absolute value as you are used to seeing in Math.
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So, this would be the absolute value squared, so 9 + 16 = the distance squared, so we get 25, which is going to wind up becoming 5.
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And we would technically get +/- 5, but when we take the square root, we know there is no such thing as a negative value in distance, so we know that we got to have a positive value.
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We can just forget about the negative when we are doing this.
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And finally, what distance did we travel? We noticed that there is a difference between the distance from beginning to end, a path a bird might take, versus the path that we actually took with our feet.
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In this case, if we measure the feet path, we have to go 3 km North, to the first point, and then change to 4 km East.
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So, we would be 3 + 4 =7 for the total km, for the total distance traveled.
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Each one of these are being different things.
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Displacement is the vector that says how do you get from where you started, to where you ended. Which path you have to take.
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You have to, no matter how you do it, no matter what path you wind up taking, you could walk like this, then walk like this, then walk like this, then walk like this, and then walk like this, and get to the point, the same starting point.
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That is going to wind up being up being the displacement vector of zero.
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Because you did not ultimately displace yourself.
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In this case, we did ultimately displace ourselves, we displaced ourselves 4 km to the East, 3 km to the North. Compare that to the distance.
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We wound up being in a different location, we are now away from our original location by 5 km.
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And finally, the feet steps, we had to actually walk using our feet.
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How far we actually traveled by foot, is going to be the total distance we traveled, 3+4 = 7 km.
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Three very different ideas, but important to remember that we can talk about each one of these things, and each one of these is going to get used, at different times.
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Speed and Velocity.
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Just like before, speed is how fast, it is just how fast you are going some time.
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So, it is that length that you have traveled, divided by the time it took you to do that travel.
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But velocity is based on displacement. It asks how you got there, not just how fast you moved to get there, but how you got there.
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Did you go there at this angle, did you go there at this angle, did you go there at this angle.
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Speed, since they are all the same length, if they were all the same length, that would be the exact same speed, but traveling in three very different directions.
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We are going to have different meeting.
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If you are traveling 60 km/h, to the North, that is very different from if you are traveling 60 km/h to the East.
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They would be the same speed, in the pedometer in your car, but they are going to be very different velocities, because you are traveling in a different path, you are traveling in different way, different direction.
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Velocity is based off displacement. Velocity is a vector as well.
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Velocity is the displacement divided by the time, so **v** = change in displacement/time, Δ d/t.
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Let us go back and look at the example we just had, we started in a house, we traveled 3 km to the North, and then we travel 4 km to the east.
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We do this in 2 hours.
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If it is 2 hours, our displacement, is equal to (4,3) km.
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So, what is our average velocity? Average velocity is the change in our displacement.
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Displacement is the change in the locations, we denote with **d**.
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We can talk about displacement, as **d**, like this, but we could also talk as the change in location.
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Once again, we have this thing where location, displacement, sometimes they get used for the same letter, so there is little bit of confusion here.
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But we know we need to talk about how far we traveled.
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We traveled (4,3) km as a vector.
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If we want to find out what the velocity is, velocity = (4,3) km / 2 hours, so velocity is going to be equal to (2,3/2) km.
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Now, if we want to know what our average speed was, we need to see what distance did we travel.
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Distance = 5 km .
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So the average speed is going to be, we are going to look at the magnitude, the size of that speed vector and it is going to be 5/2 = 2.5 km/h .
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Here is an important note. Speed from velocity.
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As we just saw, on the previous example, if you know something's velocity, we can easily figure out its speed.
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You could figure out how far it traveled total and divide it by the amount of time to get there.
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But you can even do better than that. Speed is the length of the velocity vector.
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We are going to figure it out by looking at how long is the velocity vector.
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If we want the speed, we are going to look at the distance that we traveled and divide it by the time, or we could look at the velocity vector, which is displacement/time.
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So, if we just look at the length of the velocity, we already wound up dividing by the time, and it is going to work out.
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Last time, we got that the average speed for that trip was, 2.5 km/h.
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But velocity vector, **v** = (2,3/2) km/h .
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So, if we want to make it a little bit faster, we just need to see what the magnitude of this is, and it is going to wind up being the exact same as this right here.
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Let us double check that.
