AP Physics 1 & 2 Defining & Graphing Motion

Hi everyone and welcome back to Educator.com.0000

Today I want to talk about defining and graphing motion.0003

Our objectives are going to be to understand the difference between position, distance, and displacement.0007

Our second objective is to understand the difference between speed and velocity.0013

Our third objective is to solve problems involving average speed and velocity and calculating distance displacement, speed velocity, and acceleration.0016

Our fourth objective will be constructing and interpreting graphs and diagrams of position, velocity, and acceleration versus time.0024

And finally, we will be determining and interpreting slopes and areas of motion graphs in order to help us understand what these quantities are.0030

So, with that, let us dive right in. Let us first define position.0039

An object’s position in a one-dimension -- you can assign it to a variable on a number scale.0043

Very simply, if we make this an x-axis, for example, we can put an object’s position anywhere on that x-axis and it describes where that is at a specific point in time.0048

Ha! There it is. You can assign whatever you want to be the 0-point as well as the positive and negative directions.0060

Now the symbol for positioning in one-dimension is x.0067

If we look at distance when position changes, an object has traveled some amount of distance.0071

The more position changes, the more distance is traveled.0077

Distance is a scalar. It has a magnitude only -- no direction.0080

And its standard measurement is in meters (m). Distance is a scalar -- no direction.0086

So let us take a look at an example here.0095

We have a deer that walks 1300 m East to a creek for a drink.0097

The deer then walks 500 m West to the berry patch for dinner.0100

Then it runs 300 m West when startled by a loud, fierce raccoon.0104

What total distance did the deer travel? Well, in order to do this, let us just take a look.0108

The deer traveled 1300 m East, then 500 m West, and then 300 m West.0114

So the total distance traveled, 1300 + 500 + 300, should be right around 2100 m.0123

Very straightforward. No direction required.0132

Displacement, on the other hand, is a vector which describes the straight line distance from where you start to where you end.0137

It includes a direction. It is a vector.0144

Displacement, final x position minus initial x position or we also call that Δ (delta) x is also measured in meters.0147

So let us take a look at an example now with displacement.0157

Same basic story, slightly different take on it.0161

Now our deer walks 1300 m East to the creek, then walks 500 m West to the berry patch, and then runs 300 m West when startled by a loud raccoon.0164

What is the deer's displacement? Well to find the displacement, we go from a starting point to our final point.0180

He went 1300 m East and then 500 West and 300 West.0187

Well, the total from the starting point to its final point, must be 500 m East.0192

So the deer's displacement, 500 m East.0200

Displacement is a vector, therefore, it must have a direction as well.0206

Now average speed tells you the distance traveled divided by the total time it took to travel that.0212

And we give it the symbol v with a line over it. The line over it meaning average.0218

So that is the distance traveled divided by the time.0223

Speed is a scalar -- S-S and its measured in meters per second (m/s).0226

So let us take our same example and let us look at average speed.0233

The deer walked 1300 m East, then 500 m West, and then ran 300 m West.0237

It did that entire trip in 600 seconds (s).0242

If that is the case, to find the average speed for the entire trip, v-average, average speed equals total distance traveled divided by the time it took.0246

We already found the total distance traveled is 2100 m and the time it took, 600 s.0257

Just a little bit of math and I find out that the average speed was 3.5 m/s.0266

Moving on, average velocity is the displacement divided by the time interval.0276

Since it is a displacement and displacement is a vector, average velocity is also a vector.0282

It is also measured in m/s, but it includes a direction.0288

Average velocity, again same symbol, v with a line over it, so we have got to be careful.0292

It is x - x initial over the time it took or Δx/T.0298

Let us look at that in the context of our same example as well.0304

Deer walked 1300 m East to the creek, 500 West for dinner, and then ran 300 m West.0307

What is the average velocity, the vector, if the entire trip took 600 s?0313

Once again, we are going to start off with the same sort of math.0319

The average is going to be equal to the displacement divided by the time.0323

But now our displacement is 500 m East. The time it took was 600 s.0328

500/600 is going to be 0.833 m/s and since it is a velocity -- it is a vector -- it must have a direction to go along with it.0335

Let us see if we cannot put all of that together.0352

Here is our problem with our dear friend, Chuck, the hungry squirrel.0355

Chuck, the hungry squirrel, travels 4 m East and 3 m North in search of an acorn.0358

