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

### Describing Motion II

- An objectâ€™s position is its location at a given point in time.
- The vector from the origin to the objectâ€™s position is the position vector, r.
- The change in an objectâ€™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
- Objectives
- Special Case: Constant Acceleration
- Deriving the Kinematic Equations
- Problem Solving Steps
- Example IX: Horizontal Kinematics
- Example X: Vertical Kinematics
- Example XI: 2 Step Problem
- Example XII: Acceleration Problem
- Example XIII: Particle Diagrams
- Example XIV: Particle Diagrams
- Example XV: Quadratic Solution
- Free Fall
- Air Resistance
- Acceleration Due to Gravity
- Objects Falling From Rest
- Example XVI: Falling Objects
- Objects Launched Upward
- Example XVII: Ball Thrown Upward
- Example XVIII: Height of a Jump
- Example XIX: Ball Thrown Downward
- Example XX: Maximum Height
- Example XXI: Catch-Up Problem
- Example XXII: Ranking Max Height

- Intro 0:00
- Objectives 0:09
- Special Case: Constant Acceleration 0:31
- Constant Acceleration & Kinematic Equations
- Deriving the Kinematic Equations 1:28
- V = Vâ‚€ + at
- âˆ†x = Vâ‚€t +(1/2)at²
- V² = Vâ‚€² +2aâˆ†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

### 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 II

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

*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 Ã— time.*0047

*The final position is the initial position + the initial velocity Ã— time + half the acceleration Ã— *0054

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

*2 Ã— the acceleration Ã— 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 = Î” V / T which is our final velocity - our initial velocity Ã· time.*0101

*Or I can rearrange that to say that the final velocity = initial velocity + acceleration Ã— 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 â€“ V initial.*0152

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

*Î” x is the area just Â½ base Ã— height for our triangle + length Ã— width our rectangle down here*0174

*which is going to be Â½ , our base is T and our height of our triangle is V - V initial + our height V knot Ã— 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 Î” x = Â½ T Ã— V â€“ V initial, we are just going to replace with at + V knot T which implies then that Î” x must equal V knot t + Â½ atÂ².*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, Î” x = Â½ T Ã— V â€“ 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 Â½ Vt â€“ Â½ V knot t + V knot t which implies then that Î” x = Â½ Vt + Â½ V knot t*0265

*which implies then that Î” 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 Ã· 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 = Î” x / t which implies then that Î” x = V average Ã— t which we just said is V + V knot / 2 Ã— 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 Î” x = we have our V + V knot / 2.*0363

*For our t, we have V â€“ V knot / a which implies then if we multiply through here Î” x is going to be VÂ² - V initialÂ² / 2a*0372

*which implies then that VÂ² - V initialÂ² = 2a Î” x or solving for VÂ² V finalÂ² = V initialÂ² + 2a Î” 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 / sÂ² that must be your acceleration.*0462

*What is the car's speed?*0473

*We are looking for V final after it has traveled 200m Î” 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 Î” 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 Î” x is 200 m, acceleration is 4.9 m / sÂ², 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 VÂ² = V initialÂ² + 2a Î” x.*0532

*VÂ² = 0Â² + 2 Ã— 4.9 m / sÂ² Ã— Î” x 200m or VÂ² = 1960 mÂ² / sÂ².*0542

*I do not want VÂ², 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, Î” y, a, and t.*0603

*We know V initial is 0, we know Î” 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 Î” y = V initial t + Â½ atÂ².*0631

*V initial is 0 that term becomes 0.*0641

*This becomes a little simpler Î” y = 1/2 atÂ² which is Â½ .*0645

*Actually we are trying to solve for a so this implies then that a = 2 Î” y / tÂ² which is 2 Ã— 6 m / 2.7 sÂ² or about 1.65 m / s or 1.65 m / sÂ².*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, Î” 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 Ã· time, therefore the distance traveled is going to be V average Ã— time or 18 m / s Ã— 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 Ã· 12 s.*0789

*It is just going to be 0.5 m / sÂ², we know a = 0.5 m / sÂ².*0806

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

*Let us choose Î” x = V initial t + Â½ atÂ² which will be 15 m / s Ã— 12 s + Â½ Ã— the acceleration 0.5 m / sÂ² Ã— 12 sÂ² is going to give us 216 m again.*0821

