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Lecture Comments (26)

1 answer

Last reply by: Professor Selhorst-Jones
Sun Jun 15, 2014 4:20 PM

Post by Mitrica Dragos on June 15, 2014

The displacement formula can be proved by taking a time interval t-t0 and divide it into k intervals Δt. t-t0=N*Δt. We look at every second.
at t=t0 we have v1=v0
at t=t0+Δt we have v2=v0+aΔt
at t=t0+kΔt we have vk=v0+k*aΔt
The speed increases in an arithmetic progression, so the average speed is going to be the arithmetic mean of all speeds.
The average speed is vm=Δd/Δt
so (v0+1/2*a*Δt)Δt=Δd

3 answers

Last reply by: Professor Selhorst-Jones
Sat Nov 9, 2013 3:52 PM

Post by Abdi Khalif on November 4, 2013

For the kinematics Formula #2 part how come you took the                     [d(t)=1/2 at^2 + v_i t + d_i] formula and started plugin in other formulas into it?

3 answers

Last reply by: Professor Selhorst-Jones
Tue Aug 13, 2013 10:08 AM

Post by Lavanya Kanneganti on August 12, 2013

Okay, so i understand how you use the formulas to solve the problems and how the math works, i just don't think that i would be able to solve the problem myself if it was in front of me.
I don't think i could be able to approach the problem and decide which formula to use to solve the problem and consider all of the factors that effect the object.

1 answer

Last reply by: Professor Selhorst-Jones
Mon Aug 12, 2013 2:33 PM

Post by Lavanya Kanneganti on August 12, 2013

I've always been told that gravity is equal to 9.8 m/s^2, never negative.
Is this also acceptable?

1 answer

Last reply by: Professor Selhorst-Jones
Mon Jun 3, 2013 8:17 PM

Post by Melinda Hernandez on June 3, 2013

Why are the lecture slides not coming up?

1 answer

Last reply by: Professor Selhorst-Jones
Wed May 1, 2013 2:25 PM

Post by Christopher Barnes on April 30, 2013

Can't you also derive these equations by looking at the area under the velocity curve which I believe equals the change in displacement? It's a great way to show where the equation comes from without using calculus.

1 answer

Last reply by: Professor Selhorst-Jones
Mon Jan 28, 2013 12:55 PM

Post by Lauren Long on January 8, 2013

When I first learned this concept my teacher taught my class the big four. I was wondering how could I apply these formulas to your example number two? One of these formulas is a derivative of the dt=1/2at^2+vit+di

Δx= 1/2(vi+ vf)Δt

vf= vi+ aΔt

Δx= vi(Δt)+1/2a(Δt)^2

vf^2= vi^2+2ax

1 answer

Last reply by: Professor Selhorst-Jones
Wed Dec 26, 2012 8:35 PM

Post by Ali Hashemi on December 19, 2012

The formulas need to be explained in much more detail, I do not need to know how they are derived using calculus but I do need to know which formula to use in certain scenarios. This would be understandable for someone who has already taken physics yet for someone learning this material for the first time, the use of each formula need to be explained in greater detail. A slower pace and better detail about why the specific formula is being used would be very helpful.

1 answer

Last reply by: Professor Selhorst-Jones
Sat Oct 6, 2012 5:40 PM

Post by Mohamed Aden on October 5, 2012

I am not totally satisfied with the way you explained the Formulas, it seemed you were rushing through it, it sounded a kind of like a review...

3 answers

Last reply by: Professor Selhorst-Jones
Wed Dec 26, 2012 8:20 PM

Post by Kristine Penalosa on September 16, 2012

why is the unit for gravity m/s^s now, when in the previous equations it was just m/s?

