High School Physics Frames of Reference

Welcome back to educator.com, today we are going to be talking about frames of reference.0000

So far we have talked about motion from the point of view of a still observer.0005

Now, we also really have not really talked about what the idea of being still means.0009

We will talk about that a little bit later.0012

But first let us talk about what happens when the observer is moving.0014

If you have got somebody in a car driving along, watching something happen on the side of the road, what is he seeing in terms of the kinematics involved.0018

Consider this example of a train car moving along.0026

Imagine there is a train car moving along with a single person standing on top of it.0031

There is another person outside of the train car just standing on flat ground, not moving at all, watching that train car moving.0034

What does the person on the train car see when he looks at his feet.0041

To him, his feet are moving.0046

To him, the train car is still.0048

From this person's point of view, the train car is actually not moving, it is this person who is moving backwards.0050

However, from this person's point of view, he is moving backwards, and his own feet is still.0057

When you look at a cup of water in your car, if you are driving along in a car and the cup of water in it.0065

The cup of water is, relative to you, not moving at all, so you consider it be a still object.0070

But relative to a person on the side of the road, it is actually moving along quite quickly.0075

Let us just make the things little bit more complicated.0080

What if the man on the train car were to throw a ball?0084

Say the train is moving at 100 m/s.0087

The man on the car throws a ball at 15 m/s, relative to himself.0090

The ball is moving at 15 m/s, and the train car is moving at 10 m/s.0095

Relative to this man, the car actually has a speed of zero, and the ball has a speed of 15 m/s.0102

But what is the speed of the ball relative to this person?0110

Well, relative to this guy, the standing still guy, it is going to be the 10 m/s of the train car, plus the 15 m/s of the ball, it is going to be equal to a total of 25 m/s, for the ball moving in his frame of reference.0113

We call it a frame of reference, because it is what you are referring to around you, it is the way you frame the world to look at.0131

If you are driving in the car, going back to the car example, your frame of reference is the inside of the car.0138

The car seat is not moving in your mind.0143

But, for somebody on the side of the street, it is whipping along at 60 miles an hour potentially.0145

So it is moving really quickly.0150

But for you it is not moving at all.0151

'Still' is something relative to you not moving.0153

Given a particle P and two observers, two different frames of reference, A and B.0156

If we know what the velocity of the particle is, in B's frame of reference, and we have v_BA, which is the movement of frame of reference B versus A, or we could also think of it as the movement of B from A.0164

Then, the speed of the particle in A's frame of reference is going to be the two of those added together.0185

If you have got something moving along, and then shoot something out of it, for a person outside of the thing that is moving along, he is going to see both of them combined.0192

What if we want to talk about acceleration?0202

Acceleration is just the change in velocity.0204

It is going to be the past formula that we just looked at, but now modified that each thing is acceleration.0207

The acceleration of the particle in A's frame is going to the same as acceleration of the particle in B's frame, plus the acceleration difference between those two.0211

If one frame is accelerating with respect to another frame, we are going to have to include that in our calculation for the, what the acceleration of the other frame is.0220

There is a special case of what happens when the two frames of reference have a constant velocity difference between them.0228

If there is a constant velocity, then there is no acceleration there.0237

So, the acceleration between the frames is going to be zero.0240

So, if you have constant velocity, that means that your reference frames are going to be observe both the same accelerations.0243

Acceleration will be the same if you have constant velocity difference in your reference frames.0252

So, person driving at 60 on the highway is going to see the same thing as somebody standing still on the side of a road, if they are looking at a rock fall off of a cliff,0255

because they both have constant velocity relative to one another, so they will have no acceleration relative to one another.0264

The only thing that is going to be accelerating is the thing that is falling off the cliff.0270

Expanding it to the multi-dimensional case, it is really easy because now we are just looking at each dimension act independently.0274

Each dimension does not have an impact on the other dimensions, so we can just change everything to vectors.0283

The same formulae that we had before, they are all the same, but now using vectors.0288

Just like before, in our special case, when we have got constant relative frames of reference velocities, we are going to have them see the same relative acceleration between them.0292

Some special notes.0303

First off, we have not really done a good job of defining what 'still' means to us in doing Physics problems.0307

That is pretty much OK, because you have got a really good idea what 'still' means just based on having lived for like the last 15 or more years.0313

If you are old enough to be listening to these lectures, you have got a good idea of what 'still' means.0322

When you are driving in your car, you do not really think of your car being still.0327

You think of the road as being still.0330

When you are thinking about the solar system as being a thing, you do not think of the Earth as being still anymore, you think of the sun as being still.0332

