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

### Retarding & Drag Forces

- When the frictional force is a function of an object’s velocity, the frictional force is known as a drag, or retarding, force.
- Typically we assume the drag force takes the form F=bv or F=cv^2, where b and c are constants.
- Once an object reaches its maximum velocity, when the net force on the object is zero, we say the object has reached its terminal velocity.
- To find velocity as a function of time, write your Newton’s 2nd Law equation in the form of a differential equation. Standard bodies falling through the air with air resistance can typically be solved using the method of separation of variables.

### Retarding & Drag Forces

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 0:07
- Retarding Forces 0:41
- Retarding Forces
- The Skydiver 1:30
- Drag Forces on a Free-falling Object
- Velocity as a Function of Time 5:31
- Velocity as a Function of Time
- Velocity as a Function of Time, cont. 12:27
- Acceleration
- Velocity as a Function of Time, cont. 15:16
- Graph: Acceleration vs. Time
- Graph: Velocity vs. Time
- Graph: Displacement vs. Time
- Example I: AP-C 2005 FR1 17:43
- Example I: Part A
- Example I: Part B
- Example I: Part C
- Example I: Part D
- Example I: Part E
- Example II: AP-C 2013 FR2 24:26
- Example II: Part A
- Example II: Part B
- Example II: Part C
- Example II: Part D
- Example II: Part E

### 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: Retarding & Drag Forces

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

*I'm Dan Fullerton and in this lesson we are going to talk about retarding and drag forces.*0003

*You have got it we are actually going to get to air resistance.*0008

*Our objectives are going to be to find the terminal velocity of an object.*0011

*To describe the motion of a particle under the effect of a retarding a drag force in terms of its displacement, velocity, and acceleration.*0015

*Use Newton’s second law to write a differential equations for the velocity of the object.*0024

*Derive the equation for velocity for Newton’s second law using separation of variables.*0029

*Determine the acceleration for an object falling under the influence of drag forces.*0035

*Let us start off by talking about these retarding a drag forces.*0040

*Sometimes the frictional force is a function of how fast an object is going.*0044

*Its velocity and air resistance is a great example of this.*0048

*These forces are called drag or retarding forces.*0052

*Now, as we go to our analysis of drag this is going to be our first lesson in a while that is going to get pretty heavy into the map and calculus.*0056

*Probably, seeing some things if you are taking calculus concurrently where you are seeing it in physics before you might in calculus.*0064

*That is ok, it is not uncommon.*0071

*Do your best to stick with it and then come back to this lesson later on once you are a little bit more solid*0073

*on the calculus and some of the integration we are going to be doing.*0079

*That is not uncommon at all.*0082

*If you leave this lesson a little fuzzy that is ok.*0084

*Alright, let us start by talking about a skydiver.*0090

*Assume we drop our dear friend Alex from an airplane.*0094

*Typically, the drag force on a freefalling object takes the form where the force of drag is some constant × velocity or constant × square of velocity.*0097

*Or sometimes even something in between that where b and c are constants.*0104

*For the purposes of this problem, let us assume that the drag force is to equal to a constant times the velocity.*0111

*Let us draw a free body diagram for our dear friend Alex who are about to push from an airplane.*0118

*We have the weight of Alex pulling down and the drag force backup.*0123

*We will call down the y direction since that is the way Alex was going to start his motion and finishes motion.*0132

*As I analyze this, the net force in the y direction using Newton’s second is going to be mg - the drag force*0139

*and that is all equal to mass times the acceleration in the y direction.*0150

*We just said that our drag force is equal to bv.*0155

*We can write that mg - bv = Ma.*0162

*Let us take a look at what happens when we first push Alex from the plane.*0173

*Initially at time T = 0 Alex’s velocity is 0, therefore, the drag force is 0 mg = Ma and a =g.*0178

*The acceleration = acceleration due to gravity.*0194

*Therefore, we can write at that point acceleration is equal to g.*0197

*As T increases as we head toward infinity eventually Alex reaches a maximum or terminal velocity.*0205

*Terminal velocity we will abbreviate that Vt and at that point acceleration is equal to 0.*0228

*At that point, all the forces must be balanced if there is no acceleration.*0242

*The drag force must equal mg at that point.*0246

*In Newton’s second law equation we have Mg – bv = ma with the condition of terminal velocity, acceleration is 0 and V = Vt.*0251

