AP Physics C: Mechanics Describing Motion I

Hello, everyone, and welcome back to www.educator.com.0000

I am Dan Fullerton and in this lesson we are going to start our study of kinematics, looking at the descriptions of motion.0003

Our objectives include understanding the general relationships among position, velocity, and acceleration for the motion of a particle.0011

Using kinematic equations to solve problems of motion that constant acceleration and that will be carried over into the second half of our lesson.0020

Writing an appropriate differential equations and solving it for velocity in cases when acceleration is a specified function of velocity and time.0027

Position vs. Displacement.0040

An objects position is its location at some given point in time.0042

The vector from the origin of the coordinate system to the objects position is known as the position vector which is r.0046

Sometimes it is written as rs we are going to use interchangeably in here.0053

If an object moves, its position changes.0057

This change in position is called displacement δ r or δ s.0060

Position and displacement are both vectors.0066

They have a direction as well as a magnitude.0069

In one dimension position is given by the x coordinate and the displacement oftentimes written as δ x.0071

A couple examples about the differences between these.0080

A deer walks 1300m E to a creek for a drink.0083

The deer then walks 500m W of the berry patch for dinner.0087

Before running 300m W it is startled by allowed angry fierce nasty evil raccoon.0091

What distance did the deer travel?0097

The distance the deer traveled 1300 + 500 + 300 = 2100m but what is the deer's displacement?0101

That is where how far it is from that starting point.0112

It went 1300m E and 500m W now it is only 800m E and then it went 300m W.0115

It is 500m E from where it started.0121

Δ x is its displacement is 500m E and because it is a displacement, it is a vector it needs a direction as well.0124

Average speed is the distance traveled divided by the time it took to travel that distance.0136

¯V oftentimes depicts average speed which is distance travel ÷ time.0141

Average speed is a scalar and is measured in m/s.0148

Remember that speed S is a scalar S.0153

If we look at velocity on the other hand, the velocity is the rate at which position changes0159

and position as a vector so the rate at which it changes is also a vector velocity.0163

Now average velocity is the displacement during a time interval divided by the time interval, not the distance.0169

The displacement is average velocity.0174

Average velocity is a vector, it has a direction but it also has units of m/s.0177

They are awfully easily to confuse taking the same symbol V.0183

Velocity V is a vector, speed S is a scalar S.0187

Let us take a look.0195

The deer walks 1300m E to the creek, 500m W to the berry patch before running 300m W when it is startled by that loud evil angry nasty raccoon.0197

The entire trip took 600s or 10 min.0207

What is the deer’s average speed?0211

V average is distance over time it traveled 2100m was its distance / 600s.0215

The average speed was 3.5 m/s.0224

The deer’s average velocity however average velocity is δx /t which was 500m E.0229

Its displacement divided by the time 600s or 0.83 m/s E.0239

Notice how subtle these are in their differences.0251

Average speed you are worried about distance.0254

Average velocity you need displacement.0256

Let us do an example with our dear friend Chuck the hungry squirrel.0263

Pork chop travels 4m E and then 3m N in search of an acorn.0267

The entire trip takes him 20s.0272

Find how far Chuck travelled.0274

That is easy.0277

The distance travelled is 4m + 3m which is 7m.0278

Chuck's displacement is however is a little trickier.0284

Chuck traveled 4m E and 3m N.0288

Displacement for a straight line distance from where you start to where you finish so that is a 345 triangle that must be 5m.0295

Chuck’s displacement is 5m NE and if we wanted to we can find the angle to be even more specific0305

that would be the inverse tan for calling that angle θ of our opposites over the adjacent 3m /4 m which is about 36.9°.0312

Chuck's average speed or average speed is going to be distance ÷ time or 7m /20 s which is 0.35 m/s.0326

Chuck’s average velocity is a little bit different δx/t how far his displacement divided by time?0340

It was 5m NE/ 20s which can be 0.25 m/s and it is a vector.0347

It needs a direction NE.0357

Subtle difference is you really have to know what you are talking about and read carefully0361

when you get in the distance displacements being velocity considerations.0364

Let us take a look at acceleration that is the rate at which velocity changes.0370

Acceleration is change in velocity overtime.0375

It is a vector and it has a direction and the units of acceleration are m/s or m/s².0378

That can be confusing to a lot of folks the first time you see it.0386

If velocity changes you may go from a velocity of 10 m/s to 20 m/s.0390

If that time it takes you to change your velocity by 10 m/s is 1s you went from 20 m/s to 10 m/s to 20 m/s in 1s.0395

Your acceleration was 10 m/s every second or 10 m/s².0406

As we are talking about the relationship to velocity and acceleration is the derivative of velocity with respect to time or would be written as V prime.0413