One Dimensional Kinematics

  • 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

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
  • Position 0:06
    • Definition and Example of Position
  • Distance 1:11
    • Definition and Example of Distance
  • Displacement 1:34
    • Definition and Example of Displacement
  • Comparison 2:45
    • Distance vs. Displacement
  • Notation 2:54
    • Notation for Location, Distance, and Displacement
  • Speed 3:32
    • Definition and Formula for Speed
    • Example: Speed
  • Velocity 4:23
    • Definition and Formula for Velocity
  • ∆ - Greek: 'Delta' 5:01
    • ∆ or 'Change In'
  • Acceleration 6:02
    • Definition and Formula for Acceleration
    • Example: Acceleration
  • Gravity 7:31
    • Gravity
  • Formulas 8:44
    • Kinematics Formula 1
    • Kinematics Formula 2
    • Definitional Formulas
  • Example 1: Speed of a Rock Being Thrown 14:12
  • Example 2: How Long Does It Take for the Rock to Hit the Ground? 15:37
  • Example 3: Acceleration of a Biker 21:09
  • Example 4: Velocity and Displacement of a UFO 22:43

Transcription: One Dimensional Kinematics

Hi, welcome back to Today we are going to be talking about one-dimensional kinematics. 0000

Kinematics in a single dimension. What does kinematics mean first of all?0004

Kinematics just means study of motion.0009

Something that is going to talk about how does something move.0012

That is what we are going to be learning about today. 0015

How do we talk about motion! How do we talk about things moving around! That is clearly an important part of Physics.0017

First idea, position. Position is simply the location of an object at a given moment in time. 0022

For this one, we could have something like this, and BOOM!, it would be a 3. Whatever 3 means. That is a really important point to bring up. 0027

We are the ones who have to impose the coordinate system.0035

Nature does not come with an inherent coordinate system already told to us. 0038

We have to say, okay, this place is zero and then we are going to measure metres this way.0042

So, we get 3 metres, or 3 whatever it is that we have measured.0048

But it is up to us to decide where we are going to put that zero and how we are going to measure off it. 0052

We impose the coordinate system. It is us who are assigning position value ultimately. 0057

It is coming from nature, but where our starting place is, that is on us, it is on us to figure out how we are going to orient things.0062

That is a really important thing in Physics sometimes.0069

Next idea, distance. Distance is just a measurement of length.0073

For example, let us say you start at your house.0076

And then you walk a 100 m to the North. 0079

And then after that, you feel like walking a bit more, so you walk another 50m to the North. How far would you have traveled? 0082

Well, you just put the two together, and BOOM!, you have got a 150 m of travel to the North. 0088

What if instead, we want to talk about displacement. 0093

Displacement is change in position. Displacement is very similar to distance, but sometimes they can be very different. 0097

Consider the following idea: Once again, you start at your house. And you walk a 100 m to the North.0103

But then, you keep feeling like walking, but you do not feel like walking to the North anymore, you decided to walk to the South. 0109

Now, you walk 50 m to the South. What is your displacement? 0115

What is your change in position from where you originally were. 0120

Well, your change in position from where you originally were is, that much.0123

You are now 50 m to the North of your house. So, your change in displacement is 50m North. 0127

But what is the distance that you traveled!0135

The whole distance that you traveled, you wound up going up here, and then you turned back and then you walked another. 0137

So, you walked a 100, and then you walked a 150 total.0142

You walked a 100 up, and then you walked a 50 down, so the total amount of distance you traveled was a 150 m. Very different numbers here. 0146

Important to think about this. Displacement is your change in position. 0156

How did we get from where we are now to where we end up, versus distance, which is what was the total length we took.0159

Just to compare them once again, distance: how far we traveled, how far an object travels, where displacement is the change between the start and the end points. 0166

Notation. When we are talking about these ideas, we want some kind of shorthand variable to denote them.0176

We are going to solve for this stuff eventually. 0180

It might seem confusing at first, but we are going to wind up denoting them all with just simply 'd', 'd' would be the letter we use to denote it. 0182

Location, distance and displacement, they are all going to get denoted with 'd'. 0189

And it is going to wind up getting obvious that, over time, we know what we talking about.0194

We know we are talking about distance, sometimes we know we are talking about displacement. We are going to get it contextually. 0198

We are going to get it based on how we are working with things. 0203

We will know what we mean by 'd', do not sweat over the fact that it currently means three different things. 0205

It will make sense when we are actually working on problems.0210

Speed. Speed is an idea I am sure you are all already familiar with. It is just how fast something moves. 0214

If we want to know how fast something moves, it is going to be the distance that it traveled, divided by the time it took to travel it.0218

You can travel a 100m, but there is a big difference in it if you travel it in 1 s, or if it took you a 1000 hours to travel it.0224