*Our equation becomes Mg – b Vt = 0.*0268

*Therefore, we can solve for terminal velocity and say terminal velocity is just going to be Mg divided by the constant b.*0275

*Our formula for terminal velocity based on that constant in our f drag = bv equation and we also know initially T =0.*0285

*Velocity 0 is equal to the acceleration due to gravity g.*0294

*I think that will set us up to get into our math using Newton’s second law of equation.*0299

*If we go back to our Newton’s second law equation which has velocity and velocities derivative acceleration in it,*0306

*we are really have 2 different forms of the same variable in the equation.*0312

*The variable itself and its derivative and that is called a differential equation.*0318

*We are going to solve that using a strategy known as separation of variables.*0322

*Let us start on the next page to give ourselves all kind of room here.*0329

*We will begin with our Newton’s second law of equation mg - bv = ma.*0334

*Like we said, acceleration is the derivative of velocity a = dv dt*0343

*We have mg - bv = m dv dt.*0349

*Our differential equation with V and its derivative in the same equation.*0357

*You can take entire courses on solving differential equations.*0361

*We are only to have to deal with a couple types, couple simple ones here in the course.*0364

*This one we can solve with that separation of variable strategy.*0368

*Let us walk through and see how we do this.*0372

*If mg – bv= m dv dt I can divide the whole thing by b so we would have mg / b - V = m/b dv dt.*0374

*Mg/b we just defined in the previous page that is what we called terminal velocity.*0391

*We can maybe pretty this up just a little bit by saying that VT = mg / b*0397

*Therefore our left hand side becomes Vt - V= m/ b dv dt.*0403

*Which implies then I will try to get all the variables V’s and derivative of these on the same side.*0416

*I have dv / Vt – V = b / m dt.*0423

*It is just an algebraic rearrangement.*0432

*I have dv / Vt – V I rather have V there than have V – Vt.*0436

*I am going to multiply both sides by -1 to get the left hand side as dv / V – Vt = - b / m T.*0441

*We have got all of V on one side and their variable in terms of T on the right hand side.*0455

*What we are going to do is we are going to integrate both sides.*0461

*As I integrate the left hand side I am going to integrate dv / V – VT.*0465

*My variable of integration is that velocity.*0472

*I am going to integrate from some initial velocity v= 0 to some final velocity V.*0475

*I have to integrate do the same thing to the right hand side so that will be the integral of - b / dt.*0481

*My variable of integration is T so we are going to integrate from T = 0 to some final value t.*0495

*How do I integrate those?*0504

*The left hand side fits the form du / u where if I said that u = v - Vt the differential of u would just be dv.*0506

*In the formula maybe you have learned or you have not got in there yet in the rule of d/u is the natural log of u + that constant of integration.*0523

*Since, we have the limits here we do not have to worry about our constant integration.*0533

*That means that our left hand side is going to become the natural log of u V - Vt evaluated from 0 to V.*0537

*In the right, inside -b / m that is a constant that can come out of the integral sign we are just integrating dt from 0 to T*0549

*That is just going to be T and this becomes - b / MT.*0557

*The left hand side here becomes log of V - VT substituting in V - VT - log of 0 – Vt plugging in 0 for V*0568

*That is - Vt = - b / Mt.*0583

*We are going to use the identity of the difference of two logs is the one of the quotient.*0590

*We can use a log a – log b = log a/b to say that the left hand side is log of V - Vt /- VT.*0598

*In the right hand side still - b / MT.*0613

*A little bit more manipulation to do here.*0619

*Implies that the log of V – Vt /- Vt.*0623

*Let us see if we can rearrange them a little bit and take that negative to the top.*0631

*We can write that as the log of c VT - V / dt = -b/MT which implies that the log of b1- V/VT = -b/MT.*0636

*If I want to get rid of that nasty log I can use this as a power we raised e2 on both sides to state then that 1 - V / VT*0660

*Even the natural log is whether you happen to have there = e ^{b/MT}.*0672

*Alright, next up.*0682

*We are trying to get V all by itself so I could take and rearrange this add V / VT to that side and subtract that from the other and come up with V/ VT = 1 – e ^{b/MT},*0684

*which implies then that V = VT × 1 – e ^{-b/MT}.*0703

*Or we said VT was mg / b so if we wanted to we can put that back in those terms as well that V =MG / b × (1-e ^{-b/MT}).*0716

*Those are equivalent statements.*0734

*We found velocity as a function of time.*0738

*Alright, now knowing VT we can solve for the acceleration and I'm going to give ourselves more room for that.*0743