Acceleration problem.0427

Monty the monkey accelerates from rest to velocity of 9 m/s and a time span of 3s.0428

Find Monty’s acceleration.0434

Acceleration is change in velocity ÷ time = final - initial velocity over time which is 9 m/s - 0 m/s all in 3s or 3 m/s².0438

Let us take a look at the position vector.0460

If we have a half in space and we have a particle that is moving along that path we could define its position of various points in time.0462

Its first position at some time t1 we can define its position by a vector from the origin to that point.0473

Let us call that the position vector at time t1.0482

A little while later, it is over here at time t2 so we can define the position vector at t2 as such.0486

There is its position vector at time t2.0495

What happened between t1 and t2?0500

That was when we had a change in position δr which would be that position at t2 - the position at t1.0503

If we could define S going from that point to that point there is δr.0514

While we are doing this we can look at the position function.0524

The function of time and realize that our x value changes as a function of time so we have x(t)0527

and the I ̂ direction and the unit vector direction along the x axis + y value is a function of time in the y direction, the unit vector in a y direction J ̂ .0537

If you prefer you could write this as x is a function of time for the x coordinate, y is a function of time for the y coordinate.0550

Of course you can extend at the three dimensions as well.0559

Then our average velocity vector is just going to be change in r over some time interval.0563

If we take that and go further in the average velocity, we have a particle traveling along the path find the average velocity the between 1 and 6 seconds?0574

The average velocity of the change in position divided by time before looking in the x that is going to be x - x is 0/t or we are just traveling in one dimension.0584

It is a time vs. distance traveled graph which is going to be our final value to about 5m - 2.5m/ 6s it is about 5 or 6s – 1s or 2.5/5 which is going to be about 0.5 m/s.0596

Interestingly though we could also look at the slope here.0628

If we go when we try and take the slope for those two points, we will pick a couple points on our graph.0632

It looks like an easy one to pick will be 0, 2 and we will also go over here to 6s and say that we are at 5.0644

Our slope is rise over run is going to be 5m -2m/6s -0 s or 3m/6s is 0.5 m/s.0650

The slope of the position time graph gave us the same thing as velocity it does give you the velocity.0670

Looking at instantaneous velocity.0679

Average velocity observed over an infinitely small time interval.0682

As you make that time interval smaller and smaller until it becomes infinitesimally small you get the instantaneous velocity at that exact point in time.0686

Instantaneous velocity is the derivative of position with respect to time.0696

Velocity as a function of time is the derivative of position with respect to time or you could write that S prime.0702

We wanted to know what the absolute instantaneous velocity was here 3s,0712

we would find tangent to the curve the slope right at that point in time.0717

That is 2 m/s is our slope that is the instantaneous velocity at that point in time.0722

Now we also have the xt graph, the area under velocity time graph is the displacement during that time interval.0729

If we make this instead we go to a velocity time graph.0739

This shows our velocity is a function of time.0743

The area under it, if you integrate that integral of velocity or just find the area that your change in position for that time interval.0746

Acceleration is the rate which the velocity changes.0758

Acceleration is the limit is Δ t goes to 0 and Δ v/Δ t.0762

Usually make that time interval shorter and shorter or the derivative of velocity with respect to time.0766

Since velocity is a derivative of position that is the 2nd derivative of the x respect the time.0773

And note that we write that as d² x/dt² really squaring anything this is talking about the 2nd derivative.0780

The derivative and then to take the derivative of the first derivative.0787

The average acceleration is Δ v/t so the velocity time graph we could take the slope here and I get something that looks kind like this I think take the slope at that point.0792

If we take the slope of that line, the slope of the velocity time graph at a specific point that would be about -.18m/s².0810

Which means that the acceleration at time t= 4s right where that point lines up would be -.18 m /s².0822

The slope gives you the acceleration at that point in time.0833

We can also look at some graph transformations.0838

If you have a position time graph and you take the slope you can get a velocity time graph.0841

Take the slope of the velocity time graph you can get an acceleration time graph.0847

Or going the other way start with acceleration time graph, if you take the area under the graph you get the change in velocity.0852

If you have a velocity time graph and you take the area of the integral you get the change in position.0860

You can go from one graph to another based on what you are trying to find using slopes and areas, derivatives, and integrals.0866

Alright the velocity, acceleration in two dimensions.0876

Our velocity vector is the limit as Δ t approaches 0 with a time interval gets infinitesimally small.0880

Δ r/Δ t or we wrote that as the dr dt.0890

The derivative of r with respect to time and if r is in multiple dimensions and that would be the x component derivative of the x component in the x direction0897

+ derivative of the y component in the y dimension for however many the dimensions you might have.0909

