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Lecture Comments (10)

3 answers

Last reply by: Professor Dan Fullerton
Mon Jan 4, 2016 6:52 AM

Post by Sohan Mugi on January 3 at 07:32:08 PM

Hello Mr.Fullerton! I have one question.So how would you know how to determine change in velocity and acceleration from a velocity-time graph?

1 answer

Last reply by: Professor Dan Fullerton
Mon Jan 4, 2016 6:15 AM

Post by Yuhuan Ye on January 3 at 02:09:15 PM

Hi Mr.Fullerton,
Will we have a formula sheet that has all the equations(which includes those kinematic equations) when we take the exam?
Your videos are very helpful.
Thank you!

1 answer

Last reply by: Professor Dan Fullerton
Thu Dec 4, 2014 12:33 PM

Post by Brian Bartley on December 4, 2014

Are we not able to download the lecture slides?

1 answer

Last reply by: Professor Dan Fullerton
Sun Sep 21, 2014 8:04 PM

Post by Nick Cadogan on September 21, 2014

Do we need to know how to derive the kinematics equations as shown at 6:50 for the ap physics c exam?

Describing Motion II

  • An object’s position is its location at a given point in time.
  • The vector from the origin to the object’s position is the position vector, r.
  • The change in an object’s position is called displacement.
  • Velocity is the time rate of change of displacement: v=dx/dt.
  • Acceleration is the time rate of change of velocity: a=dv/dt.
  • The slope of the position-time graph is the velocity. The slope of the velocity-time graph is the acceleration.
  • The area under the acceleration-time graph gives you change in velocity. The area under the velocity-time graph gives you change in position.
  • For cases of constant acceleration, you can utilize the kinematic equations to solve for unknown quantities.
  • Objects under the force of gravity only are said to be in free fall.
  • The acceleration due to gravity on the surface of Earth is 9.8 meters per second per second toward the center of the Earth.

Describing Motion II

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
  • Objectives 0:09
  • Special Case: Constant Acceleration 0:31
    • Constant Acceleration & Kinematic Equations
  • Deriving the Kinematic Equations 1:28
    • V = V₀ + at
    • ∆x = V₀t +(1/2)at²
    • V² = V₀² +2a∆x
  • Problem Solving Steps 7:02
    • Step 1
    • Step 2
    • Step 3
    • Step 4
    • Step 5
  • Example IX: Horizontal Kinematics 7:38
  • Example X: Vertical Kinematics 9:45
  • Example XI: 2 Step Problem 11:23
  • Example XII: Acceleration Problem 15:01
  • Example XIII: Particle Diagrams 15:57
  • Example XIV: Particle Diagrams 17:36
  • Example XV: Quadratic Solution 18:46
  • Free Fall 22:56
    • Free Fall
  • Air Resistance 23:24
    • Air Resistance
  • Acceleration Due to Gravity 23:48
    • Acceleration Due to Gravity
  • Objects Falling From Rest 24:18
    • Objects Falling From Rest
  • Example XVI: Falling Objects 24:55
  • Objects Launched Upward 26:01
    • Objects Launched Upward
  • Example XVII: Ball Thrown Upward 27:16
  • Example XVIII: Height of a Jump 27:48
  • Example XIX: Ball Thrown Downward 31:10
  • Example XX: Maximum Height 32:27
  • Example XXI: Catch-Up Problem 33:53
  • Example XXII: Ranking Max Height 35:52

Transcription: Describing Motion II

Hello, everyone, and welcome back to

In this lesson we are going to continue to talk about describing motion and kinematics.0003

Our goals understand the relationship among position, velocity, and acceleration for the motion of the particle.0010

Use the kinematic equations to solve problems of motion with constant acceleration 0016

and write appropriate differential equations and solve for velocity in cases in which acceleration is specified function of velocity and time.0021

Let us start talking about constant acceleration.0030

For cases where we have a constant acceleration, where velocity is always changing by a constant rate or not changing at all,0034

we can derive a set of kinematic equations that will allow us to solve for unknown quantities.0040

These pre key equations are the final velocity = the initial velocity + acceleration × time.0047

The final position is the initial position + the initial velocity × time + half the acceleration × 0054

the square of the time that has elapsed or the square of the final velocity = the square of the initial velocity + 0061