Whatever that you kind of centre your ideas around, and you think of other things moving relative to, then the 'still thing' is not the box, because we talked about it moving.0339

The still thing is going to be the ground that the box is on.0349

This is something that you will have a really good intuition to because you are just used to thinking this way.0352

If you are on a bike headed towards a tree, you do not think the tree as coming towards you attacking you, you think, 'oh, I better turn, so I do not hit that tree',0357

because you are the thing that is going to hit that tree, because you think of the tree as being still.0366

There is no way to decide which is the still one or which is not the still one.0369

There is no definite thing here.0374

It is just a matter of different frames of reference.0376

But, we know what things are able to move and what things are not able to move, and so when we work on Physics problems, it is going to be really important for us to chose the things that do not move, as where you view things from.0378

If we view things from other points of view, it can cause some weird problems to happen.0391

Right now, in all the kinematics equations we have seen, everything would be fine, in the force stuff that we will be seeing soon, everything would be fine,0396

but things will really get looking weird when we start looking at energy, so it is in our best to look at it this way, and it is also the easiest way to solve our problems.0402

So, in general, just try to do thing that you would intuitively think of as 'still'.0411

One another thing, just want to warn you, things get really weird once you get close to light speed.0416

Do not worry about that, this is a basic high school Physics course, we are getting into understanding Newtonian Physics, if we want to talk about0422

light speed, we are going to have to talk about relativity, and that is a later course in the future.0430

But, you do not think that you can have something going at a real real large fraction of the speed of light, and that everything will be the same that we are used to.0434

Speed of light, really really fast, is not going to be a main issue for the problems that we are looking at, but it is something to keep in mind.0441

Let us start looking at some simple examples.0447

A train car is moving forward at 11 m/s.0450

A man on the train car is now walking forward at 3 m/s, and he throws a ball backwards at 14 m/s.0455

What will be the velocity of the ball from the frame of reference of the outside observer?0463

That stationary outside observer is just standing on flat ground, not moving, he is still in the way we want to think about still.0469

He is going to see this ball, be thrown backwards, from an object that is moving forward.0478

How fast is the man moving forward?0484

What is the frame of reference of B moving at?0486

The velocity between B and A = 11+ 3, the twp put together because the man is moving forward and we want to know what his frame of reference is, not the train car on its own,0489

because we are looking at the ball in reference to the man, so, those two is 14.0504

We know that the velocity of the particle from B's point of view = -14 m/s, since going to the right is taken as positive.0510

We add them together, we get the particle from A's point of view is going to be 14 m/s + (-14 m/s) = 0 m/s .0530

From A's point of view, the ball actually does not go anywhere, it just sits there in space, and then drops down, as soon as gravity gets a hold of it.0547

In B's point of view, it whips out of his hand, and goes backward really quickly.0555

But from A's point of view, because he is used to that already moving forward, it just feels like an object has been dropped directly down the train.0559

Example 2.0566

Consider we have got a river that is 96 m wide, from river bank to river bank.0568

We have got a South flowing current in that river that is moving at 2 m/s.0577

If you just drop a little box in, it would flow down the river at 2 m/s.0581

We have got a boat that start on the West bank and will be propelled by the wind, Eastward at 8 m/s.0587

We have to keep in mind that wind is moving at 8 m/s, but as soon as in the open water, as soon as it starts moving across those 96 m, it is going to be affected by the river.0594

So, what would be the speed of the boat?0603

The speed of the boat is going to be those two things put together.0605

Like normal, going to the right is going to be positive, going down, although really in this case it is more South, is going to be negative.0608

Our vector is going to be equal to, we move right at 8 m/s, and we move South at -2 m/s; ACTUALLY we move up down vertically at -2 m/s, or we go 2 m/s South.0616

You can either tell the magnitude and the direction or just the number.0633

If we have got a velocity of 8 m/s, and -2 m/s, that tells us what the velocity is in component form.0637

If you want to know what its speed is, we need to take the magnitude of that.0646

We do that, because that is the hypotenuse, so we take, 8² + (-2)² = 68, the square root of which is 8.25 m/s.0649

Now, we have got how fast the boat is moving.0671

When the boat lands, what would be its displacement?0673

First we need to figure out when the boat lands.0676

The boat going to go this way, but it is also going to be knocked down this way.0678

So, its real path is going to be something like this.0684

We do not know what that is yet, but we need to figure out when it lands.0687

When it lands, it will have no effect on how up or down it is because the river bank is 96 m apart throughout.0690