The entire trip takes him 20 s. Find the distance he travels, his displacement, his average speed, and his average velocity.0364

Well distance traveled, if he went 4 m East and 3 m North, how far did he go?0374

He went 7 m -- distance --7 m.0381

What is his displacement? Well to do that we need to figure out how far he went from his starting point to his ending point.0383

He started off down here. He went 4 m East, then he went 3 m North.0389

His displacement then, is going to be the vector from the starting point of the first to the ending point of our last.0400

There is our displacement. That is a 3/4 5 triangle or you could find the magnitude of that by the Pythagorean Theorem.0406

You should still come up with 5 m. So his displacement, Δx must be 5 m and direction -- he went Northeast.0414

How about his average speed? Again, average speed is going to be distance divided by time.0425

His distance we just found was 7 m. It took 20 s, so that is going to be 0.35 m/s for his average speed.0433

Average velocity, on the other hand, is going to be his displacement divided by time.0447

Displacement was 5 m to the Northeast in 20 s or 0.25 m/s Northeast.0453

It has a direction, it is a velocity.0464

So note very carefully here, average speed and average velocity do not have to have the exact same numerical value, even when you are talking about very similar paths or the same paths.0467

All right -- acceleration. Acceleration is the rate at which velocity changes.0480

If everybody just kept going a constant velocity, the same speed, it would be a pretty boring world.0487

So when velocity changes, we call it an acceleration. It is the change in velocity divided by the time.0492

And Δv or Δanything is always the final value minus the initial value.0498

Acceleration, too, is a vector. It has a direction and its units are meters per second per second.0504

What that means is if your speed changes -- if your acceleration is 5 m/s/s, your speed increases by 5 m/s every second.0511

I should say your velocity changes 5 m/s, every second.0520

We also call that a meter per second squared (m/s ²).0525

All right. Let us take a look at an example with acceleration.0529

Monte, the monkey, accelerates from rest, so his initial velocity is 0, to a velocity of 9 m/s.0534

V = 9 m/s in a time span of 3 s. Find Monte's acceleration.0543

Now acceleration is change in velocity divided by time or final velocity minus initial velocity divided by time.0548

That is going to be 9 m/s - 0/3 s or 3 m/s/s, which we would typically write as 3 m/s ².0559

Very straightforward. A simple problem in acceleration, but hopefully it starts giving you the idea.0577

Now, we have to talk a little bit about average values versus instantaneous values. They are not the same thing.0582

Average values take into account an entire time interval.0591

Instantaneous time values tell you the rate of change of a quantity at a specific instant in time at that specific point.0594

They are not always the same. So let us take a look at a problem with average velocity.0602

A motorcyclist travels 30 km in 20 minutes. 30 km in 20 minutes -- there is part of our trip -- at a constant velocity.0610

He takes a 10 minute break, then travels 30 km in 30 minutes at a constant velocity.0617

Find the cyclist's minimum instantaneous velocity, maximum instantaneous velocity, and average velocity.0626

It is pretty easy to see the minimum instantaneous velocity is going to occur here where they are taking a break.0634

Position is not changing -- velocity is 0. So the minimum instantaneous velocity is 0.0640

Where are we going to have the maximum instantaneous velocity?0648

Well, that is going to be when we travel the biggest distance in the smallest amount of time, or up here we are traveling 30 km in 20 minutes.0651

So the max is going to be 30 km/20 minutes or 1.5 km/min.0660

Now to find the average velocity though, we have to take into account the entire trip.0675

Average velocity is going to be the total displacement, 60 km, divided by the total time.0682

20 + 10 + 30 is 60 minutes, which will be 1 kilometer per minute (km/m).0691

And note here that the average velocity is between the minimum and maximum -- always going to be either between or equal to the minimum or maximum.0699

It cannot be outside that and be a real average velocity. A great check on your problem solving.0708

Let us take a look at particle diagrams.0715

It is kind of similar to the effect you might see if you had an old leaky car that has an oil leak and every second, every specific instance in time, it leaks one drop of oil.0718

It is a consistent drop every specified unit of time.0727

So if we were traveling down the road to the right in our car at a constant speed, we would see oil drops hit the ground evenly spaced.0732