2 × the acceleration × the change in position.0068

When you know any of these 3 kinematic quantities with constant acceleration you can use these formulas to solve for the other 2.0072

The trick is always going to be finding and understanding what 3 of these are before you go on to get the rest of them.0080

How did we come up with these equations?0088

Let us take a minute and see if we can derive them.0089

Here we are showing a velocity time curve for something that is steadily increasing its speed.0092

It is accelerating at a constant rate.0098

To find our kinematic equations I'm going to start with the definition of acceleration a = Δ V / T which is our final velocity - our initial velocity ÷ time.0101

Or I can rearrange that to say that the final velocity = initial velocity + acceleration × time.0113

There is our first kinematic equation.0119

For our second, we are going to take a look at this in a little more detail and pull out my ruler here.0126

I'm going to look at this where we have this as a function of time.0132

We are going to call this for some final value time T and I'm going to break up our shape here into a triangle and 0138

a rectangle that is going to come in handy in just a minute.0146

As we look at our graph, if this is our initial velocity, this is our final velocity, then this distance here is V – V initial.0152

If we want to take a look at the change in position Δ x that is the area under the velocity time curve.0165

Δ x is the area just ½ base × height for our triangle + length × width our rectangle down here0174

which is going to be ½ , our base is T and our height of our triangle is V - V initial + our height V knot × T.0187

But we already know from equation 1 that V =V initial + at.0203

Therefore V - V initial must equal at.0209

This implies then that Δ x = ½ T × V – V initial, we are just going to replace with at + V knot T which implies then that Δ x must equal V knot t + ½ at².0214

There is equation 2.0243

For our 3rd equation, we are going to come back up to the formula we have there in green, Δ x = ½ T × V – V knot + V knot t we have right here.0247

We are just going to manipulate it a little bit differently.0264

We are going to say then that ½ Vt – ½ V knot t + V knot t which implies then that Δ x = ½ Vt + ½ V knot t0265

which implies then that Δ x / t = V + V knot / 2 which is another formula not exactly whenever kinematic equations but useful.0295

That left hand side is V average and V average is V final + V initial ÷ 2 for something that is accelerating that has a constant acceleration.0309

We are going to use that here in a little bit, I believe.0319

Let us keep going with derivation now.0323

V average = Δ x / t which implies then that Δ x = V average × t which we just said is V + V knot / 2 × t.0325

Another tricky little substitution here if V = V initial + at, we can solve that for t to say that t = V - V initial / a.0350

We can then substitute in to say that Δ x = we have our V + V knot / 2.0363

For our t, we have V – V knot / a which implies then if we multiply through here Δ x is going to be V² - V initial² / 2a0372

which implies then that V² - V initial² = 2a Δ x or solving for V² V final² = V initial² + 2a Δ x.0381

There is that 3rd kinematic equation.0412

It is fairly easy to derive these.0418

As we start dealing with a bunch of these kinematic problems, there are couple problem solving steps 0423

when you can use kinematic equations for constant acceleration problems that will help things out.0429

First is to label your analysis for horizontal and vertical motion.0434

Choose a direction to call positive and stick with that throughout the problem.0439

Typically it helps if you call the direction the object is moving initially as the positive direction.0443

Create a motion analysis table.0448

Fill in your givens and once you know 3 items in your table you can always solve for the unknowns using your kinematic equations.0450

Let us take a look at an example here.0460

A race car starting from rest V initial = 0 accelerates uniformly at a rate of 4.9 m / s² that must be your acceleration.0462

What is the car's speed?0473

We are looking for V final after it has traveled 200m Δ x.0475

Let us pick a direction to call positive and we will call to the right positive in this case.0482

Horizontal motion problem and we will list our items of interest, V initial, V Δ x, a, and t.0487

We will fill in what we know, V initial is 0, we are trying to find V.0500

The final Δ x is 200 m, acceleration is 4.9 m / s², and we do not know time.0505

What I am going to do now is I'm going to look at what I have and what I'm trying to find 0515

and see if I can find a kinematic equation that has all the items that I want in there.0521

I do not care about t, what I am looking for a formula that has those things in it.0527

