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Physics: Position, Velocity, Acceleration cont'd

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  • Type: Video Tutorial
  • Length: 10:15
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 110 MB
  • Posted: 07/01/2009

This lesson is part of the following series:

Physics (147 lessons, $198.00)
Physics: Kinematics (18 lessons, $28.71)
Physics: Investigating One-Dimensional Motion (6 lessons, $8.91)

This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.

Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.

Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.

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Velocity is obviously not the same thing as position, but it is connected. Velocity tells you something about how position is changing with time. Similarly, acceleration is certainly not the same thing as velocity, but they're connected. The connections between position, velocity and acceleration are just so important. And we're going to really be making use of them so often you have to get some kind of a fluency. It's worth doing some kind of graphical exercises just to look and see how they hook up together.
Let me begin with a graph that we've seen before. This is a graph of position versus time for, in fact, a ball which was thrown up in the air. It starts at ground level. As time goes by, it goes higher and higher, comes to a stop, and heads back down again. That's x versus t. And we had written down the mathematical formula . And when we were talking about velocity, we simply took the derivative, dx dt, to find the velocity as a function of time. The derivative of t^2 is 2t, so you get minus ten t. The derivative of 20t is 20. So there is the formula for velocity, dx dt as a function of time. It's a straight line with a negative slope. So let's just put that graph up and take a look at it. And we've talked about the connections between these two graphs, how if you measure the slope of x versus t at any instant in time, it will tell you the velocity.
Let's take one more step. Remember that acceleration is the derivative of velocity with respect to time. So if I just take the slope of this graph; that should tell me about the acceleration graph. So let's set x aside, and let's think about acceleration as a function of time. Here's the picture. The slope of a straight line is a negative number, and it's a constant. The slope is the same here and here and here. It's a negative constant. And what's the value of that constant? You have to remember the velocity was negative ten t plus 20. So the acceleration, which is dv by dt, well what's the derivative of the function negative ten t plus 20? The derivative of negative ten t is just negative ten. The derivative of the number 20 is nothing. You take the derivative of a constant, you get zero. So the answer is just negative ten. And sure enough, that's what I plotted.
So you can kind of think about these graphs, and they connect to one another. It's sort of an interesting exercise to try to go the other way around. If I told you that acceleration was negative ten, could you go from that to this graph? It's a little bit tricky. I would have to tell you one more piece of information. What was v at some moment in time? Because all this tells me is v versus T has to be a straight line with slope negative ten. It doesn't tell me whether it's this line or this one or this one. It could be any line with slope negative ten. So I do have to give you one more piece of information if you want to work backwards.
Let's do a slightly more involved example just to practice with this a little bit. Let's just start up at the top with x versus t. Let's focus our attention on it. Position versus time, and it's some much more complicated motion. First of all we start of with negative x. So we're over here on the left-hand side of the number line, and we're heading even more negative. We're heading toward more and more negative positions. At this point in time, what's happening? We were getting more negative. And now we're about to get more positive. So we must have come to a halt. And then we're moving forward, and you'll notice that this stretch of the curve, suddenly at this time, something has changed. Here the path is curvy. And all of a sudden at t equals to one unit, we're going on a straight-line path. So we're still off in negative x land, but we're moving forward. X is getting less and less and less negative. At this point, we're very close to the origin. And now, once again, something interesting happens at t equals to two units. The interesting thing that happens is that the line goes from straight to curvy again. This line isn't nearly so curvy as it was over here. That's the qualitative description of x.
Now let's just think about what v should be. You could do it graphically just by looking at the slope of the tangent. At first it's a fairly steep negative tangent. That means that our velocity is negative. We're moving to the left, and that's plotted as a negative point here, some big negative velocity. But as time marches by, we're still moving left, but not as quickly. Remember at this point the curve is flat. That means we've reached our most negative point. We've come to a halt. And you'll see that the part where this is flat lines right up with where v equals zero. We're stopped. We're continuing to move. And now we're moving back toward the right. Our velocity has become positive. I've drawn a straight line here. What does that mean? If the velocity is a straight line, this must have been a parabola. That's not so easy to see with your eyes, but--actually I tried to sketch the best parabola that I could, and velocity is just increasing constantly.
Something happened here. Remember? The velocity, all of a sudden, did it go to zero? No, just because you have a constant slope of x versus T doesn't mean you've stopped. It means you have a constant velocity. Delta x over delta T, throughout this straight-line region, is a positive constant. You're moving to the right. You're off in negative x land marching toward the origin to your right. So that's a positive constant velocity. It's tricky business. You've really got to think about the motion, and what you mean by velocity. Then it all makes sense.
What happens here? Well, we're still moving to the right, but the slope of the line is slowly getting less and less and less. In fact, right over here at the far end, it looks like we've almost reached our maximum positive value. Our velocity is still toward the right, but it's getting less and less and less positive. And right where the x curve goes flat, that means x has reached it's maximum value. That's right where v becomes zero. We've come to a halt again. And once again, I was trying to draw a parabola, but a parabola that's not quite so curvy. This slope of v is not quite so steep as it was over here.
Finally, let's go from v of t to a of t. I can almost fit this all in one picture. Forget about x. We're interested in a, . You don't really have to go all the way back up to the top. It's an important point. Many times you will see x versus t, and someone will ask you a question about acceleration. You can't do it in one step. You've got to take two steps. Use x to find v. Take one derivative. Then take another derivative to get a versus t. Given v versus t, it's straightforward. What's the slope of this straight line? It's positive and it's constant, a positive, constant acceleration. So we are accelerating toward the right. Remember we were marching to the left but slowing down? That was our description of the motion. Our velocity was negative, but getting less and less. That's a positive acceleration. Notice that we come to a halt, turn around, but nothing particularly interesting is happening in the velocity world. It's just marching along going from negative to positive. And absolutely nothing interesting happens in the acceleration world at this point. It's the same old positive constant acceleration.
All of a sudden something happens. When interesting things happen in the real world, something interesting is happening in the acceleration world. Boom! Acceleration went from positive constant to zero acceleration. Does that mean we freeze? No, it means we've stopped changing our velocity. We were moving forward. We had a positive velocity. We're still moving forward with a constant zero slope, not changing velocity. That means that x is increasing, but in a nice straight-line fashion.
Finally another interesting thing happens. All of a sudden our acceleration is negative. We were heading toward the right, but our acceleration is negative. You can see that from this curve. Velocity is a straight line. Its slope is negative and constant. And you could kind of make sense of a negative acceleration. Remember we were walking to the right but going more and more slowly, and coming to a halt. If you extend time a little bit further, this velocity curve is just going to continue on down in a nice straight line, which means that x will turn around and you'll start heading left again going more and more and more rapidly.
So it's a nice example. You'll see lots of examples like this as we continue along in physics. It's a tricky business. It really requires some thought. Given just this top graph, x versus t, it's all in there. X versus t tells you everything you need to know, but it's not always obvious. It's not manifest in this graph. It takes some thinking to go to its derivative, and then even more thinking to go to the second derivative. Each of these pictures is nice to draw. They tell you different stories. You think about what's going on. It's the same process, you're just thinking about them in a different way. And sometimes this is a useful language, and very often this is a useful concise language.
Kinematics
Investigating One-Dimensional Motion
Another Look at Position, Velocity, and Acceleration Page [2 of 2]

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