Because they are evenly spaced, you know the car is moving at a constant velocity.0744

So just by looking at the oil drops you would be able to go take a look and see exactly what is happening.0746

Now when you looked at these, there are a couple of questions that might remain in your mind.0753

You are pretty certain the car is traveling at constant velocity, but can you tell if it is traveling to the right or to the left just from the oil drops.0757

Now assuming that you were not looking on which side of the road they were on, you really cannot tell.0765

But you do know that regardless, the acceleration is 0.0770

It is moving at constant velocity because these are all evenly spaced -- same spacing between each of the drops.0777

On the other hand, what if our car was accelerating to the right?0782

Our particle diagram is now non-uniformed. We have smaller spacings over here and bigger spacings over here.0786

Drops are getting further apart, velocity is changing, the car is accelerating.0795

Can you think of a case in which the car could have a negative velocity and a negative acceleration, yet speed up?0801

Think about it for a second.0810

Here is our car -- put some wheels on it and make it a nice, pretty little car. There it is.0811

If it has a velocity to the left and we have called to the left negative, that would be a negative velocity.0819

If it is accelerating in that direction, it is going faster and faster and the velocity to the left is getting bigger and bigger in magnitude, but that is a negative acceleration.0825

Negative acceleration does not mean slowing down. It is not decelerating.0836

Negative acceleration just means that you are accelerating in whatever direction you have called negative.0841

If you have velocity and acceleration in the same direction, the car will be speeding up.0847

Its speed will be increasing.0854

Let us take a look at some motion graphs.0858

One of the most popular types is a position time graph. It shows an objects position as a function of time.0861

So let us assume that we have some cute little dog that wanders away from our house at a constant 1 m/s.0866

So the dog does that -- starts at time 0 -- wanders away from the house for a little bit, so the position is getting further and further away from this origin -- the house -- until here the dog decides it has had enough and takes a five second rest.0872

It bops down in the grass in the backyard for 5 s. Its position does not change.0885

Then the dog returns to the house at 2 m/s.0890

So it is going back the opposite direction and we end up with a little bit steeper slope because it is coming back to the house faster.0894

There is a basic position time graph. We can learn an awful lot from this graph though.0902

The slope of this graph tells you the dogs velocity at any given point in time.0909

For the first part of the trip, if we take the slope of the graph -- right here -- the slope there is going to be rise over run.0913

We rise 1, 2, 3, 4, 5 meters in a time of 1, 2, 3, 4, 5 seconds or 1 m/s. That is the dog's velocity. Remember?0927

So the slope of a position time graph, gives you the velocity.0942

Let us take a look here when the dog's taking a rest.0947

Slope of a flat line is 0. The dog's velocity was 0.0950

Let us take a look here coming back to the house over here on the right.0954

The slope there again, rise over run. Our rise now is 1, 2, 3, 4, 5, but it is in the opposite direction.0959

It is going down, so our rise is -5 meters and the time it took goes from 10, 11, 12, 12 1/2 -- 2.5 seconds.0969

So I come up with a slope of -2 m/s.0979

The velocity of the dog is -2 m/s -- where that negative sign -- all that is telling you is that the negative is indicating the direction.0984

So position time graphs can be a very useful tool for describing the motion of an object.0998

We could also make a velocity time graph. It shows the velocity of an object as a function of time.1004

It is related to the position time graph by the slope.1012

So here is the position time graph we were just talking about for the dog wandering away from the house and back.1015

Down below we have the velocity time graph for basically the same information.1021

Initially, the dog's velocity, the slope here was 1 m/s, so our velocity down here for that same time interval of 5 s is 1 m/s.1027

Then the dog took a rest for a couple of seconds -- velocity was 0 -- slope was 0.1038

Here we have a value on our velocity time graph of 0 for that same time interval.1044

Now to end the story, the dog came back to the house -- our slope was -2 m/s.1051

Our velocity down here is -2 m/s for the same time interval, so these graphs are very closely related.1057

We can look at the area under the velocity time graph to tell us the change in the displacement of the dog even.1068

Here is how that works.1073

As we look at the area under the graph of our velocity time graph -- if we take this area, the velocity under the graph, the area between the 0 line and where our graph is, we get that rectangle.1075