There is one that would be V² = V initial² + 2a Δ x.0532

V² = 0² + 2 × 4.9 m / s² × Δ x 200m or V² = 1960 m² / s².0542

I do not want V², I want V.0560

V is going to be equal to the square root of that which is + or - 44.3 m / s.0563

Common sense tells me of course I'm looking for the positive so I can get rid of the negative there.0571

My answer is 44.3 m / s.0576

Let us take a look at a vertical problem.0584

An astronaut is standing on the platform on the Moon drops a hammer.0587

If the hammer fall 6 m vertically in 2.7 s what is its acceleration?0591

This is a vertical problem and since the hammer goes down first that is called down our +y direction.0597

Then the list of our information of interest V knot, V, Δ y, a, and t.0603

We know V initial is 0, we know Δ y is 6m and there time is 2.7 s.0613

What we are trying to find? Acceleration.0621

Our 4 items that would be nice to have all a nice formula are right there.0625

We can do that using the formula Δ y = V initial t + ½ at².0631

V initial is 0 that term becomes 0.0641

This becomes a little simpler Δ y = 1/2 at² which is ½ .0645

Actually we are trying to solve for a so this implies then that a = 2 Δ y / t² which is 2 × 6 m / 2.7 s² or about 1.65 m / s or 1.65 m / s².0653

How about a problem that is a little bit more involved, a 2 step problem?0683

We have a car traveling on the straight road at 50 m / s and it accelerates uniformly to speed of 21 m / s in 12 s.0688

Find the total distance traveled by the car in that 12 s interval.0695

We will call to the right our positive direction and list our items of interest V knot = 15 m / s, V final is 21 m / s, Δ x is what we are trying to find.0700

a we do not know and time is 12 s, that is what I'm after.0718

However, it does not look to me like we have a very straightforward path to getting that as it currently stands.0730

What formula has those 4 things in it I do not see it.0738

What we could do know is we could remember the average velocity is V initial + V final / 2.0743

That is going to be 15 m / s + 21 m / s / 2 or 18 m / s.0749

Average velocity is distance traveled ÷ time, therefore the distance traveled is going to be V average × time or 18 m / s × 12 s which is 216 m.0757

One way you can go about solving that.0773

We do not actually have to do it that way especially if you do not remember that equation.0775

We could also go and we can find acceleration first.0780

We know enough to find acceleration even though we do not need it and then plug that back into another kinematic equation.0784

Let us try that a = change in velocity / time which is final velocity - initial velocity / t or 21 m / s - 15 m / s ÷ 12 s.0789

It is just going to be 0.5 m / s², we know a = 0.5 m / s².0806

We can choose any kinematic equation we want now because we know the 4 other items as long as it has Δ x in it.0814

Let us choose Δ x = V initial t + ½ at² which will be 15 m / s × 12 s + ½ × the acceleration 0.5 m / s² × 12 s² is going to give us 216 m again.0821

Or we could have picked even a different kinematic equation, V final² = V initial² + 2a Δ x.0851

Solving for Δ x that is going to be V final² - V initial² / 2a which will be (21 m / s² -15 m / s² ) / 2 × 0.5 m / s² or 216 m.0860

A lot of different path to a solution.0881

Something we are going to see coming up again and again and again in physics.0883

There may not be any correct single path, there may be multiple paths to an answer.0886

Some may be considerably easier than others or more straightforward but oftentimes 0891

there more than one solution to a problem that will lead you to a correct answer.0895

Taking a look at acceleration.0903

How long must a 5 kg kitty cat accelerate at 3 m / s² in order to change its velocity by 9 m / s?0905

That looks like a horizontal problem again, V initial, V Δ x, a, and t.0914

What do we know?0923

We want to have a change in velocity of 9 m / s.0924

Δ V = 9 m / s and acceleration is 3 m / s².0928

We can do this quite simply if we wanted to recognizing that acceleration is change in velocity /time.0935

Therefore time = Δ V / a or 9 m/s / 3m/s² which is 3s.0942

Let us take a look at some particle diagrams.0959

A spark timer is a little device that shots at regular intervals and what you do is 0962

you have a paper go through it and it makes dots on the paper.0967

Those dots correspond to that time interval so you can examine the dots to help you see what is going on with the motion of an object.0972