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If we do sqrt(2^2 + (3/2)^2), we get sqrt(4 + (9/4)), i.e. sqrt(25/4), which is equal to 5/2, exact same thing.
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So, it is just a question of do we wind up figuring out the distance, and then divide it by the time, or do we wind up figuring out the velocity, which already involves dividing by the time.
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And then just figure out how long that vector is on its own. So, two ways to do it.
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Normally it is going to be easier if we want to find out the speed of something, we know its velocity vector, we just toss it in to this.
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(**v**x)^2 + (**v**y)^2, we can easily figure out what our speed is from that.
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Acceleration just comes from velocity. Since velocity is vector, acceleration must be a vector.
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Acceleration = change in velocity / time .
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So, if we have two different velocities in two different times, we find out what the difference between them is, and then we divide it by the time.
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Gravity is going to be just as the same as before, but we remember it is only going to be effective only in one axis.
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There is no lateral gravity on the Earth.
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If we jump in to the air, you do not get shifted to the right or to the left by gravity.
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So, we are not going to have any x-axis shifting and since when we normally talk about coordinate systems, this tends to be down, this tends to be up, and these are right and left.
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Occasionally we will also look from the top-down when we talk about North and South and East and West.
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But if we are talking about something that is lateral and moving vertically, then we are going to normally do it as up and down being the y-axis.
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If up and down is the y-axis, then we need -9.8 m/s/s, because we are going down at 9.8 m/s/s .
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There is not going to any real changes from the formulae before.
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This is the exact same as it was before, except now are winding up looking at it with vectors.
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So, everything is now working in terms of the vectors here.
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Displacement, the location at time t = 1/2 × **a** t^2 + **v**it + the initial location.
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This one is a little bit special, in that we are going to wind up having to break it down into its components.
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Remember, before we had, vf^2 = vi^2 + 2a × displacement.
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That was what happened when we were looking in one dimension.
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But if we are now looking in two dimensions, this equation right here applies in each dimension, so we wind up applying it over the x's and then over the y's.
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So we just need to separate it and break it into each one.
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We cannot talk about a vector squared, that does not mean anything.
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We cannot just scale it, because who is going to multiply who?
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Instead we have to break it into its components, then we easily square the components of that vector.
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So, we just keep the x axis, the x components separate from the y components, the y axis.
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They each do their own thing, as long as the acceleration and the distance, they are not changing at all.
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As long as you got a steady acceleration, then we are safe, and we can talk about it this way.
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Acceleration = change in velocity, so now we are just working in terms of vectors, and same as velocity = change in location, or the displacement.
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Lets us look at some examples:
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If we have a car driving North at 30 m/s, and then it turns right, so this is vi, turns right, and the final, goes East at 30 m/s.
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If the car takes 5 s to complete the turn, what is the average acceleration that the car has to have on it for the duration of the turn?
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The first thing to note here, is we have to be talking in multiple dimensions.
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We are talking in multiple dimensions, because the car is moving North and then East, so we doing in two totally different directions.
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We cannot just do this in one coordinate axis.
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So, what we really want to do is, we want to convert these directions and lengths into actual vectors.
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First we are going 30 m/s to the North, so that is going to be positive, so we get, (0,30) m/s is **v**i.
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**v**f = 30 m/s to the East.
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If we want to find out what the acceleration is, acceleration = (change in velocity vectors)/(time involved).
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That is, (final - initial)/time = ((30,0) - (0,30)) / 5 = (30,-30) / 5 = (6,-6) m/s/s , because it is changing, for every second it goes, it winds up changing either 6 m/s faster to the East, or -6 m/s 'faster' to the North.
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What that really means is, 6 slower.
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-6, we mean as we are now being accelerated to the South, which means since it is already moving to the North, it is going to lose some of its speed to that southern acceleration.
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Second example is going to be a long one, it is going to create a bunch of different ideas that we are all going to hook together to help us understand how the stuff works.
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We have a ball being thrown out of a window from a height of 10 m, with an initial speed of 20 m/s angled 30 degrees above the horizontal, we can ask some questions.
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What is the initial velocity vector **v** for the ball?
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Then, ignoring air resistance, how long does it take the ball to hit the ground?
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Finally, ignoring air resistance once again, what is the ball's displacement from its starting point, and what is its distance?
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We have got some building, and a window, somebody throws a ball out of that window.