*If we wanted to do that let us take a look here.*0748

*A = dv dt which would be the derivative of what we just found from our velocity which is VT - VT e ^{-b/MT}.*0753

*I will expand that out so that is going to be equal to the derivative with respect to T of the derivative of the constant is going to be 0.*0766

*We will just have – VT e ^{-b/MT}.*0779

*We can pull our constant out -VT derivative with respect to T of e ^{-b/MT}*0785

*which implies then that a = - VT × derivative e ^{u} is e^{u} du.*0796

*We have our e ^{u} e^{-b/MT} × du is going to be another – b/m*0806

*which implies then that a is going to be equal to - Vt was mg /b replace that there we have - b/m and we still have e ^{-b/MT}*0815

*which implies then that a = we can do some simplifications.*0835

*We got some m there and there.*0838

*B there and there our negative signs.*0840

*I'm just going to get that a = g × e ^{-b/MT}.*0843

*You are going to see forms of the solutions like this for differential equations quite regularly where*0854

*you have a constant times e ^{-b/MT} raised to some power times T.*0860

*Or a constant ×1 – e ^{negative power}.*0866

*It will come up again and again.*0870

*These are all of the form where you have 1 – e ^{-T} / some time constant.*0872

*Here time would be m/b or something times e ^{-T} / time itself.*0879

*It is going to come up here.*0886

*We are going to see it in the electricity and magnetism course.*0887

*When we talk about capacitors and get into inductors that form keeps coming up again and again.*0890

*It is really nice that once you start to get a feel for that form you can almost guess the answer to these problems*0896

*before you go all the way through the math to actually prove it.*0901

*Or look at the initial and final conditions to help draw graphs of these before you actually go solve them.*0906

*Let us take a look at how these would look graphically.*0913

*I'm going to draw a couple graphs here and we are going to draw the acceleration time graph, velocity time graph,*0917

*and there is our position or displacement time graph.*0927

*Let us see what all these are going to look like.*0932

*We will start up here in acceleration time graph.*0951

*We will do a velocity time graph and also a y displacement time graph.*0955

*We said initially as far as acceleration goes to moment Alex left the airplane when velocity was 0 the acceleration of Alex was g.*0966

*We are going to start at a maximum value and over time as Alex goes faster and faster the acceleration*0974

*is going to decrease and decrease until reaching terminal velocity when there is no more acceleration.*0981

*We start at g and we have to decay down here to 0 when it is an exponential decay so we can radiate like that as we approach that asymptote.*0987

*That should be right on the line there.*0996

*As far as velocity goes, we know when we first push Alex out of the plane the velocity is 0*1000

*We know that point and after a long time we know we eventually get to some value of terminal velocity which is mg / b.*1005

*We will write that in there as an asymptote and we will have an exponential increase there following that same basic shape.*1014

*For displacement, displacement begins at 0 and increases as speed increases*1024

*until reaching a constant rate of increase when the velocity reaches VT.*1029

*As far as displacement goes we are going to start at 0.*1035

*Increase and increase until we get to some point where it is just going to be linear as*1038

*we have a constant velocity when we are running it just terminal velocity.*1046

*There would be graphs of acceleration, velocity, and displacement as functions of time that correspond to those calculations we just did.*1051

*Let us take a look and see how this would look in a free response problem.*1063

*You can download the problem yourself if you want 2005 free response 1 from the mechanics exam.*1067

*A link to it there where you can google it and will take a few minutes look it over and give it a try and come back here and let us see how you did.*1073

*2005 mechanics 1 looking at part A, we have a ball that is thrown vertically upward at some initial speed.*1083

*It has some air resistance given by - kv the positive direction we are going to call up.*1092

*Is the magnitude of the acceleration of the ball increases or decrease or remain the same as it moves upward?*1098

*We are calling up the positive y direction and if I were to draw a free body diagram of the ball*1105

*as it moves up we have its weight down and we have that drag force kv down.*1110

*Net force in the y direction is just going to be - mg - kv is equal to MA y which implies that Ay was just going to be -g - kv/M.*1119

*If we look at that as V goes down, as it slows down, as it gets higher and higher, V gets smaller.*1134

*It looks to me like our acceleration, the magnitude of our acceleration in the y direction must decrease based on our formula.*1142

*There is A must decrease.*1152

*Let us take a look at part B.*1157

*Write but do not solve the differential equation for the instantaneous speed of the ball in terms of time as it moves upward.*1162