The area of that rectangle is length times width.1089

Our length is 5 s. Our width is 1 m/s, which is 5 m.1094

By the time we get to 5, our area is 5 m, but a look at our position time graph, right at that point, the position is 5 m.1103

If we keep going through our graph and say "Hey over here at 8 s, what total area do we have?"1112

All the area to the left of that 8 s is still 5 m, so over here at 8 s, our position is still 5 m.1119

And if we keep going -- if we wanted to say what is the dog's position over here at 12 1/2 s -- well -- now we have another area to take into account.1128

We also have this rectangle and since it is below the line -- although in math they may tell you there is not officially a negative area, there is a meaning to negative areas on the graphs in physics -- that area, which is going to be length times width -- we have 2 1/2 s width and we have -2 m/s.1138

So 2 1/2 x -2 is going to be -5 m.1162

So all of the area added up to the left of this red line -- we have +5 and -5 gets us 0.1167

But look, down here at exactly that point in time. 12 1/2 s, just like we have here, our position is back to 0.1173

Now the slope of the VT graph, the velocity time graph, can also give you a lot of information.1186

It can give you the acceleration. So with a problem like this, we can also look and say, "What is the acceleration of the car at T = 5 s?1192

Well at T = 5 s -- what we have is a slope of 0.1203

So the acceleration at that point -- 0 m/s ².1209

How about the total distance traveled by the car during the 6 s interval?1215

Well, to get the total distance traveled, we need to take the area under all of this.1220

How do you take the area of that? A couple of ways to do it, but probably the easiest in my mind is I would break this up into a couple of shapes.1230

Over here we have a triangle and on the right hand side we have a rectangle.1237

Let us add up their areas to find the total distance traveled.1243

Total distance traveled is going to be the area of our red triangle, 1/2 base times height, plus the area of our blue rectangle.1248

One-half our base is 4 s. Our height is 10 m/s and our rectangle -- our length is from 4-6 -- 2 s, and a height of 10 m/s.1261

So 1/2 x 4 x 10 -- that is going to be 20 m + 2 x 10 -- 20 m -- for my grand answer of 40 m traveled.1279

All right -- how we can use slopes and areas.1298

Now, acceleration time graphs -- you get by taking the slope of the velocity time graph.1304

So, if you take the area under the acceleration time graph, you get the objects change in velocity.1311

We have a pattern here. We started with position time graphs.1316

If we took the slope, we got velocity.1320

If we have a velocity time graph and we take the slope, we get acceleration -- or the other direction -- if we take the area under the acceleration time graph, we get the change in velocity.1322

From the velocity time graph, if we take the area, we get the change in position.1332

You can keep going with this pattern forward and backwards to find whatever information you need to based on these motion graphs.1336

Let us take a look at another example with a position times, sometimes distance time graph.1345

Which graph here best represents the motion of a block accelerating uniformly down an inclined plane?1350

Well let us think about what is going to happen if we have some inclined plane or a ramp and we have a block that is accelerating down the ramp.1358

Initially it is going to be at position 0 and over time it is going to get a larger and larger distance traveled.1369

So, right away, we can eliminate number one on our choices.1378

Now, as it goes faster and faster, it is going to cover more and more distance.1383

We would also think of that as seeing that. . . .1389

If we looked over here at a velocity time graph, we could make a velocity time graph and say "You know, it probably starts at some 0 velocity and goes faster and faster and faster."1392

Well if that is the case, we also need to look at something where the area is getting progressively bigger, therefore the distance traveled must be getting progressively greater for the same time interval.1403

The correct answer must be that one.1415

And we could also look at it from the opposite direction. Right here, the slope is 0, so we have a velocity of 0.1418

Here we have a bigger slope. We have a bigger velocity.1424

Here we have a very big slope. We have a very big velocity as time increases.1428

Therefore, that would be the graph that best represents the motion of a block accelerating uniformly down an inclined plane.1434

Let us take a look at a little bit of a challenging one. See if we cannot push a little bit.1444

Bobbie dribbles the basketball on the ground.1448

Draw the position time graph, the velocity time graph, and the acceleration time graph for the basketball as it travels down from Bobbie's hand -- bounces back up to her hand.1450

We will assume the floor is position Y = 0.1460

So let us start by making a graph of position Y versus time.1464

Let us make one for velocity versus time and let us make one for acceleration versus time.1472

