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The position of a car in a drag race is measured each second, and A the results are tabulated below.$$\begin{array}{|l|c|c|c|c|c|c|}\hline \text { Time } t(\mathrm{s}) & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \text { Position } x(\mathrm{m}) & 0 & 1.7 & 6.2 & 17 & 24 & 40 \\\hline\end{array}$$Assuming the acceleration is approximately constant, plot position versus a quantity that should make the graph a straight line. Fit a line to the data, and from it determine the approximate acceleration.

$$3.153 \mathrm{m} / \mathrm{s}^{2}$$

Physics 101 Mechanics

Chapter 2

Motion in a Straight Line

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

University of Washington

Hope College

McMaster University

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If you plot the position versus the square of the time, you will indeed get the following. You can plot the points versus the, uh on a graph and do a best fit, and you will get a slope of about 1.5765 And if you double that float, you can indeed get the acceleration. So these were the data points, which I have written out here, and I used a computer in my case. You can do this by hand if you know the formulas for best fit, and you can use graph paper and you do get a best fit line with this slope and really this equation and you do get the acceleration as twice the slope and I'll show why it's twice the slope in just a second. But it's a pretty good fit. The points don't scatter too much, and you see how nicely in scrolling up and down. We can just kind of look at this that the points lie really close to the best fit line all the way to zero, and the horizontal axis is time squared in seconds. Units of second squared vertical axis is distance and meters now notice that there is a small why intercept of about 0.365 and that can just be attributed probably to error. There's a small degree of air in the data now. One might wonderful. Why would you have the square of time as the horizontal axis? And why would you have, um, twice the slope as your acceleration? But you have to think about position as a function of acceleration as well as time. And if you right X of T, which is position is a function of time. You have the following. You have position the initial position and let's assume we have zero initial velocity so we don't have that germ. And let's just say that we have an initial position and we have an acceleration term which has the square of the time. If we make t squared are independent variable, then we would have a linear plot. This would then be our slope 1/2 times acceleration. This would be our slope T squared. Would equal are independent. Variable and X not would equal our y intercept. So it's possible that the object was not exactly at the origin when the data was taken or perhaps the measurements were not performed. If the object has a certain length, perhaps the measurements were not performed at a certain point along the objects length. And so therefore the Y intercept is not zero, even though the position is recorded as being at zero. So there's some possible air associated with that, so this could be attributed somewhere to some error related to that, so that could account for the whiner set not being zero. However, it is small. Now when we look at the slope, it's 1.56 or rather 1.5765 So if we take that slope and we write it down right here, 1.5765 and we said it equal to 1/2 a. This is why the acceleration will be twice the slope. We multiply both sides by two and we get an acceleration equal to twice the slope now, and the acceleration then becomes three points 153 Now the units need to be written in terms of the quantities that we plotted. So it's always the vertical access units over the horizontal axis units. So we will right the vertical units is meters divided by the horizontally access units which your second squared. And indeed we do get acceleration units, second squared, go on the bottom. So we do have an acceleration of three 0.15 We can just go to two decimal places if we like, and just a 3.15 meters per second squared. And if you do this by hand, although it would be tedious, I would suggest a calculator or computer. You can use the following equations for the best fit line and those air given by Well, first of all, you need the mean of all your data Points X mean equals the sum over the points where you have in data points. In our case, we have six and you divide this some over in. You have the some or rather the mean of your wide data points the vertical axis. In our case, we would have in equal six. We would identify the vertical axis with X, which is positioned Position X, and we would identify Let me take that bar off. I mean, the bar just means averaging. So this is why bar would be the average over the positions. X would be identified with T squared Times Square, basically and X bar would be the average over the time squares, basically the times where values and again any in is equal to six. To calculate the slope of the best fit line, you would have this complicated quantity, which can be quite tedious to calculate. This is why you want to use a computer or calculator. You take each data point of Times Square, which would be X and subtract the mean from it. But you're doing this in a some and you multiply these by each position minus their mean values and divide by the following quantity, which is each time squared minus their means but squared, their quantity squared. And this some appears divided by this. Some down here in On top of that, you need their Y intercept, which is given by be equals. Why average minus in which we just had up here times their X average. So this is the equation for a line. Rather, that's why intercepted the best fit line, I should say, then the the line itself, I would just be why equals in X plus Capital B This is the equation of the line that you want, But you identify X. If you wanted to do this problem with this method, you would identify X with t squared. And why with position and I am with this the M that you see in this equation of the line is identified here with him. This capital B is identified with The camp will be here, etcetera. So it's a very tedious process, but by hand, that's how you would calculate the best fit. And if you do this by hand, which I did for comparison, you'll find that the slope in is equal to 748 0.583 over 474 0.583 which then equals 1.5765 and that capital B Y intercept is equal 2.365 So now if we look and we compare these with what we obtained using a computer in this graph, we see that they are indeed an agreement with the plot from the computer

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