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We know, if we were to set a horizontal straight line, this is going to be an angle of 30 degrees, and this ball comes out at 20 m/s.
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Down here is the ground, and as time goes on, the ball is going to fly forward, and then gravity is going to take more and more of a share of its velocity, and it is going to eventually land, hit the ground somewhere.
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We want to figure out how long is it going to be flying through the air, how long is it going to take before its y location hits zero.
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Once we know that, we can figure out, how far is it going to make it to the right.
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First, we are going to have to know, what is its initial velocity vector.
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If we have got, something that is 20 long on this side, and 30 degrees here, we can figure out what the other sides have to be, just use trig.
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So, this side, since it is the side opposite, is going to be sin(30) × 20.
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This one, would be cos(30) × 20.
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So, cos(30) × 20 = sqrt(3)/2 × 10 = 17.3 m/s, approximately.
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And sine is going to be 10 m/s.
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That is going to give us initial velocity vector of (17.3,10) m/s. (right, up).
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Now we are ready to move on, because now we have got a vector, and we can make use of this vector, and break each component off, and work with each component separately.
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If we want to figure out how long does it take the ball to hit the ground, we do not care what its x component is, it is the same thing if the ball hits the ground here, or here, or here.
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All that matters is, when does it have that zero.
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When does the ball make contact at the ground, when is the y axis location zero.
00:21:56.000 --> 00:22:13.000
If we have got, d(t) = (1/2)at^2 + vit + initial location, we can then just break this into looking at the y's throughout.
00:22:13.000 --> 00:22:18.000
We just look at the y components throughout, and we can figure out when it is going to hit the ground.
00:22:18.000 --> 00:22:23.000
What is its acceleration?
00:22:23.000 --> 00:22:43.000
The acceleration is, -9.8, so (1/2) × (-9.8) × t^2 + 10t + 10 .
00:22:43.000 --> 00:22:46.000
We want to solve this for when does its location becomes zero.
00:22:46.000 --> 00:22:55.000
We want to make that left side set to zero, and we want to see what 't' will allow that zero to appear on the left side.
00:22:55.000 --> 00:22:59.000
So, -4.9t^2 +10t +10.
00:22:59.000 --> 00:23:03.000
How do we solve something like this, we got a couple of choices.
00:23:03.000 --> 00:23:06.000
Right here, we got a polynomial.
00:23:06.000 --> 00:23:10.000
If we have got a polynomial, one of the first things we can do is factor it.
00:23:10.000 --> 00:23:12.000
To me, that does not look really easy to factor.
00:23:12.000 --> 00:23:20.000
The easiest thing after that, is to chuck it into the quadratic formula, make that machine go through it.
00:23:20.000 --> 00:23:22.000
What is the quadratic formula?
00:23:22.000 --> 00:23:34.000
(-b +/- sqrt(b^2 - 4ac)) / 2a , and that is going to be the solutions when time is going to make it equal to zero.
00:23:34.000 --> 00:23:43.000
That gives us one way to do it, one other way, you could just plug that into a computer or a powerful calculator.
00:23:43.000 --> 00:23:47.000
But let us go through the quadratic formula, because we could all just work through it that way.
00:23:47.000 --> 00:24:15.000
The answers, t = (-10 +/- sqrt(10^2 - 4 × (-4.9) × 10)) / (2× (-4.9))
00:24:15.000 --> 00:24:50.000
Keep working through that, we get, (-10 +/- sqrt(100 + 196)) / (-9.8) = (-10 +/- sqrt(296))/(-9.8) = (-10 +/- 17.2) / (-9.8)
00:24:50.000 --> 00:24:53.000
Now, we are going to have to ask ourselves, are we going to go with the plus, or are we going to go with the minus.
00:24:53.000 --> 00:25:11.000
Because mathematically, both of those are correct answers, both of them are times when this equation is going to be fulfilled, it will be fulfilled, that equation will be fulfilled both at the plus and the minus.
00:25:11.000 --> 00:25:20.000
But, we can see logically that the ball only hits the ground one side, it does not have both a forward time and a negative time.
00:25:20.000 --> 00:25:28.000
So, what we want to do, is we know our answer can only be a positive time, because this equation is not true at t < 0.