All right. Initially the ball is going to start off in Bobbie's hand.1484

So position-wise, it is going to start up here at some positive value.1488

We also know after a little while, it is going to hit the ground -- position 0 -- and as it comes back up to Bobbie's hand, it is going to come back to where it started.1493

So we have those points for position.1500

Now for velocity, let us assume that Bobbie drops it as opposed to really pushing it, just to make it a little simpler.1505

That means its initial velocity is going to have to be 0.1510

As it is traveling down towards the ground, it is going to go faster and faster.1514

Then the moment it hits the ground, its velocity switches its direction, but keeps roughly the same magnitude.1519

So we are going to have to have this jump back up here and as it comes back up to Bobbie's hand, it gets slower and slower and slower until its velocity is 0 right at her hand level.1526

So we are going to have to have something like that and we will fill in some of the other points in a minute.1535

As far as acceleration goes, the entire time the objects in the air and nothing's touching it, it is acceleration is the acceleration due to gravity here on Earth.1541

It is constant. It does not change.1550

It is -9.8 m/s ², so it is going to be constant -- then that ball is going to come in contact with the floor.1551

Its acceleration is going to change very quickly and then it is going to be in -- we call free-fall -- again.1559

It is going to be in the air with no other forces on it.1565

It is going to have a constant 9.8 m/s ² again.1568

All right. How can we start to fill these in?1572

Well, when the ball hits the ground, its acceleration is going to be positive for a second.1575

It has to be in that direction to change its velocity.1582

So we are going to have to have a spike in our acceleration time graph.1584

For our velocity graph, it is going to start at 0 and it is going to go faster and faster and faster.1591

Then, when it hits the ground it is going to have a spike.1596

It is going to have a very high velocity as it sways back up, slowing down, slowing down, slowing down -- stopping.1599

And finally, as we look at the position of the basketball -- if we take a look, we have to have some sort of path that allows the ball to do that to come back up to Bobbie's hand.1608

So that is a pretty in-depth example and much more complicated example of how you can put position, velocity, and acceleration all together to make one complete story for what is happening to an object.1621

All right, let us do another one.1633

Draw the velocity time graph for a ball tossed upward which returns to the point from which it was tossed.1636

Well, I am going to start off by making my axis again. This is going to be a velocity time graph.1643

If we toss something upwards -- like throw it up -- the moment it leaves my hand, it has its biggest velocity. Right?1653

It's positive -- slowing down, slowing down, slowing down, slowing down -- stops for a split second, switches directions, speeds up, speeds up, speeds up, speeds up, speeds up, but in the negative direction.1661

So if it starts off with its biggest velocity -- it could be there -- a little bit later at its highest point, for a split second it stops, then it goes faster, and faster, and faster in the opposite direction.1672

So the velocity time graph for that situation would look something like that.1685

All right, let us take a look at one last example.1693

How can we get displacement from a velocity time graph?1696

The graph below shows the velocity of an object travelling in a straight line is a function of time.1700

Determine the magnitude of the total displacement of the object at the end of the first 6 s.1705

So we have a velocity time graph -- we want displacement. Right away you should be thinking area.1711

Velocity time graph wants displacement -- you need to take the area.1717

So at 6 s, we will draw our line there.1720

We need the area of everything under the graph to the left of that.1723

Again, a couple of ways you can do this -- but the easiest way I see it off the top of my head is to break this up into a triangle and a rectangle.1729

The area under that should give us the total displacement.1741

So we have the area of the triangle, 1/2 base times height or 1/2 times our base 2 s times our height of 10 m/s is going to be 1/2 x 2 x 10 -- 10 and seconds versus seconds in the denominator -- meters.1745

And the area of our rectangle, length times width, or from 2 to 6 s is 4 s times its height -- 10 m/s -- seconds over seconds cancel out -- 40 m -- so the total displacement then, I just add those two up, 40 + 10 -- 50 m.1767

Hopefully, that gets you started with some of these quantities that describe motion and motion graphs, particle diagrams, position time diagrams, velocity time diagrams, acceleration time diagrams -- gets you started, gets you going.1791

I definitely recommend some more practice on your own.1804

Thanks for watching Educator.com. We will be back soon. Make it a great day!1807