*From our free body diagram again Ay = -g - kv/ m which implies then since our A is the derivative of velocity with respect to time.*1170

*We can write it as dv dt = - g –kv /m.*1185

*That is a differential equation that would probably give us credit but let us clean it up just a little bit.*1194

*I'm going to write that in a manner that will be a lot more useful that is going to be M dv dt = - mg – kv*1200

*but either one of those you have your different equation down.*1215

*For C, find the terminal speed of the ball as it moves downward.*1219

*For C, looking at terminal speed.*1224

*At terminal velocity we know the net force =0*1227

*which implies then that our free body diagram is going to have something like this kv,*1235

*Let us be careful it is M according to this problem Mg which implies that terminal velocity kv terminal must equal Mg or a terminal velocity Mg / k.*1244

*Let us move on to check out part D.*1269

*For part D, we are asked does it take longer for the ball to rise to its maximum height*1273

*or to fall from its maximum height back to the height from which it was thrown?*1278

*A tricky question.*1283

*Let us see.*1284

*On the way up, friction brings the ball to a stop quickly.*1286

*This helps bring it to a stop more quickly than if they were no fiction.*1300

*On the way, down the friction slows the ball down so it has more time in the air on the way down.*1308

*That means that average velocity on the way up has to be greater than*1323

*the average velocity on the way down because it happen more quickly.*1328

*And because the distance is constant distance traveled on the way up and the way down*1334

*the time to go down is greater than the time to go up because T = d/V.*1341

*I would say then that it takes longer to fall.*1349

*Some sort of explanation like that to go along with your answer.*1360

*Part E, on the graph, sketch a graph of velocity vs. Time for the upward and downward parts of the ball's flight.*1365

*We are going to need a graph here.*1377

*Here is our velocity, there is our time, and we are looking at what happens for the upward and downward parts.*1387

*It starts at some initial velocity V0.*1402

*It is going to cross the axis here to have a velocity of 0 at its highest point.*1407

*It is going to take less time than it does on the rest of the trip because it is going to take longer to fall.*1412

*On the way down, we are going to have some value of terminal velocity.*1418

*We will draw an asymptote in here for our V terminal.*1427

*I would think that our graph would probably look something like this where it is approaching terminal velocity.*1434

*Something like that final velocity.*1446

*Something like that is your approach of final velocity at time Tf.*1451

*I think that covers that one.*1454

*Let us take a look at one more free response problem.*1462

*Let us go to the 2013 APC mechanics exam.*1466

*You can find it at this address or google it.*1470

*It will take a few minutes to look it over and print it out, give it a try, and come back here and hit play.*1474

*We will see how it worked for you.*1478

*In this problem, we have a box of mass M at rest in the constant applied force being applied*1483

*there is a frictional drag force proportional to kv where V is the speed of the box,*1489

*k is some positive constant, and we are given a dot to draw on label the forces.*1495

*Draw our free body diagram actually.*1499

*Let us draw our free body diagram first.*1501

*There is our box.*1504

*We know we have the normal force, we have the weight of the box, the force of gravity.*1507

*We have some applied force which we are calling Fa.*1513

*We must have our frictional force our drag force kv.*1518

*For part B, it asks us to write but do not solve the differential equation that can be used to determine the speed of the box and that sounds familiar.*1528

*We have done that sort of thing.*1536

*Let us take a look net force in the x direction is going to be the applied force - kv assuming we are calling to the right positive.*1539

*All that must be equal to Ma but as a differential equation A = dv dt therefore Fa – kv = M dv dt.*1550

*That will work.*1571

*There is our different equation.*1572

*We have velocity and its derivative in the same equation.*1574

*Moving on to part C, determine the magnitude of the terminal velocity of the box.*1583

*At terminal velocity acceleration is 0, f net is 0 therefore we know that the applied force in the x direction must equal kv in magnitude.*1591

*Therefore, the applied force = kv terminal or solving for V terminal that is just going to be our applied force divided by k.*1606

*Part D, use the differential equation to derive the equation for the speed.*1624

*Alright, we have to do some math and let us give ourselves some room.*1630

*Starting with our equation f - kv = M dv dt, we are going to do the separation of variables again.*1636

*This implies then that dv / f - kv = dt / M.*1646

*Let us see how it would integrate that.*1657

*The integral of the left hand side dv / F - kv must equal the integral of dt / M.*1659

