The most basic level of this section of physics, mechanics, is known as kinematics; essentially this is just the physics behind calculations of speed and distance travelled. Here we introduce a lot of terminology used ad nauseam in mechanics, as well as give you some formulae used in mechanics. Its important to note that, although mathematically things like velocity and speed will be the same number, velocity will give you both the speed and direction while speed will only give you half of the answer, the speed. Once youre comfortable with the math and concepts of vectors and scalars, we can move on to the same topic but in multiple dimensions.
Kinematics is the study of motion and how to mathematically describe it (without concern for the forces causing it).
Position is a location. It is up to us to assign a coordinate system.
Distance is a measurement of length. We can talk about the distance between two points or the distance traveled. Notice that these aren't always the same thing.
Displacement is the change between start and end points. While distance is always positive, displacement can be negative.
Speed is how fast something moves. The distance it travels divided by the time involved.
Velocity is speed along with a direction. It is the displacement of an object divided by the time involved. While speed is always positive, velocity can be negative.
Acceleration is the rate at which a velocity changes. The change in velocity divided by the time involved.
Delta: We use the Greek capital letter delta (∆) to indicate "change in." We can talk about change in velocity as ∆v = vf − vi.
Now that we have delta, we can express velocity as v = [(∆d)/t] and acceleration as a = [(∆v)/t].
Gravity is a constant acceleration on Earth, always pointing down. Because of this, we make it negative: g=−9.8 [(m/s)/s].
The most important formula for kinematics connects all these ideas together: d(t) = [1/2] a t2 + vi t + di.
Another formula allows us to connect final velocity and initial velocity without needing time: vf 2 = vi 2 + 2a(∆d).
One Dimensional Kinematics
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Δ (Displacement) divided by time, Δ (Displacement) divided by time for velocity. 0354
If you want to talk about speed, it would be Δ (Distance) divided by time. 0359
Acceleration is the rate at which velocity changes. 0364
It is a big difference to be in a truck that accelerates really slowly, it gets from 10m/s to 30 m/s over a course of 10s, versus an incredibly fast sports car, that gets from 10m/s to 30 m/s in 2 s.0367
So, acceleration is how fast we manage to change velocity, or in our new letter, our new formation, that we can use Greek letters, new way of talking, we have got acceleration = Δ v / t .0383
So, if you are anywhere on Earth, you are going to experience an acceleration of -9.8 m/s/s.0461
The reason that you are not currently falling through the floor to the centre of the Earth is, you got something resisting that.0467
If you jumped to the air, for that brief moment, before you eventually landed on the ground again, you would be accelerated down towards the centre of the Earth at -9.8 m/s/s. 0471
That negative sign denotes that we are always pointing down, that we are always going down with it, because generally we talk about up as the positive direction.0480
So, -9.8, that negative is there to denote we are going to go down.0489
As a special note, some people denote acceleration as m/s/s, others denote it as m/s2, as we are dividing by second twice.0494
I prefer m/s/s, because it gives you a little bit more of an idea of what is going on with the acceleration, it is how many are m/s changing for every second.0502
A 5 m/s/s means that 1s later you are going 5 m/s more.0511
m/s2, same meaning mathematically, but we lose a little bit of the inherent sense of what we are getting at. 0517
Formulae. So, the most important formula for all kinematics is what shows us all of these different ideas interact.0525
The distance, your location, location at some time, t is equal to 1/2 times the acceleration the object is experiencing times t squared plus the initial velocity of the object times t plus the initial location di. ( dt = 1/2 at2 + vit + di)0531
We cannot actually explain where this formula is coming from, because it comes out of calculus.0546
The way we get this formula, it does not take much difficulty, if we had a little bit of calculus under our belt, we will be able to pick it up fast, but at the moment we do not have calculus.0553
So, We cannot understand where this is coming from quite yet, but we can just go ahead and accept this on a silver platter, so this is a really important formula, we will be taking it for right now, and we will be running with it.0560
That di, we are not going to need it, we are going to move it over to the other side for now, so we get df - di. 0653
And then, while we are at it, let us multiply in that 'a', on that one, so we got, 1/2 times, well, it is going to cancel out, the squared of the a2 on bottom, so we are going to get, vf, let us change this up, so we have a fraction that split now.0659
