Physics 1-21-09 Wednesday Daily Homework Statistics 118 Responses Mean: 0.944 Median: 0.96 Do we want to turn the front lights off? This okay? A friend of mine used to visit a psychology class back in the day. This class was so easy that everyone was getting ninety something percent on every homework. My goal is not to do that. If you are over.8 you are doing fine. If the curve is so high, you'll be fine. Anyone having trouble accessing online homework? Student: We have to buy a new access code? I'm afraid you have to buy a new one. Talk to the bookstore about that. Student: They'll take it back. There's a few more days on the free online access. Acceleration Remember last Friday we talked about velocity. We decided that velocity was equal to the change in displacement, x, for unit time t. If we took the average over time this will give us an average velocity. If you take a limit as deltat approaches zero, you end up with the calculus limit which is the instantaneous velocity. Acceleration has the same relationship to velocity that velocity has to position. Acceleration is to change is velocity over time. Acceleration can be negative. If it is, the velocity number is going down. You can chain these together. Student: Do we take the derivative of acceleration? This is the part you need to know. I'm going to wander off into fun stuff here. Quantity Jerk: how fast you are changing the acceleration. Acceleration is the last derivative that starts to make physical sense. You'll see that when we talk about forces.
Let's look at acceleration as a function of time. We're going to use meters per second squared. This acceleration curve is continuous. If you take the average acceleration between two points, you are going to end up with an average value. If you want to figure the instantaneous acceleration, you need to take the tangent curve to a certain point. How are these three related? Let's start with a set of values for acceleration and assumption for initial position of velocity with should be able to derive velocity and position as a function of time. All you know is the change in velocity. Velocity is relative. We have a positive and constant velocity, which gives you a straight line. The acceleration becomes zero. The change in velocity becomes zero. We have a low magnitude, so the slope is downward. Third step- position versus time. Where did we start? We are going to assume the starting position is equal to zero. You don't actually know. You have to choose a given point. We are going to say the initial x value is zero. So point at the origin. If the acceleration is a positive constant value, the velocity starts at zero and ends high. The position versus time curve is not linear. Linear changes in velocity lead to parabolic changes. The slope here should be a straight line. This tangent slope will continue up here. Over time the velocity is going to decrease. It will end up at a positive number. Final step, we have decreasing velocity, we have decreasing slope. Student: Since its not starting at zero, wouldn't it be steeper than that? The slope doesn't matter what happened in here, it only matters what this value is. If this were the same velocity, they should have the same slope. I think you are right, its a little low. We can go the other direction. If you are given a position versus time graph, you can take this the other way. Looking at the velocity changes, you can tell by the parabolic curve we have an increasing velocity. This period of constant slope implies the velocity is constant for that period of time. Here the shape is concave downward. A linear change in velocity. Anytime you see parabolas opened upward, a positive second derivative, or a curve that opens downward, you are looking at negative acceleration. This is an important point. At the end, the curve is opened upward, implying positive acceleration. Acceleration versus time is constant, or is a stepwise function. Who has seen this already? Who has not? Let's think about the graphs more concretely- what this means if you have a particle or a physical thing moving as a function of time.
Motion Diagrams Imagine you are in a laboratory. You have a single toy car that you are moving as a function of time. You are looking at this from the side. Here is the car. Its a bug- that makes it easy to draw. AS you move the car in time, it is moving to the right. You are taking a picture at constant intervals. A camera is taking thirty frames per second, you are looking at every tenth frame. We're looking at position as a function of time. You have to infer the time. You are seeing just a picture. It is critical that this be a strobe light at regular intervals or take a picture. The time gap has to be the same in every case. What is the velocity versus time? They are equal distances apart. The distance between them is displacement, Δx. The distance between these is the velocity. Here the velocity over time is changing. The difference between the size of the arrows as a function of time, is positive and nonzero. We are looking at acceleration on the bottom here. The figure the acceleration, you need to figure out the velocity, and then in order to figure the acceleration, figure the difference between each velocity element. The difference between each velocity is zero. This is an uninteresting acceleration curve, this first one. What do we do to figure out what is going on here? Look at the distance between the points, that's the displacement Δx. The time between them is all the same. The difference between those is an acceleration of a large magnitude in the opposite direction. If you were in this car, as a function of time, you would be slowing down relative to the earth. Imagine you are on an LA freeway, its a nightmare. Everyone is moving relative to you at the velocity of about zero. If someone slams on the brakes, they accelerate in a negative direction relative to you. When you imply deceleration, it assumes what you are measuring everything relative to. In this class we aren't going to talk about deceleration. It implies what you are measuring things to. This could be a rocket out in space: There is no solid earth to measure relative to. Both of these are accelerations. Student: We can't see it over here, because of that wall. I'll try to write larger. Who has looked at the class website lately? I am starting to post the notes taken by our note taker here. If you miss a day or can't keep up, take a look at the website. These are unofficial notes that you may or may not find useful as we go through homeworks. We are going to discuss the problem of how the equations behave if you are talking about a specific case where acceleration is always constant.
If the acceleration is constant, you would expect the velocity versus time curve to be a line. The equation for a line is y=mx+b. If we are talking about velocity and its a line, velocity equals: We have a positive acceleration. V 0 is the velocity at which this crosses the y axis. What is the equation for acceleration as a function of time? What does the position look like as a function of time? Its parabolic, and it opens upright because the acceleration is positive. We are going to use the equation for velocity average. In the case where the acceleration is constant, if you want to know the average velocity, you take the mean, the sum divided by two and look in the middle. This is the definition of average velocity: Vf +Vi over 2. We'll plug this into the equation. Recall from last Friday that position is equal to: 14 here We plug in this average velocity that exists if the acceleration is constant. Student: isn't there a t in there somewhere? Yes. Thank you. This describes the position of a particle when the acceleration is constant. This is an important equation. If you are looking at position as a function of time, what does this look like? Its parabolic. V 0 t controls where the parabola bottoms out. In the case of constant acceleration, time always looks like an open parabola. Agree? So last few minutes I'm going to talk about one specific case you'll see in the homework for Monday. There is a weekly due homework, there are seven problems. The one day homework is a motion diagram. Tonight is the first recitation section in this room. He's going to talk about how to do these constant acceleration problems. It'll be helpful with the homework. Remember, its due Monday. For constant acceleration due to gravity, objects are falling in the vicinity of earth's surface. Its actually 9.81 meters per seconds squared.
In the case of gravity, if y is the position above the earth's surface as a function of time: v 0 is initial y velocity. On Friday we are going to do problems. Come ready.