Or in bracket notation the dx dt for the x, dy dt for the y.0916

Acceleration then is the derivative of velocity with respect to time which would be the 2nd derivative of x with respect to t in the x direction0927

+ the 2nd derivative of y with respect to t in the y direction or in bracket notation again t ⁺2x/dt², 2nd derivative of y with respect to t.0941

You can keep expanding upon that for however many dimensions you need.0961

Typically we are going to be working with 2 and 3 dimensions in this course.0965

Let us talk a little bit more about derivatives.0970

If we have some function x as a constant × t to some exponent N, the derivative of x with respect to is that power N × our constant × t to the N – 1.0974

The basic polynomial derivative formula.0987

We are going to be using that a bit so let us practice for second.0990

The position of the particles as a function of time is 2 -40 + 2 t² -3 t³.0994

Find a velocity and acceleration of the particles as a function of time.1002

Velocity is a function of time is the derivative of x with respect to the time or you could write it as x prime.1007

You might even see that written as x with dot over it.1015

That is going to be, I am going to start at this side just because I like the bigger exponent first that will be -9t²+ 40 -4.1019

To find our acceleration as a function of time that is the derivative of velocity with respect time.1035

Or the 2nd derivative of x with respect to time or we could write this as V prime or V with the dot,1043

or x double prime or x with two dots they all mean the same thing.1054

Eventually we take the derivative of our velocity and I would come up with -18t + 4.1061

Alright more examples, an object moving in a straight line has a velocity V in m/s that varies with time t.1072

According to this function 3 +2t², find the acceleration of the object in 1s.1080

Acceleration is just the derivative of the velocity with respect to time which is going to be derivative to the velocity is going to be 4t.1087

Since we know in this problem that T = 1s acceleration is just going to be equal to 4 × 1 or 4m/s².1098

For part B, determine the displacement of the object between t = 0 and t = 5s.1114

We want to know the displacement Δ x that is going to be the integral from T = 0 to 5s.1120

We can start using definite integrals here because we are given some limits on the time.1127

Our velocity with respect to time which will be the integral from T = 0 to 5s of 3 +2t² Tt which will be 2t³ /3 + 3t all evaluated from 0 to 5s.1131

Remember what this means, that means we are going to plug the 5 in for the t first so we would get 2 × 5³ /3³ /3 + 3 × 5 – 0³ + 3 × 0 = 0.1162

What I come up with here was Δ x is going to be 5 × 5 = 25 × 5 = 125 × 2= 250 ÷ 3 + 15= 98m.1183

There we go the displacement of the object between 0 and 5s.1209

Taking a look at another example.1215

The velocity of time curve for the tortoise and hare traveling the straight line is shown below.1216

I will color the tortoise here in orange and our hare in blue.1222

What happens at time T= 30s?1227

Interpreting these can get a little tricky so let us take our time and go right through it.1231

At T=30s it looks like the tortoise and the hare have the exact same value for speed.1235

A tortoise and hare have the same speed.1244

How do you know the two have travelled the same distance at time T = 60s?1256

Let us see.1262

At T=60s if we look here we are given the velocity time graph.1263

If we want to know distance travelled that is the area under the graph.1269

For the tortoise, that would be the area of this rectangle which is 60 × 4 =240m.1272

For the hare, it is the area of this triangle which you can probably see visually is the same or use ½ base × height you can find that the area for the hare is 240m.1283

At T =60s, the area under each curve is the same so Δ x must be the same.1295

What is the acceleration of the hare at T= 40s?1307

What I would do is I would go over here to T=40s and take a look and say the slope of the line there should give you the acceleration.1322

Acceleration is slope or change in velocity over change in time that looks like we are going from the slope of that line is -8 m/s / 60s is -4/30 – 2/15m/s².1334

That would be the hare’s acceleration.1355

One last one, which of the following pairs of graphs best shows the distance travelled vs. time in speed1360

and speed vs. time for our car accelerating down the hill from rest?1367

Let us take a look at the first one.1373

If distance and time, it looks like it is moving the same amount every time and the speed is constant.1374

If it is accelerating that does not make any sense, it cannot be A.1382

It looks like the further it goes, its distance travelled for each of the time is getting bigger and bigger.1387

The slope of this with the different points gives you your speed graph getting bigger at a constant rate.1392

B looks like it is accelerating your speed is constantly increasing at a constant rate.1397

This does not make sense because the slope of this does not match your speed curve.1403

Same here the distance is increasing linearly while speed is going up quadratically.1408

The answer must be B.1413

Hopefully that gets you a start on describing motion.1416

We will go a little bit further with this in our next lesson describing motion part 2.1420

Thank you so much for watching www.educator.com and make it a great day.1424