For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Related Articles:

### Describing Motion I

- 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 I

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
- Objectives
- Position / Displacement
- Object's Position
- Position Vector
- Displacement
- Position & Displacement are Vectors
- Position & Displacement in 1 Dimension
- Example I: Distance & Displacement
- Average Speed
- Average Velocity
- Example II: Speed vs. Velocity
- Example III: Chuck the Hungry Squirrel
- Example III: Chuck's Distance Traveled
- Example III: Chuck's Displacement
- Example III: Chuck's Average Speed
- Example III: Chuck's Average Velocity
- Acceleration
- Example IV: Acceleration Problem
- The Position Vector
- Average Velocity
- Instantaneous Velocity
- Instantaneous Velocity
- Instantaneous Velocity is the Derivative of Position with Respect to Time
- Area Under the Velocity-time Graph
- Acceleration
- Graph Transformations
- Velocity and acceleration in 2D
- Polynomial Derivatives
- Example V: Polynomial Kinematics
- Example VI: Velocity Function
- Example VI: Part A - Determine the Acceleration at t=1 Second
- Example VI: Part B - Determine the Displacement between t=0 and t=5 Seconds
- Example VII: Tortoise and Hare
- Example VIII: d-t Graphs

- Intro 0:00
- Objectives 0:10
- Position / Displacement 0:39
- Object's Position
- Position Vector
- Displacement
- Position & Displacement are Vectors
- Position & Displacement in 1 Dimension
- Example I: Distance & Displacement 1:21
- Average Speed 2:14
- Average Speed
- Average Speed is Scalar
- Average Velocity 2:39
- Average Velocity
- Average Velocity is a Vector
- Example II: Speed vs. Velocity 3:16
- Example II: Deer's Average Speed
- Example II: Deer's Average Velocity
- Example III: Chuck the Hungry Squirrel 4:21
- Example III: Chuck's Distance Traveled
- Example III: Chuck's Displacement
- Example III: Chuck's Average Speed
- Example III: Chuck's Average Velocity
- Acceleration 6:11
- Acceleration: Definition & Equation
- Acceleration: Units
- Relationship of Acceleration to Velocity
- Example IV: Acceleration Problem 7:05
- The Position Vector 7:39
- The Position Vector
- Average Velocity 9:35
- Average Velocity
- Instantaneous Velocity 11:20
- Instantaneous Velocity
- Instantaneous Velocity is the Derivative of Position with Respect to Time
- Area Under the Velocity-time Graph
- Acceleration 12:36
- More on Acceleration
- Average Acceleration
- Velocity vs. Time Graph
- Graph Transformations 13:59
- Graphical Analysis of Motion
- Velocity and acceleration in 2D 14:35
- Velocity Vector in 2D
- Acceleration Vector in 2D
- Polynomial Derivatives 16:10
- Polynomial Derivatives
- Example V: Polynomial Kinematics 16:31
- Example VI: Velocity Function 17:54
- Example VI: Part A - Determine the Acceleration at t=1 Second
- Example VI: Part B - Determine the Displacement between t=0 and t=5 Seconds
- Example VII: Tortoise and Hare 20:14
- Example VIII: d-t Graphs 22:40

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: 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 changes*0159

*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 specific*0305

*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 carefully*0361

*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 direction*0897

*+ 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 direction*0927

*+ 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 speed*1360

*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

1 answer

Last reply by: Professor Dan Fullerton

Fri Jun 9, 2017 6:25 PM

Post by K Lee on June 9 at 04:59:18 PM

For 3.5Q on the AP Physics C Companion: Mechanics book, I think you meant t=1s and t=6s, not t=1s and t=5s, unless I am misreading.

I didn't know where else to post this. Thanks!

1 answer

Last reply by: Professor Dan Fullerton

Tue Dec 1, 2015 8:25 AM

Post by Jim Tang on November 30, 2015

In Example 7, Q2, how come you took "distance traveled" to mean displacement, even though it's understood somewhat. I thought we were supposed to distinguish between them, as you pointed out earlier in the lecture.

1 answer

Last reply by: Jesse Lefler

Wed Sep 9, 2015 6:11 AM

Post by David Schaller on September 8, 2015

At 19:24 why is it 2tcubed over three? I don't quite follow.

1 answer

Last reply by: Professor Dan Fullerton

Mon Sep 7, 2015 5:09 PM

Post by Shehryar Khursheed on September 7, 2015

Where are the practice problems at the end of the lesson? That is, problems that I can do individually rather than you going over it in the lectures.

1 answer

Last reply by: Professor Dan Fullerton

Thu May 7, 2015 5:47 AM

Post by Joshua Bowen on May 6, 2015

Hey i can not see the lecture it says error 2302

1 answer

Last reply by: SH L

Sun Mar 15, 2015 11:40 AM

Post by Lily Lau on March 15, 2015

At 12:19 for instantaneous velocity. Should the graph have the axis with respect to V instead of X ?

1 answer

Last reply by: Professor Dan Fullerton

Sat Mar 14, 2015 11:43 AM

Post by Hlulani Rikhotso on March 14, 2015

where can i get ap physics c past question papers

1 answer

Last reply by: Professor Dan Fullerton

Thu Jan 8, 2015 12:46 PM

Post by Isaac Martinez on January 8, 2015

Download lectures?

1 answer

Last reply by: Professor Dan Fullerton

Sat Dec 27, 2014 12:39 PM

Post by Jaime De Vizcarra on December 27, 2014

In the example VII: Tortoise and Hare, shouldn't the acceleration of the hare at 40s be a negative value since it is decelerating? {(2.5(m/s)-8(m/s)}/(40s)= -0.1375 m/s^2

1 answer

Last reply by: Professor Dan Fullerton

Sun Dec 7, 2014 3:12 PM

Post by Dawud Muhammad on December 4, 2014

hey professor,other than these vids, what else can i do to sharpen my skills..??

1 answer

Last reply by: Professor Dan Fullerton

Sat Sep 20, 2014 9:08 PM

Post by hasan lopez on September 20, 2014

Can you post some practice problems so we can apply what we learned