A spark timer is used record the position of a lab card accelerating uniformly from rest.0980

We probably tied this piece of paper onto a lab card and that card go through it and had the card move 0985

as it went through the timer it makes a dot and set time intervals that gives you a feel for what is going on.0990

Each 0.1s the timer marks a dot on a recording tape to indicate the position of the cart.0995

If that is a 0s and that means this must been 0.1 s, that dot must have been at 0.2 s, 0.3 s, 0.4 s, you get the idea.1002

Find its placement of the card at time t = 0.3 s.1016

I will do that and just go to 0.3 s to recognize from my ruler that must be 9 cm, Δ x = 9 cm.1021

Find the average speed of a cart from 0 to 0.3 s.1031

Average speed is displacement ÷ time so that would be 0.09 m or 9 cm ÷ 2.3 s to get there which is 0.3 m / s.1035

Let us go a little further with this problem.1055

Find the acceleration of the cart.1058

To find the acceleration of the cart we could use Δ x = V knot t + ½ at² realizing that the cart began at rest.1062

It says it is accelerating from rest V knot 0.1078

Therefore we can rearrange this for a to say that a is going to be 2 Δ x / t² or 2 × it went 0.09 m here at 0.3 s ÷ 0.3 s² 1082

going to give us an acceleration of about 2 m/s².1104

On the blank diagram draw at least 4 dots indicating a cart moving at constant velocity.1110

I will probably have something like that constant velocity they should be evenly spaced.1117

When you do these problems there are times when you can come into quadratic situations.1126

Arnie, the aardvark accelerates at the constant 2 m/s² from an initial velocity of 1 m/s.1130

How long does it take Arnie to cross a distance of 50 m?1137

Horizontal problem will start with our known information V initial as 1 m/s.1141

We do not know V final, displacement is 50m, acceleration is 2 m/s².1146

We do not know the time and we are trying to find that.1157

As I go to solve this I look for my items of interest, we have got those 4.1161

We are worried about that one time.1170

We have got a formula that has those in there.1173

Δ x = V initial t + ½ at² which implies that Δ x 50 = V initial is 1.1177

T + ½ a and ½ × 2 is going to be 1.1190

T² which implies then that t² + t - 50 = 0.1196

It is a quadratic equation and we can use the quadratic formula.1206

Let us do that for practice t = -b + or - the square root of b² – 4ac / 2a which is going to be -b is going to be -1 + or – 1210

the square root of b² 1² - 4 × a which is 1 × c which is - 50 / 2a which is 1.1225

That is going to be equal to -1 + 14.18 / 2 or -1 -14.18 / 2 which implies that t = 6.59 s or -7.59 s.1248

Here we got to use a little bit of common sense and know which one of those does not really fit here?1271

Of course that is the negative time, the correct answer must be 6.59 s.1276

That is a lot of work.1281

Typically when you see a quadratic formula pop up in a kinematic equation like this you can get around it by being a little bit clever.1284

An alternate solution, imagine instead we want to solve for the final velocity first.1293

We could do something like final velocity² = the initial velocity² + 2a Δ x which is going to be equal to 1² + 2 × acceleration 2 × displacement 50, 1298

which implies that V² is going to be equal to 201 m / s² or V final is about 14.18 m / s.1313

Maybe a little bit familiar.1327

Once we have that we could go to another kinematic equation V = V initial + at1329

therefore t = V - V initial / a which is 14.18 m / s - 1 m/s ÷ 2 m/s².1336

All looks familiar perhaps which gives us 6.59 s same answer but we did not have to go to use the quadratic formula.1351

And really is just another way of getting to the same thing.1362

If you do see you got a quadratic and you are not a big fan of using a quadratic formula solve for something else first 1366

and then you can go use a different kinematic equation.1372

Free fall when the only force acting on an object is the force of gravity the objects weight we refer to the motion of that object as freefall.1378

That we are going to ignore the force of friction or air resistance.1386

If it drops something it is in free fall, that includes objects that have a non 0 initial velocity.1391

If you throw something down as long as the only force acting on it is the force of gravity we will call that freefall.1397

If we drop a ball and a sheet of paper of course it is obvious there not going to fall at the same rate.1406

If you can remove all the air from the room however you would find that they do follow at the same rate.1411

