WEBVTT physics/ap-physics-1-2/fullerton
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Hi and welcome back to Educator. com. I am Dan Fullerton.
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This lesson is going to be about relative motion.
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Our objectives are going to be to talk about the concept, what it means to say motion is relative and also to calculate the velocity of an object relative to various reference frames.
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So what is a reference frame? That describes the motion of an observer.
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The most common reference frame we are used to is the Earth.
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The laws of physics we study in this course are going to assume we are in an inertial, non-accelerating reference frame.
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That is not completely accurate because of the rotation of the Earth and other things having to do with the way we rotate around the sun.
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But it is close enough for our purposes.
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Now there is no way to distinguish between motion at rest and motion at a constant velocity in an inertia reference frame.
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Imagine for example, you are in an airplane.
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And once the airplane takes off, they put the windows down, the shades down on the window and it is an extremely smooth airplane. No turbulence, whatsoever.
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As long as it is moving at a constant velocity, you cannot tell whether you are on the ground or whether you are in the air flying at a constant velocity.
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There is no experiment you can do without having outside influences from inside the airplane that is going to tell you whether you are traveling at constant speeds or you are completely still.
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My physics perspective, they are really the same thing.
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Now, to talk about motion is relative, let us use an example.
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Imagine you are sitting back in the lawn chair watching a train travel past you to the right at 50 meters per second (m/s).
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From your reference frame, a cup of water that you would see through the trains window is moving at 50 m/s along with the train as well.
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However, if you were sitting on that train right beside the cup of water, the cup of water would appear to be at rest.
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Motion is relative. The velocity of that cup. The motion of that cup depends on where you are and what you are doing.
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Now imagine, you are on the train, staring out the window, watching a student sitting in a lawn chair.
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From your reference frame, the cup of water on the train remains still, but from your vantage point, that student sitting in the lawn chair is moving to the left at 50 m/s.
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Again, motion is relative. So, caculating relative velocities.
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If we consider two objects, A and B, sometimes calculating the velocity of A, with respect to reference frame B can be straightforward.
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Such as, what is the speed of a car with respect to the ground. We are pretty good at that.
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Or walking on a train, what is the speed of the person with respect to the train.
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All of those are pretty straightforward.
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But what happens if we have more objects involved?
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Well here is a formula or a procedure that helps me understand how to do these.
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What we are going to do, is we are going to do this in terms of an example.
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We are going to call Object A, our cup. Object B, is a train and Object C is the ground.
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If we want to know the velocity of Object A with respect to C, the velocity of the cup with respect to the ground, we can figure that out if we know the velocity of the cup with respect to the train plust the velocity of the train with respect to the ground.
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So this would be the velocity of the train with respect to the ground. VAB would be the velocity of our cup with respect to the train and we're trying to find the velocity of the cup with respect to the ground.
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A simple example, but pretty easy to do.
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And the key here is as long as you look at the velocity of one object with respect to another, you can change any other velocities you want to as long as the middle letters keep matching.
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A and B, B and C, what you are going to end up with is a velocity A to C.
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So for example, if we wanted to extend this, the velocity of A with respect to E, whatever those objects are, we could find by taking the velocity of A with respect to B, plus the velocity of B with respect to C, plus the velocity of C with respect to D, plus the velocity of D with respect to E.
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As long as these middle letters match, B to B, C to C, D to D, what you end up with is velocity of A with respect to E.
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A little bit easier to see with some more concrete examples, so let us take a look at some of those.
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A man travels at 60 m/s to the East with respect to the ground.
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A business man on the train runs at 5 m/s to the West with respect to the train.
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Let us find the velocity of the man with respect to the ground.
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And the first thing I am going to do is I am going to identify what my different objects are.
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I will call the train, Object A, the business man, let us call him B, and ground -- C.
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So if we define East as positive, then the velocity, we want the velocity of the man, B, with respect to the ground.
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Velocity of B with respect to C, must be the velocity of B with respect to A, plus the velocity of A with respect to C.
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A is going to match up and we will get B and C.
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So what is the velocity of B with respect to A.
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The velocity of the business man with respect to the train is 5 m/s to the West or -5 m/s.
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The velocity of the train with respect to the ground, well, it is 60 m/s to the East.
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Add those vectors up, I get 55 m/s and it is positive so that must be to the East.
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All right. Let us take another example.
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An airplane flies at 250 m/s to the East with respect to the air.
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The air is moving at 15 m/s to the East with respect to the ground.
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Find the velocity of the plane with respect to the ground.
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All right, well once again, let us identify our objects.
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We will call the airplane, P, for plane. Let us call the air, A, and we have the ground here too, G.
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We want the velocity of the plane, P, with respect to the ground.
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We know the velocity of the plane with respect to the air is 250 m/s and we know the velocity of the air with respect to the ground is 15 m/s.
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So, if we are trying to find the velocity of the plane, with respect to the ground, that is the velocity of the plane with respect to the air, plus the velocity of the air with respect to the ground.
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Let us take a look -- our A's match up -- we will be left with BPG.
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That is going to be 250 m/s plus 15 m/s or 265 m/s.
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All right. Let us try one where we look at a couple of dimensions.
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Now we have an airplane flying at 250 m/s to the East with respect to the air.
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The air is moving at 35 m/s to the North with respect to the ground.
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We want the velocity of the plane with respect to the ground.
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Same thing we did before, the velocity of the plane with respect to the ground is still going to be the velocity of the plane with respect to the air, plus the velocity of the air with respect to the ground.
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Again, those are all vectors, but what we have to remember now is they have direction.
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So, the velocity of the plane with respect to the air, VPA, that is 250 m/s to the East.
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The air is moving 35 m/s to the North with respect to the ground.
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So, there is velocity, air with respect to the ground is 35 m/s.
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We have to add these up in vector fashion in order to get the velocity of the plane with respect to the ground.
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We have our vectors lined up tip to tail, so VPG, go to the starting point of the first, to the ending point of the last.
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There is velocity of the plane with respect to the ground.
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To find out what it is quantitatively -- the magnitude of the velocity of the plane with respect to the ground is going to be -- can use the Pythagorean Theorem -- the square root of 35 ² + 250 ² or about 252 m/s.
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If we wanted to know our angle here, θ -- θ is going to be the inverse tangent of the opposite over the adjacent, 35 over 250 or about 7.97 degrees.
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So, a two dimensional problem. Let us do one more.
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An oil tanker, let us call it T for tanker, travels East at 3 m/s with respect to the ground, while a tugboat, B, pushes it North at 4 m/s with respect to the tanker.
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Find the velocity of the tugboat with respect to the ground.
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So we want velocity of the tugboat with respect to the ground -- that must be the velocity of the tugboat with respect to the tanker, plus the velocity of the tanker with respect to the ground.
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Now what do we know? The velocity of the tugboat, VBT, with respect to the tanker is 4 m/s.
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The velocity of the tanker with respect to the ground is 3 m/s to the East.
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So, VBG, is going to be the vector sum, right there, just like we have done before, where that is a 3:4:5 triangle, so that is going to be 5 m/s.
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All right a brief introduction to relative motion.
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Hope that gets you started and on a good path.
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Thanks for watching Educator. com. Make it a great day.