That is where speed comes in.0230

If a car travels 100 m in 5 s, What would its speed be? 0232

Well, speed, simply the distance traveled, 100 m divided by 5s, and that is going to get us 20 m/s.0236

It travels 20 m/s, because every second it goes 20m.0246

This is an important point. m/s is the standard S.I. metric unit for motion, the 'm' and 's' being the units, which brings this into begin with.0252

Velocity. Velocity, is very much like speed, but instead we are going to be talking about displacement.0264

Velocity is equal to the displacement divided by the time.0269

So, it's how much the position or the location has changed from beginning to end, not the total distance that it travels over.0272

A race car can travel around a circular track 200 times in a couple of hours, but what was its velocity on the average!0280

Well, it did not leave the track, it just eventually parked in the same spot it left from, so it had no average velocity.0287

But it would have had a very great average speed because it whipped around that track very quickly.0293

We denote velocity with a 'v'.0298

New idea. Delta, Δ.0302

It is the Greek letter Δ. Δ is a capital Greek letter pronounced Delta.0305

It is used to indicate 'change in'. 0310

If we want to indicate a change in location, we can say Δ d, which would be the starting location subtracted from the ending location.0313

Final location, dfinal - dinitial, which is what displacement would be, right?0321

To figure out what your displacement would be, it is going to be where you ended minus where you started.0327

So, Δ, the idea of 'change in', lots of time, we are going to have to consider not what the absolute value is, what its whole value is, but what was the change from the beginning to the end.0333

Really important idea.0342

Get this in just a simple, this nice Δ symbol.0344

For velocity, v = the Δ of the distance.0350

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

Change in velocity divided by time. 0396

If we had a car moving with a velocity of 20m/s, and accelerated at a constant rate of 40 m/s over 10s, what would be the car's acceleration?0400

Well, acceleration, is equal to Δ v, so Δ v/t, 0408

Δ v is vfinal - vinitial, divided by time, we plug in the numbers we have, vf = 40, ended at 40 m/s, it started at 20 m/s, and it is divided by 10s.0415

So, that is going to get us, 20/10 or 2.0438

And the units we get out of this is, we had m/s on top, and we divided by seconds on the bottom, so it is going to be m/s/s.0443

Gravity is a constant acceleration that always points to the centre of the Earth.0453

It is always -9.8 m/s/s.0457

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

We will be using it a lot.0571

Next idea is one that allows us to relate velocity, acceleration and distance without having to use time. 0573

And this one we can actually derive.0578

First of, remember, acceleration is equal to, change in 'v' over time. 0580

Well, that is the same thing as vf - vi over time. 0587

We could solve this for time if we felt like it.0592

Time is equal to vf - vi divided by acceleration. 0595

We can take this idea right here, and we could plug it into this.0601

Now, first thing is m let us state first this is actually dfinal because that is where it is at the time t that we are looking at, the final location for our purposes.0607

1/2 at2 + vit + di, because that is the starting location where the object was.0615

We plugged that into there, and we are going to get, just box this out, so we have little bit of room to work.0625

df = 1/2 a (vf - vi/a)2 + vi(vf - vi)/a + di.0631

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

So, we got 2a Δ = (vf-i)2 + 2 vi(f- vi) .0707

Let us take all this, move it over here, for ease of work.0745

So, we have got 2a Δ D = (vf)2 - 2vfvi + (vi)2 + 2vi vf - 2 (vi)2.0755

So, at this point we see, this shows up here, and here, so we cancel them out, -2v (i)2 is going to cancel out.0785

So, we are going to get 2a Δ D = (vf)2 - (vi)2 . 0790

So, we can move this over, and we have got (vi)2 + 2a Δ D = (vf)2. 0800

That is exactly what we see up here. 0816

So, if you know the final velocity, and you know the initial velocity, and you know the acceleration, then you can find the change in distance. 0819

Or if you know three of these four elements, you can do it.0824

We can do this without having to work around with 't'. Sometimes, that is a nice thing for us.0827

Remember, acceleration will have to be constant for this, and also for the problem above this, but everything else works out great. 0830

Final couple of formulae to point out. Just definitional formula.0840

So, we defined acceleration = change in velocity/time and velocity = change in distance/time. 0844

With this point, we are ready to hit the examples.0852

If we have a rock, and we throw it directly down from a very tall cliff, the initial velocity is -7.0 m/s.0854