The first thing we need to do is to figure out how long it takes.0697

It will land when it is past the 96 m, so time = (96 m) / (8 m/s) = 12 s.0701

So it takes 12 s to get across, now we want to see how much South movement does it make.0716

Its distance in the y = (-2 m/s) × (12 s) = -24 m to the vertical, or 24 m to the South.0723

Its total displacement would be how far it went across, which we did not figure out directly, but we already know that the time moment when it lands is when it has hit the 96 m which is what we solved for over here.0735

We know that it is going to have a displacement of 96 m to the East, because that it has to be when it lands, and it is going to have a South displacement of -24 m, and that is our displacement.0748

Finally, a bystander is watching an eagle fly at a height of 200 m at a speed of 30 m/s carrying a rock.0760

So, above the ground by 200 m , and the eagle is flying along at 30 m/s.0772

The eagle lets go of the rock and we are going to allow ourselves to ignore air resistance.0780

What will be the speed of the rock when it hits the ground from the eagle's reference frame?0785

This is really simple, this is just a classic- 'how fast the thing fall', because from the the eagle's frame of reference the rock is not moving horizontally, the rock is moving along with it.0790

The eagle is moving, the rock is also moving that much horizontally, so all we care about from the eagle's frame of reference is, acceleration down, just gravity to deal with in this case.0799

So, how do we figure that out?0811

We know that the v_initial, we will be looking at everything in the y-axis point of view, because in the x-axis point of view, in the eagle frame, remember this is all about the eagle's frame of reference, it is where we are doing all of our work in.0812

So, we are only going to be looking at the y-axis.0832

So, the v_initial = 0, because at first it is not going anywhere, it is just going with the eagle at its current speed.0835

The eagle lets go, we do not know what v_final is, that is what we are going to look for.0843

We know its acceleration, acceleration = g = 9.8 m/s/s .0848

And finally the distance = 200 m, that it falls.0855

We got this great formula, v_f² = v_i² + 2at = 0 + 2 × 9.8 × 200 = 3920 .0858

We take the square root of that, now it gives us, speed = 62.6 m/s .0883

It tells us what speed the rock is moving the instant it hits the ground.0899

We know that the rock from the eagle, is moving at 62.6 m/s, and if we were to put a sign on this we would say -62.6 m/s, but we also just know since we have got this arrow right here, that is moving down.0904

So, it is to us, the physicist to pay attention to what we are doing with our work here.0924

62.6 m/s is its instantaneous velocity at the moment of impact.0928

Now, we want to see what does the bystander see.0935

The bystander is going to see both of them combined.0938

He is going to have the speed of not just, because from the eagle's point of view, he was moving and the rock was moving,0941

but from the bystander's point of view, he is not moving, while the rock is moving.0951

So, we have got the rock moving at 30 m/s, once again we have got that height of 200 m.0957

Now, how do we figure that out?0963

If we have got the velocity of the rock moving at 30 m/s, we know that when it lands it is going to hit with 62.6 m/s.0965

Notice that these two different reference frames, the eagle and the human, they both have no y-difference.0979

So, eagle versus human, no y-velocity relative.0987

They are not moving in terms of height, the eagle is flying along parallel to the ground.0997

So, the eagle is flying along parallel, and it drops the rock, then the rock is only going to be accelerated vertically, so the human, the human does not see any horizontal change.1002

The horizontal velocity is going to remain the same, until something else acts on the rock, which will be the ground.1014

So, as long as it has not touch the ground, or just as it touches the ground, it is going to still have the same velocity that it initially had.1019

From the human's point of view, we are going to see, (we do not even have to do any Math for this, because we are doing this all by reasoning), the v_final is going to be,1026

What is the y?1036

The y will be 30 m/s, because it has not had any change to it, there is no horizontal acceleration, the only thing acting on the rock is gravity.1037

We know, in the eagle's reference frame, that when it hits the ground, it is going 62.6 m/s, down into the ground, since we are in component we got to have a negative sign.1048

That is what the human sees, he sees the same speed as before, but now in addition it has got this horizontal component to it, because there is no difference between the accelerations that the two things are seeing,1061

because the only relative difference that they have is this horizontal thing.1072

Finally, what is the acceleration for human versus eagle?1077

They are both going to have the exact same thing, because the velocity of A to B = (30 m/s, 0 m/s), and it does not change, it stays the same throughout, that means our acceleration here is, nothing.1082

There is absolutely no relative acceleration, so the only thing that they are seeing is, that they are going to see the same acceleration in both cases, which is just gravity.1104

So, that gives us an idea of what reference frames are.1111

See you next time!1113