We are going to analyze the motion of objects by neglecting air resistance that type of friction at least for the time being.1416

Eventually we are going to pull air resistance back into the mix here in this course.1422

Near the surface of earth objects accelerate down at the rate of 9.8 m/s /s we call 1429

this acceleration the acceleration due to gravity and gets a special symbol g.1436

For the purposes of the AP exam we want to make the math a little quicker you can round 9.8 up to 10.1441

More accurately g is the gravitational field strength.1448

As we move further away from earth of course the acceleration due to gravity decreases.1452

Let us take a look at objects falling from rest.1458

That object falls from rest that means its initial velocity is 0 since the objects initial motion is down 1461

it is typically useful to call down the positive y direction and acceleration is +g.1467

If we called down the y direction and the object accelerates down that would be a = 9.8 m/s².1474

G is always 9.8 m/s² whether the acceleration is +g or -g depends on your reference.1485

Let us take a look at falling objects.1496

How far will a brick starting from rest fall freely in 3s neglect air resistance.1498

Since it is going down first we will call down our y direction.1504

We will set up our table of information V initial = 0 from rest.1509

V final we do not know, Δ y we do not know.1513

Acceleration we do know it is 9.8 m/s² and let us round that to 10 m/s² and that is positive because we call down positive and time is 3s.1517

Using our kinematic equations to find how far Δ y = V initial t + ½ ay t² 1531

and since V initial is 0 that term becomes 0, Δ y = 1/2 a 10 × t² 3².1541

9 × 10, 90 × ½ 45m1552

Alright how about objects launched upward?1562

In this case you must examine the motion of the object on the way up and the way down.1565

Since the objects initial motion is up typically it is useful to call that the positive direction.1570

If we call up the + y direction the acceleration and the problem would be -g or -10 m/s² 1575

because it is pointing in the opposite direction of what we called positive.1584

It is nice when things coming up and back down is it the highest point when it gets to its highest point, 1588

that peak for a split second its vertical motion stop since velocity there is 0 before it starts accelerating again on the way back down.1594

There is some symmetry of motion there.1601

We throw something up and we are neglecting air resistance it takes the exact same amount of time 1604

to go up as it does become down to that same level.1609

The initial velocity you threw it up with is the same as the initial velocity it comes down with.1612

Theoretically if you neglect air resistance and you shoot a bullet straight up, the velocity of the bullets as it comes out of the gun 1617

is going to be the exact same as the velocity when it is coming down just in the opposite direction.1625

In reality it does not work that way because of air resistance but I think you get the idea there.1631

A ball thrown upward, a ball thrown vertically upward reaches a maximum height of 30m above the surface of the Earth, 1638

find the speed of the ball and its maximum height.1644

Let us call up the positive y direction and we are going to have to recognize as the object goes up and comes back down at its highest point the vertical velocity is 0. 1648

Therefore the speed of the ball at its maximum height 0 m/s.1660

A basketball player jumps straight up to grab a rebound, if she is in the air for 0.8 s which is pretty good jump how high does she jumped?1669

Let us call up our +y direction and realize that she comes up and back down and all of that happens in 0.8 s 1679

which means she hits her highest point at 0.4 s half of that.1691

If we wanted to we could analyze her motion on the way up or the way down that breaking this in half is going to make it a lot easier1697

because if you want to know how high she jumps we want to know what is happening halfway through that motion.1704

Let us look at the way up first.1709

Up at y direction we know now V initial, we do not know we know final velocity is going to be 0, Δ y is what we are trying to find.1713

A must be -10 because we call the positive in the time to go up is 0.4 s.1727

Let us see do we have an equation that will give us Δ y write off?1735

I do not see anything easy.1739

Let us solve for initial velocity first.1741

If a V = V knot + at and we could say that V initial = V - at which is 0 - a -10 × 0.4 s is 4 m/s.1745

Our initial velocity must have been 4 m/s.1763

We can use an equation with Δ y we want to find how high she goes.1767

Her highest point Δ y is V initial t + ½ at² which is going to be V initial is 4 m/s × 1772

our time 0.4 s + ½ × acceleration -10 × time squared 0.4²1782

Complies that Δ y is about 0.8 m that is a pretty serious vertical leap.1793

That was one way to do it.1801

We could have also looked at what was going on by analyzing our motion on the way down instead of the path up.1803