It is going to be just 'a' on the bottom, because it used to be a2, but we hit it with an 'a'.0675
vf - vi, but this part is still squared, plus vi × (f - i)/a .0679
At this point, we got an 'a' on both of our sides, we got this 2 showing up here.0696
So let us multiply the whole thing by 2a, and let us also realise, df - di is another way of saying that it is just 'change in', Δ.0700
What is acceleration? Acceleration = change in velocity / time .0875
Since there is nothing else acting on it, it is just a rock falling, we do not even have to worry about air resistance, acceleration is just equal to gravity, so we got -9.8 = change in velocity / time.0881
But, it is up to us to pay attention and to go, 'Oh yea, that would not make sense for that to be a..'. 1242
Math, it is a really useful tool when we are solving Physics, but it is up to us to keep it rained in.1249
It is just going to give us the answers, it is just going to do its own thing, because we are using it to model real world phenomena.1254
It is up to us to pay attention to how we are applying it.1260
So, in this case, we will get two answers out of it, but only one of them makes any reasonable sense. So, that is the answer we have to choose. 1263
Third example: We have got a person on a bike, traveling 10 m/s.1270
The person begins braking, and then comes to a stop 10 m later. What was its acceleration then?1274
In this, did we say what the time involved was? NO, We did not!. We could probably work it out.1279
We could solve for it. But, it would be easier if we did not have to. 1285
So instead, we have got that formula (vf)2 = (vi)2 + 2a × (change in distance) .1288
Let us say, what he started out was just at zero, because we are setting it arbitrarily.1299
It does not mean to us where we start, and our change in distance is that 10 m.1303
Final can be 10, but the important thing is, wherever he started, he stops 10 m later, so the change in distance is 10. 1307
The acceleration, we do not know the acceleration, that is what we wanted to know. 1314
We know what he started at, he started at 10 m/s .1317
Do we know what he stopped at? Yes, we know that he came to a full stop, so final velocity must be zero. 1320
We plug these in, we get 02 = 102 + 2a ×10 , 0 = 100 + 20a . 1325
So, we have got -100 = 20a, we have got -5 m/s/s , because that is what acceleration comes in, equals our acceleration, which makes a lot of sense, because if this person is moving forward, for them to come to a stop, they are going to have to have a negative acceleration, they are going to have to be slowing themselves down.1345
They are going to have to be opposing the movement they already had.1360
Final example: We have got a UFO that is currently 500 m away from the surface of the earth.1364
Here is the Earth, and here is our little UFO, hovering in space, and it is currently 500 m above the Earth.1370
And at this moment, it has velocity of 50 m/s, so it is not really sitting there.1384
We take a snapshot and in that first snapshot, it is 500 m away from the surface of the Earth.1389
And in that snapshot, it currently has a velocity of 50 m/s.1394
It also has a constant acceleration, of a 100 m/s/s.1402
So, what would be the velocity of the UFO in 10 s? 1411
If we want to know the velocity of the UFO, the acceleration = change in velocity / time.1413
We know that the acceleration is 100, change in velocity divided by, we are looking 10s later, so we have got 1000 m/s = change in velocity, so our final velocity, is going to be, (our initial velocity, let us denote it as vi).1418
vf is going to be those two added together, so we have got 1050 m/s.1439
If we want to know what its location is, how far is the UFO from the surface of the Earth in 10 s, we need to set up that same equation we used before, distance (t) = 1/2 at2 + vit + initial starting location.1446
We plug those in, we are looking for its distance at 10 s.1467
Is equal to 1/2 times, what is the acceleration?, interesting point to look at, is gravity affecting this, yes, gravity is affecting this, but we do not need to subtract for it, because we are told in the problem that it has a constant acceleration of 100 m/s/s, away from the Earth.1471
So, whoever gave us the problem, however we got this information, we know the acceleration.1486
There is some forces involved, gravity is pulling, and it probably has got thrusters or something, causing it to move away from the Earth.1491
But, we do not have to worry about that, because we are told what its acceleration is precisely.1496
So, we are good with that. So, 100, (and its positive because it is moving up), times the time squared, we do know the time, plug that in, 102 plus, what is initial velocity, 50 times the time, 10, plus its initial location, 500.1500
After plugging things in, 50 × 100, 1/2 times 100 becomes 50, 102 becomes a hundred, plus 500, plus 500, we get 50 × 100 becomes 5000, plus 500 + 500 becomes 1000.1523
So, we get in the end, it is 6000 m above the Earth.1546
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.