We ignore air resistance. What would be the speed of the rock 4 s later? 0866

Well, gravity is equal to -9.8 m/s/s.0869

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

What is the time? 4 s . 0895

We multiply both sides by 4, we are going to get -39.2 m/s = change in velocity. That is not quite enough, we started at something. 0897

This is our change in velocity. We have to take -7, our initial velocity, and add it to our change in velocity.0910

So, that is going to be equal to -7 + (-39.2) = -46.2 m/s .0921

Next problem. Second example: 0936

We got that same cliff, but now we are going to say, it is precisely 200 m tall.0939

Once again, we chuck a rock down it, at a speed -7 m/s .0944

Ignoring air resistance, how long will it take that rock to hit the ground below?0953

Gravity = -9.8 m/s/s, and what formula we are going to use, we are going to use location, based on time, the final location at some time, 't', is equal to 1/2 at2 + vit + the initial location.0958

For this one, what is our initial location, 200 m above the ground.0979

Let us make the ground, zero.0983

That is our base location. 0986

So, our initial location is, positive 200 m (+200 m) above that ground.0989

Our initial velocity is -7 m/s towards the ground.0993

And our initial acceleration, the only acceleration we have throughout, the constant acceleration we have throughout is, -9.8 m/s/s, the acceleration given by gravity. 0996

We plug all those things in, what are we going to want to solve for, we are going to want to solve for the time one were at zero.1006

So, d(t) = in general, for any time it is going to be 1/2 t2 × (-9.8) = -4.9 t2 + (-7t) + 200 .1012

So, we want to know our location at any time 't', we just chuck in that time, and we will find out what our location is.1040

At least until we hit the ground, or if we go back before zero.1046

Because this equation is just Math, it is supposed to tell something, assuming we are using it right. 1050

But if we go past the time it hits the ground, it does not know where the ground is, it is just a Math equation.1054

So it starts giving us negative numbers.1059

It is going to operate like a problem, it is up to us to be careful of how we use some of these equations. 1061

But, we know what we are doing here.1066

So, what final location do we want to look for? We want to look for the zero. 1068

We want to find out when is that rock at the ground.1072

So that is the case, -4.9 t2 -7t + 200 . 1075

We want to solve this equation for what time, it will give us the location, the final location of zero.1082

How do we solve something like this?1088

We got a couple options. One, we could look to factor, but that does not look very easy to factor to me.1090

Two, we could put it into a numerical solver or some sort of good calculator, and we could figure it out.1095

We could graph it with a graphing calculator, and look for the zeroes, or we could put it into the good old quadratic formula: (-b +/- (b2 - 4ac)1/2)/2a. 1100

In this case, what is our b?1113

It's -7. So, we got (7 +/- ((-7)2 - 4 × (-4.9) × (200))1/2) / 2 × (-4.9) .1116

It does not look very friendly, that is not super friendly and easy, but we can definitely punch that out if we work through. 1147

So, equals (7 +/- (49+3920)1/2)/(-9.8) = (7 +/- (3969) 1/2)/(-9.8), 1153

We are going to wind up getting two different answers out of this, the two possible answers for our time, this quadratic formula would give us, are -7.1 s and 5.7 s, because of that plus/minus. 1176

We are going to get two optional answers depending on if we work with plus, or if we work with the minus.1190

Because it is two different answers that will work. Both of these answers are going to solve this equation right here. 1194

But which one is the right one? Well, this equation does not apply negative time.1200

We start the time, we want to start the clock when we throw it, that is when we set our initial distance, that is when we set our initial velocity, that is the moment of throwing.1206

So that is when we need to set our time as zero.1214

So that is our zero time, just as we throw the rock. 1216

So, that is the case, we are going to have to go with the positive answer because it is the only one that make sense.1220

-7.1 s, that works because what we have described here is, we have described a parabola.1225

We are looking for when does that parabola cross the zero.1232

When does it hit the x-axis. What are the zeroes, the roots to this parabola.1236

Of course, there is two answers to this.1240

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

It's moving pretty fast at the end there.1443

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

Hope you enjoyed that. Hope it made sense. 1553

Next time we will look at how multi-dimensional kinematics works, when we are looking at more than one dimension.1555

See you at later.1560