With the basic equations of motion and an understanding of how variables in different dimensions can be calculated, we can move on to problems that mess around with the observer rather than the object. Previously when we made calculations on a ball being thrown in the air, we assumed the person looking at the ball was standing still on the ground. We would say that relative to that person standing still the ball went up and straight down; however if someone was running past the ball after it was thrown up it would appear the ball was moving away from the runner. Think of when youre in a car on the highway and another car passes you; that car is probably going about 70MPH, yet it only looks like its going 5MPH. This is a good visualization of what we mean by different frames of reference; relative to you the car is going 5MPH, while relative to the road its going 70MPH.
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.
The same thing goes for acceleration.
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.
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
'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 vBA, 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
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
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
We know that the vinitial, 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 vinitial = 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 vfinal 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, vf2 = vi2 + 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
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
This book includes a set of features such as Analyzing-Multiple-Concept Problems, Check Your Understanding, Concepts & Calculations, and Concepts at a Glance. This helps the reader to first identify the physics concepts, then associate the appropriate mathematical equations, and finally to work out an algebraic solution.