*We are going to integrate here from our velocity V = 0 to some final value V in the right hand side from T = 0 to some final value T.*1668

*If we are going to fit this into the form du / u we would need -dv in the top and we would also need –k.*1678

*You need –k on the top.*1686

*If we are going to put –k on the top to fit that form we have to multiply by -1/k so that we maintain the same value.*1689

*We cannot just arbitrary throw things in there.*1696

*In the right hand side it looks ok to integrate.*1699

*This implies then, that we have -1/k integral of du/U is going to be the natural log of our U which was f-kv evaluated from 0 to V.*1702

*The right hand side is just going to be t/M.*1709

*Expanding out our left hand side we have if we do this log I am going to take a moment and I am going to put our –k over the right hand side.*1726

*If I multiply both sides by –k let us put –k there.*1738

*It can go away and can make this a little bit simpler to see.*1743

*Our left hand side becomes the log of f –k and I plug in V for my variable v - the log of f-0 for our V.*1747

*That is going to be – the log of f = - kt/M.*1758

*Which implies that the difference of the logs is the log of the quotient so we have on the left hand side our log of f-kv/f = -kt/M.*1768

*If I raise both sides to the e, the left hand side becomes f-kv/f = e ^{-k/Mt}.*1784

*A little bit more rearrangement here.*1798

*F – kv / f let us multiply both sides by f to get f –Kv= f e ^{-k/Mt}.*1801

*We will get V all by itself.*1813

*Let us get kv= f-fe ^{-k/Mt}.*1816

*We can factor out that f ÷ k so the velocity is going to be f/k × (1- e ^{-k/Mt}).*1824

*There is part D.*1847

*Finally for part E, on the axis sketch a graph as the speed as a function of time and label the asymptotes things like that.*1851

*We are getting pretty good at graph and these sorts of things by now.*1863

*Let us give it a shot.*1864

*We have V on the y axis, time on our x, and we know it is going to start at some velocity 0.*1877

*We can also plug that in for T in our formula.*1886

*If T is 0, e⁰ is 1.*1890

*1-1 Is 0 so the velocity would be 0.*1892

*We will start here at 0 and as T gets big that whole term goes to 0 so we have f/k as our asymptote.*1895

*Let us mark that here f/k.*1906

*The shape of our graph is something like that.*1913

*Retarding forces and drag forces, air resistance.*1920

*Hopefully that gets you a good start.*1923

*Thank you so much for watching here at www.educator.com.*1924

*I look forward to seeing you soon and make it a great day everyone.*1927

4 answers

Last reply by: Professor Dan Fullerton

Wed Oct 19, 2016 7:41 AM

Post by Brad Greer on October 10, 2016

In many of the downloadable lecture slides, several identical slides in a row appear (e.g. in Mech-10 (Retarding and Drag forces), slide Velocity-as-a-Function-of-Time_332 duplicates slide 747 and slide 916, and all have only a title. Also, in some cases, the slide is unreadable because it is distorted or partially off the page, e.g. slide 748 in this lecture. Is there any way to get better copies of the slides, that contain all the information from the lecture?

1 answer

Last reply by: Professor Dan Fullerton

Thu Aug 4, 2016 1:26 PM

Post by Peter Ke on August 1, 2016

For Example 2, at 28:18 I understand why you put -k next to the dv but how did you get -1/k on the outside of the integral?

1 answer

Last reply by: Yuhuan Ye

Sun Jan 3, 2016 9:58 PM

Post by Yuhuan Ye on January 3, 2016

Hi Mr.Fullerton,

I have a question on "velocity as a function of time", just right before you integrate the equation, where did the "d" in "b/m*dt" go?

1 answer

Last reply by: Professor Dan Fullerton

Wed Oct 15, 2014 2:14 PM

Post by Scott Beck on October 15, 2014

Is the Ap C Physics exam different from Ap physics 1 and 2? Could this course also prepare for the algebra based ap physics exams excluding optics and modern physics?

3 answers

Last reply by: Scott Beck

Wed Oct 15, 2014 1:25 PM

Post by Scott Beck on October 15, 2014

Why isn't the equation for net force in y direction Vo - mg + kv=ma? Since the ball was launched upward, and since the equation would be written Vo - mg -Fdrag but since Fdrag is -kv, to negatives make a positive turning it into Vo-mg+kv=ma?? Why isn't Vo included in the equation from the free body diagram?