*Or we could have picked even a different kinematic equation, V finalÂ² = V initialÂ² + 2a Î” x.*0851

*Solving for Î” x that is going to be V finalÂ² - V initialÂ² / 2a which will be (21 m / sÂ² -15 m / sÂ² ) / 2 Ã— 0.5 m / sÂ² 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 / sÂ² in order to change its velocity by 9 m / s?*0905

*That looks like a horizontal problem again, V initial, V Î” x, a, and t.*0914

*What do we know?*0923

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

*Î” V = 9 m / s and acceleration is 3 m / sÂ².*0928

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

*Therefore time = Î” V / a or 9 m/s / 3m/sÂ² 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, Î” x = 9 cm.*1021

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

*Average speed is displacement Ã· time so that would be 0.09 m or 9 cm Ã· 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 Î” x = V knot t + Â½ atÂ² 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 Î” x / tÂ² or 2 Ã— it went 0.09 m here at 0.3 s Ã· 0.3 sÂ² *1082

*going to give us an acceleration of about 2 m/sÂ².*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/sÂ² 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/sÂ².*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

*Î” x = V initial t + Â½ atÂ² which implies that Î” x 50 = V initial is 1.*1177

*T + Â½ a and Â½ Ã— 2 is going to be 1.*1190

*TÂ² which implies then that tÂ² + 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 bÂ² â€“ 4ac / 2a which is going to be -b is going to be -1 + or â€“ *1210

*the square root of bÂ² 1Â² - 4 Ã— a which is 1 Ã— 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 velocityÂ² = the initial velocityÂ² + 2a Î” x which is going to be equal to 1Â² + 2 Ã— acceleration 2 Ã— displacement 50, *1298

*which implies that VÂ² is going to be equal to 201 m / sÂ² 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 + at*1329

*therefore t = V - V initial / a which is 14.18 m / s - 1 m/s Ã· 2 m/sÂ².*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/sÂ².*1474

*G is always 9.8 m/sÂ² 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, Î” y we do not know.*1513

*Acceleration we do know it is 9.8 m/sÂ² and let us round that to 10 m/sÂ² and that is positive because we call down positive and time is 3s.*1517

*Using our kinematic equations to find how far Î” y = V initial t + Â½ ay tÂ² *1531

*and since V initial is 0 that term becomes 0, Î” y = 1/2 a 10 Ã— tÂ² 3Â².*1541

*9 Ã— 10, 90 Ã— Â½ 45m*1552

*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/sÂ² *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 easier*1697

*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, Î” 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 Î” 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 Ã— 0.4 s is 4 m/s.*1745

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

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

*Her highest point Î” y is V initial t + Â½ atÂ² which is going to be V initial is 4 m/s Ã— *1772

*our time 0.4 s + Â½ Ã— acceleration -10 Ã— time squared 0.4Â²*1782

*Complies that Î” 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 Î” y.*1823

*Acceleration now is 10 m/sÂ² because she is accelerating down and we called down y direction and t is 0.4 s.*1827

*Î” y = V initial t + Â½ atÂ² which is Â½ Ã— 10 m/sÂ² Ã— 0.4Â² 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

*Î” 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 Î” y, A is 10 m / sÂ² and our time is 0.7 s.*1911

*Now V = V initial + at which is 0.5 m/s + a 10 m/sÂ² Ã— 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

*Î” y is what we are trying to find, acceleration is -10 m/sÂ² 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

*VÂ² = V initialÂ² + 2a Î” y which implies that Î” y = VÂ² - V initialÂ² / 2a.*1999

*Now we can substitute in our values Î” y = 0Â² - 30Â² / 2 Ã— -10 â€“ 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 Ã— time.*2062

*Displacement is velocity Ã— time so 30 m/s Ã— 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 Rushâ€™s position = Evil Eddieâ€™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 Ã— 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 www.educator.com and make it a great day everyone.*2204

3 answers

Last reply by: Professor Dan Fullerton

Mon Jan 4, 2016 6:52 AM

Post by Sohan Mugi on January 3, 2016

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, 2016

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?