00:25:28.000 --> 00:25:34.000
Before this moment, we do not know where the ball was, the ball was sitting in some apartment before somebody picked it up, and decided to through it out the window.
00:25:34.000 --> 00:25:40.000
At that moment, that equation right here, was not true before time = 0.
00:25:40.000 --> 00:25:46.000
Once time = 0, that is the moment the ball is actually thrown, we know we can start using this equation.
00:25:46.000 --> 00:25:49.000
So, the only solution that will work, is the one where time > 0 .
00:25:49.000 --> 00:26:02.000
We look at this, and we are going to have to find a way for us to have a negative number on top, because (-10 + 17.2) will give us a positive on top, and then we divide by a negative, we wind up getting a negative thing.
00:26:02.000 --> 00:26:11.000
We are going to actually have to go with the negative answer right here, not the negative answer, but the negative of the plus-minus, because it is the only one which would work.
00:26:11.000 --> 00:26:19.000
If we were to solve it out and get both answers, we would be able to get 2 of them, one positive answer and a negative answer.
00:26:19.000 --> 00:26:27.000
But then the negative answer we know, cannot work, so we have to chose which sign is going to give us a time that is positive.
00:26:27.000 --> 00:26:37.000
We chose the negative one, so knock out the positive one, so we are going to get, 2.78 s.
00:26:37.000 --> 00:26:42.000
So that ball has 2.78 s flight time, before it hits the ground.
00:26:42.000 --> 00:26:46.000
With that, we can now figure out what its x location is, when it hits the ground.
00:26:46.000 --> 00:26:54.000
If it has got 2.78 s flight time, we have see how far it has managed to travel in the x, before it hits the ground.
00:26:54.000 --> 00:27:19.000
Its location in the x is just going to be its 17.3 × 2.78 .(velocity × time)
00:27:19.000 --> 00:27:22.000
But, do we have to worry about acceleration?
00:27:22.000 --> 00:27:33.000
The formula that we normally work with, (1/2)at^2, remember, there is no lateral gravity, we do not have anything accelerating the ball once it is thrown, so all we have to worry about is the initial velocity.
00:27:33.000 --> 00:27:53.000
We multiply those out, we get, 48.1 m is the x location, so if we want to combine that, we have got initial location vector is at where, where does it start?
00:27:53.000 --> 00:27:54.000
Does it start at (0,0)?
00:27:54.000 --> 00:27:58.000
No!, remember, the house starts up here, 10 above the ground.
00:27:58.000 --> 00:28:03.000
So we will make the x axis have it be zero at the point of the window it comes out of.
00:28:03.000 --> 00:28:08.000
So zero for the initial location, but it starts at +10.
00:28:08.000 --> 00:28:17.000
Final location, it hits the ground, and it is at 48.1, and then it has got 0 here.
00:28:17.000 --> 00:28:31.000
So the change, the displacement that it experiences between those two locations, is going to be (48.1,0) - (0,10).
00:28:31.000 --> 00:28:42.000
So we are going to get (48.1,-10), because it traveled down 10 m before it hits the ground, is the displacement it experiences.
00:28:42.000 --> 00:28:57.000
If we want to know what the distance it experiences is, remember, we throw it out of the window, and it lands over here, if we want to know what its distance is, we just have to measure, there is the displacement, so we just need to know what is this length.
00:28:57.000 --> 00:29:04.000
So that length is going to be, the size of the displacement vector.
00:29:04.000 --> 00:29:14.000
Size of the displacement vector, sqrt((48.1)^2 + (-10)^2).
00:29:14.000 --> 00:29:21.000
Punch that out, we get, 49.1 m
00:29:21.000 --> 00:29:29.000
So the ball has a displacement vector of 48.1 to the right, and down 10, so -10 on the y axis.
00:29:29.000 --> 00:29:33.000
But the distance between where it lands, and where it started moving is 49.1 m.
00:29:33.000 --> 00:29:41.000
If we want to know how far its travel was, we do not have the mathematical technology yet, we will be able to figure that out.
00:29:41.000 --> 00:29:50.000
We need to go learn some more calculus before we will be able to figure out what its travel path was, how long it had to move through to get where it landed.
00:29:50.000 --> 00:29:55.000
But we can figure what is the distance from where it landed to where it started.
00:29:55.000 --> 00:29:59.000
Hope that all made sense, see you later.