For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Frames of Reference

- An observer's motion affects how they perceive the world around them.
- The motion of an observer defines a
*frame of reference*. If a particle moves at the same velocity as an observer, the particle will seem at rest to the observer. - If we want to switch from one reference frame to another, we have to combine what the first reference frame observes and the difference between the two frames.
→v

PA= →v

PB+ →v

BA. - The same thing goes for acceleration.
→a

PA= →a

PB+ →a

BA - In the special case when the velocity between the two reference frames is constant, they will observe the same acceleration.
- The above formulas go for classical mechanics (what this course is studying). If we get close to the speed of light, however, the rules change and we have to use relativity.
- When you are working on problems, try to choose a "still" frame of reference. This will almost always make the problem much easier and more intuitive.

### Frames of Reference

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Fundamental Example 0:25
- Fundamental Example Part 1
- Fundamental Example Part 2
- General Case 2:36
- Particle P and Two Observers A and B
- Speed of P from A's Frame of Reference
- What About Acceleration? 3:22
- Acceleration Shows the Change in Velocity
- Acceleration when Velocity is Constant
- Multi-Dimensional Case 4:35
- Multi-Dimensional Case
- Some Notes 5:04
- Choosing the Frame of Reference
- Example 1: What Velocity does the Ball have from the Frame of Reference of a Stationary Observer? 7:27
- Example 2: Velocity, Speed, and Displacement 9:26
- Example 3: Speed and Acceleration in the Reference Frame 12:44

### High School Physics Online Course

### Transcription: 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 about*0422

*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, 8 ^{2} + (-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 _{f}^{2} = v_{i}^{2} + 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

1 answer

Last reply by: Claire yang

Tue Aug 23, 2016 3:18 PM

Post by Claire yang on August 23 at 03:13:23 PM

For example 3, wouldnt the initial velocity be the speed of the eagle, not 0?

0 answers

Post by Claire yang on August 23 at 02:39:10 PM

`In example 3, isn't gravity -9.8 meters per second per second?

0 answers

Post by Joy Ojukwu on March 3 at 09:10:20 AM

drawing the diagram of Pythagoras theory will make it more clear.

1 answer

Last reply by: Professor Selhorst-Jones

Mon Feb 22, 2016 12:57 AM

Post by Peter Ke on February 20 at 11:47:18 AM

Can you please explain how you got the acceleration for AB=0, I got lost in example 3.

1 answer

Last reply by: Professor Selhorst-Jones

Mon Aug 3, 2015 5:47 PM

Post by Timothy White on June 16, 2015

For most practical purposes, "still" is motionless relative to the earth.

1 answer

Last reply by: Professor Selhorst-Jones

Thu Nov 13, 2014 10:16 AM

Post by Lexlyn Alexander on November 12, 2014

the concept of relativity is still a bit confusing to me especially the vpa parts

1 answer

Last reply by: Joyce Choi

Tue Aug 13, 2013 10:13 PM

Post by Joyce Choi on August 13, 2013

For example 1, for vpa, shouldn't it be 14 m/s + (-11 m/s)?

1 answer

Last reply by: Professor Selhorst-Jones

Wed Jul 24, 2013 2:21 PM

Post by Wootae Song on July 18, 2013

For example 2, when you find t, shouldn't you divide 96 by 8.25 instead of 8?

1 answer

Last reply by: Professor Selhorst-Jones

Tue May 14, 2013 12:03 PM

Post by Goutam Das on May 13, 2013

Why is the concept of relative velocity is different in case of "Light"?