https://youtu.be/hz833P_toJU
PHYS 1101: Lecture Four, Part Three
Okay, Section 1, Location, location, location. This is the first variable that we need to define. In fact, for location there are two types of vectors that we’re going to need. The first vector is position vector. We’re going to use x with an arrow over the top to indicate that.
The other one is called the displacement vector. We’re going to use a triangle here, that’s a Greek symbol for delta, x to represent that. We’ll see that this displacement vector is always just a difference between two position vectors. It’s the difference between a position vector uniquely saying were the object at the end of the problem is or where the object ended up compared to the origin, minus where the object started out compared to the origin.
Let’s go through each term there carefully. Let’s start with the position vector. So for position vector, specifically what is it? It’s a vector that’s going to uniquely identify an object’s position and that’s only going to be unique only because we have initially set down an axis, a positive direction defined, and a physical location in space that’s going to represent 0. We could decide this origin is somebody’s house and they’re leaving their house to walk to the east for 5 miles or whatever the motion is. It’s up to us to define where 0 is. The position vector, as an example, could be at an instant in time let’s say this person, this object is right there.
Let me actually draw the person in red. I’ll try to be consistent with that, that the object whose motion we’re describing we’ll draw in red, so we always focus on it.
Let’s say this person at this instant, we glance up and they happen to be standing here, which is 2 meters to the right of the origin. The position vector is a vector whose tail’s at the origin and the head is at the location of that object, that is the vector x. At that instant, that’s the person’s position, so you would write that vector as equaling, we’re going to use our scalar component description, scalar component x multiplying by this unit vector.
Let me emphasize for you here, this is the scalar component and in our example then the scalar component would be equal to +2 meters. Remember the scalar component, the value of it here, the magnitude, will have units associated with it that physically then will match or make sense with the kind of vector that this is. The magnitude of this vector has to be in units of length, and we’re going to use the SI meter. Positive tells me that this vector points in the positive x direction.
So, here’s a quiz question for you. Question 4: What would be this scalar component if the person was 5 meters to the left of the origin?
The vector we’re going to work with a lot in this class is really made up a two position vectors, and that’s a displacement vector. The definition of this vector is that it uniquely identifies where an object ended up compared to its starting location.
So, as an example, here’s our one-dimensional motion, this person walking right to left, etc. Displacement vector, it now captures the difference between two positions, where I started compared to where I was. So let’s assume this person started out at the initial position, and that’s what this sub zero will mean, an initial position vector here that ends at the 2 meter mark. And as time went on we don’t exactly know the motion in between, but at some later time when we decide to end the problem, stop the clock, the person is now standing at this location. This is represented by a final position vector. In this class, a plain x with an arrow over the top is going to represent the position vector at the end of the problem.
Time t0 corresponds to the initial position at that instant, what the clock reads at the beginning. Then the variable t is going to represent what the clock reads at the end when the person is at this final position. Almost always in these problems we’re going to assume that we start the clock at the very beginning, so the initial time we would set at 0 seconds. Then maybe the final time is going to be 5 minutes, 10 minutes, a year, any time that is appropriate or is defined in this problem.
So with these two position vectors we can define the displacement vector. The displacement vector is the vector, it’s just a straight line arrow. The tail is at the end of the first position vector, or the tail starts where the object started and this vector ends, or the head, is located where the object ended up, straight line. This is Δx. That vector we call the displacement vector.
Okay, I want to point out an important feature of this. Let me write for you here mathematically what we’ve done. Let’s translate this equation into real life and see that it makes sense. Think about first the tail-to-tip analysis. I’m adding two vectors, they happen to be parallel, but notice that the initial vector plus this displacement vector tail-to-tip adds to the final position vector. That says physically that the final location of this object is the initial location plus whatever change in position happened during this motion.
So, let me add up here that this is, this displacement vector, it’s from where the object was at t0 to the final time, so during that time interval what was the change in position? If we add that to where we started we end up with the final correct location of the object.
So here I have this written out for you in words, what I just said, to help you translate that mathematical equation. Where the object ended up is equal to, it is where it started plus this change in location.
I want to point something out. With these vector equations, like I show you here, if we do some legitimate mathematical rearrangement then the interpretation of a rearranged version of this equation will physically also be legitimate.
Here’s what I mean — that probably sounded pretty obscure — let me show you what I mean. Let me take this equation, I want to rearrange it so I’m focused on what Δx physically represents, so it’s easier to translate and read what Δx is. To do that, I want an equation that says Δx is equal to the rearrangement of this, what? Then I can physically translate this easy. This displacement is equal to whatever mathematical expression I have on the right.