If we want to look at the way down what we know she starts at initial velocity 0 so it is called down the positive y.1809

V initial before looking at just this window where she is coming down the initial is 0.1818

We do not know V final, we do not know Δ y.1823

Acceleration now is 10 m/s² because she is accelerating down and we called down y direction and t is 0.4 s.1827

Δ y = V initial t + ½ at² which is ½ × 10 m/s² × 0.4² or 0.8 m.1838

Noticed that analyzing on the way down we only have these one formula so it was a little bit faster.1856

Either one will work but another case where there are multiple ways to solve the problem 1861

but one is typically a little more streamlined than the other.1865

How about a ball thrown downward?1870

We threw a ball straight down there with the speed of 0.5 m/s from a height of 4m.1872

What is the speed of the ball 0.7 s after it is released?1877

All right it is called down the y direction because that is the direction the ball moves initially.1882

V initial is .5 m/s we do not know V and we want to figure that out.1888

Δ y is a little tricky I will put 4 m in there but it does not say the ball travels 4m it says 1895

it was thrown from a height of 4m and does not say the hits the ground.1902

Therefore we do not know how far it is gone when it is 0.7 s.1906

From now we do not know the Δ y, A is 10 m / s² and our time is 0.7 s.1911

Now V = V initial + at which is 0.5 m/s + a 10 m/s² × 0.7 s which is going to be 7.5 m/s.1921

There is our final velocity.1942

Alright a quarter kilogram baseball is thrown upward with the speed of 30 m/s.1949

Neglecting friction the maximum height reached by the baseball is what?1955

Let us take a look and we are going to call up the y direction and as we do that the maximum height all we know V initial is 30 m/s.1960

Its highest point, its top, just by looking at just the way up we can call its final velocity 0 m/s.1971

Δ y is what we are trying to find, acceleration is -10 m/s² because we called up our positive direction and time we do not know either.1979

I'm looking for something that has those quantities in it.1993

V² = V initial² + 2a Δ y which implies that Δ y = V² - V initial² / 2a.1999

Now we can substitute in our values Δ y = 0² - 30² / 2 × -10 – 900 /-20 is going to give us 45m.2012

Another type of popular problem is called catch up problems or something like this.2033

Rush, the crime fighting super hero, can run at the maximum speed of 30 m/s while Evil Eddie the criminal mastermind can run 5 m/s.2037

If evil Eddie is 500m ahead of Rush, how much time does Evil Eddie have to devise an escape plan?2047

Let us see here, we can start by looking at their positions.2056

The position of Rush is going to be given by the speed × time.2062

Displacement is velocity × time so 30 m/s × time.2068

The position of Evil Eddie is he starts 500m ahead of Rush but adds at the rate of 5 m/s 500 + 5t.2072

How much time does Evil Eddie have?2083

He has until Rush catches him, when their positions are the same.2085

If we have let xr Rush’s position = Evil Eddie’s position and solve for time that should tell us how long Evil Eddie has to figure out how to get out of the situation.2089

That implies then that 30t = 500 + 5t or 25t = 500 therefore t = 500 / 25 or 20s.2101

Evil Eddie better do a lot of thinking in that 20s.2118

How far must Rush run to capture Evil Eddie?2121

The distance Rush covers is going to be his position, since he started at 0 that is going to be 30t which is 30 × 20 s or 600m.2127

Rush is going to run 600 m in the same time Evil Eddie runs is 100m because he started at 500.2141

Alright, one more sample here.2149

3 A model rocket of varying mass are launched vertically upward from the ground with varying initial velocities,2154

from highest to lowest rank the maximum height reached by each rocket neglecting air resistance.2161

It looks like they have different masses, from lightest to heaviest.2167

This one has the highest initial velocity, the lowest, and medium.2173

The key here is there are no mass dependents.2177

The only thing that is going to matter is that initial velocity.2180

The one with the highest initial velocity with the biggest initial velocity is going to go the highest.2183

That will be one first which is going to then go to the rocket 3 the next highest 2189

and rocket 2 will have the lowest maximum height because there is no mass dependents.2197

All right thank you for watching and make it a great day everyone.2204