Well, the algebraic rearrangement is pretty easy to do. I want to isolate this, I want to subtract this vector x0 from both sides. So then I have these canceling on the right and on left then I’m left with x minus x0 is equal to Δx. Let me just flip the order of that so I can read it left to right, as I’m more inclined to do.
So by doing legitimate mathematical operations, the integrity between these vectors is maintained, so the physical interpretation then, the physical integrity of what these represent is maintained, and now I can read that the displacement, this change in position, is equal to the final position minus the initial. Let me write that down here. My change is equal to final minus initial. We’ll see that a lot in this class, because we’re focused on motion, we’re focused on how things are changing as time goes on, how the position, the speed, the velocity of this object changes. So we’re going to work with these delta quantities a lot.
So, here’s my summary, clear, variable summary of these location variables that we need. Let me highlight these. These are what you need to memorize. When we go to work a problem if we’re going to leverage these mathematical tools that we’re developing to help us solve the problem, we have to work with these variables. We have to be able to map the problem information into these.
We’re going to be studying the motion over some time interval. We’re going to start the clock at some time t0. We’re going to watch a car driving to the right or a plane landing or taking off. There will be a final scope or a scope to the motion, meaning an initial time and a final time. Time is always going to have units of seconds.
At the start of the problem there will be a unique position of that object. It’s only unique, because somewhere we will have said what 0 means. If this has a value of -5 it means at the start of the problem the object started out 5 meters, or whatever the units are, to the left or in the negative direction, the negative x-axis direction. Then likewise at the final time the object will be at some final position coordinate. Okay, so they have units of meters and seconds.
To see if you understand what I mean by that and how specific these definitions have to be, I’ve got two quiz questions for you, Number 5 and Number 6, so pause the video here and think about these questions.
I clearly indicate for you what the start is and what the finish is. If it helps, let me jot down here that the start would be what corresponds to time t0 and the variable t represents the time at the finish.
All right, I want to give you an example of how these variables are going to show up in a problem, and how this is going to work, how we’re able to use these to describe motion.
Here’s a problem, a typical problem that you might have. A jet liner is traveling northward landing with a speed of 69 meters per second. Once the jet touches down it has 750 meters of runway in which to reduce its speed to 6.1 meters per second. Compute the average acceleration of the plane during the landing.
So the scope of this problem is from initial time when it touches down with this speed to a final time when it has slowed down to this speed, and it does that over this runway that’s 750 meters long. So, I’ve done the first step of this problem, which is, define an axis and an origin. I happened to choose positive y to be up, and I’m going to put the origin at where the place touches down. I could have chosen a different convention, but that seemed a reasonable choice for this problem.
With this definition of this origin and positive y up, you need to answer these two questions here. Question 7: what’s the correct value you’d have to assign for y0?
And then Question 8: what’s the correct value you would assign for the variable y?
I want to make one last comment on this topic of location, location, location. We also, on occasion, will have need to talk about distance. That’s distinctly different from displacement. Distance, we’re going to mean specifically is the actual length along the path, the actual distance along the path that the object travels from start to finish.
I’ve got two bullets here to talk about features of this quantity we call distance. First of all, it’s not a vector. It’s just a literal length, it’s a scalar quantity. Not a scalar component, but just a scalar only, a positive number that literally means the number of paces that the person walked.
The second important bullet to point out to you is that using distance, this specific definition of it, is only marginally useful in Physics. We are, in fact, all the problems, all the homework problems, aside from those that ask you specifically if you understand the difference between distance and displacement, all of the problems will not use distance at all. We’re really going to use these vector quantities of position vectors and displacement vectors.
Let me sketch for you and try to highlight the difference here. For this little example where the person started here and finished here, between time t0 and t — if all you know are these two snapshots between t0 and t, and you don’t have any more information in the problem, this person, for all you know, could have walked to a location at 4 meters, walked back and then circled around again to then finally end up at this position of 4 meters to the right of the origin. The displacement vector, as I emphasized above, regardless of what happened in between, it would be a vector going from the initial location to the final location, straight line as the arrow flies. For displacement, all you care about it where you started and where you ended up.
The distance, what would it be for this problem? It would be plus 4, plus 2, plus 2 again. In terms of the total distance, that would equal 8 meters for this problem, but the displacement, which is a vector, that vector Δx, which I can write as its scalar component, so this is vector quantity now. In this case that scalar component — I’m sorry this should be Δx — that scalar component for Δx is plus 3 meters. It’s positive here because this vector points to the right and its length is 3 meters. So, these are very different quantities, distance versus displacement.
All right, that concludes the specific terms we need to think about and define having to do with location.