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Inconsistency in a Velocity-Time Graph

  • #1
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Homework Statement


We're supposed to convert a position-time graph to a velocity-time graph to an acceleration time graph.

These are the values for the position-time graph:
Screenshot (294).png


Homework Equations


m = y2-y1/x2-x1

The Attempt at a Solution



I found the slopes for each time which were 5, 8.5, 17, 22, 32, 40. These would be the velocities, but they aren't rising by a constant amount like they should be when you convert a curve from a position-time graph to a straight diagonal line in a velocity-time graph. When I find the slopes of these velocities, I also don't get the same slope every time like you would get graphing an acceleration-time graph.

Any help to point me in the right direction would be appreciated. Thank you.
 

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  • #2
haruspex
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they aren't rising by a constant amount like they should
Taking the change in position and dividing by elapsed time does not give the velocity at the initial or final position, just the average velocity over the interval, which will be the actual velocity at some unknown point in between.
Instead of taking slopes from a graph, try writing a generic equation for position as a function of time (under constant acceleration) and using the data to find its coefficients.

Edit: On second thoughts, are you sure it is supposed to be constant acceleration? If not, use Chestermiller's approach in post #7.

If it is supposed to be constant acceleration then the best is to write the generic equation with unknowns for acceleration and initial velocity then use standard two variable regression analysis to get the best fit.
 
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  • #3
Ray Vickson
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Homework Statement


We're supposed to convert a position-time graph to a velocity-time graph to an acceleration time graph.

These are the values for the position-time graph:
View attachment 230158

Homework Equations


m = y2-y1/x2-x1

The Attempt at a Solution



I found the slopes for each time which were 5, 8.5, 17, 22, 32, 40. These would be the velocities, but they aren't rising by a constant amount like they should be when you convert a curve from a position-time graph to a straight diagonal line in a velocity-time graph. When I find the slopes of these velocities, I also don't get the same slope every time like you would get graphing an acceleration-time graph.

Any help to point me in the right direction would be appreciated. Thank you.
Your equation
$$m = y_2 -\frac{y_1}{x_2} - x_1$$
makes no sense; that IS what you wrote!

If you mean
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
then use parentheses, like this: m = (y2 -y1)/(x2-x1). When parsed according to standard rules for reading mathematical exressions, something like a-b/c means ##a - \frac{b}{c}## while a+b/c+d means ##a + \frac{b}{c} + d##. So, if you wanted ##\frac{a-b}{c}## in the first case you need to break the priority between division and subtraction by inserting parentheses: (a-b)/c. If you wanted ##\frac{a+b}{c+d}## in the second case, again you need to over-ride the priority of division over addition, by using brackets: (a+b)/(c+d).
 
  • #4
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I also don't get the same slope every time like you would get graphing an acceleration-time graph.
Who says the acceleration is constant? I think you are doing everything right. And don’t worry about fitting a function as somebody else suggested. With discrete data and no reason to expect some functional form for the underlying motion, the best you can do are these discrete velocities and accelerations.
 
  • #5
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What Physics course are you enrolled in? Because if it is Physics 1 then it more than likely just wants the average velocity for the time period however if you're in a calculus-based physics course it'll want you to find the equation for the derivative of the curve.
 
  • #6
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What Physics course are you enrolled in? Because if it is Physics 1 then it more than likely just wants the average velocity for the time period however if you're in a calculus-based physics course it'll want you to find the equation for the derivative of the curve.
Look at his velocities, which are correct, calculate the accelerations, and then you tell me what functional form you think you are going to fit to this data. While I’m sure you can fit a 5th order polynomial to the resulting five points, I’m pretty certain that isn’t what they are supposed to do.
 
  • #7
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Plot a graph of the velocities at the half-way point of each time interval. For example, velocities at 0.05, 0.15, 0.25, 0.35, 0.45, and 0.55 are 5, 8.5, 17, 22, 32, and 40. These are your best estimates of the velocities at these times. Then do the same thing again with these velocities and times to get the accelerations; the accelerations would apply at the half-way points: 0.1, 0.2, 0.3, 0.4, and 0.5.
 
  • #8
vela
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I found the slopes for each time which were 5, 8.5, 17, 22, 32, 40. These would be the velocities, but they aren't rising by a constant amount like they should be when you convert a curve from a position-time graph to a straight diagonal line in a velocity-time graph. When I find the slopes of these velocities, I also don't get the same slope every time like you would get graphing an acceleration-time graph.
Are these numbers from data you took in a lab? If so, you've discovered that the real world doesn't always match the idealized calculations you learn in class.

Your best bet is to follow haruspex's suggestion and fit a curve to the data.
 
  • #9
haruspex
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Your best bet is to follow haruspex's suggestion and fit a curve to the data.
Not so sure now... please see my edit to post #2.
 
  • #10
andrevdh
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If you plot a position time graph, jon-b you will see that it froms a smooth rising curve.
On the graph you can then construct tangent lines at various points on the curve and then determine the gradient of the tangent lines.
These gradients will then be velocities of the moving object at the times where the tangent lines touches the graphs.
 
  • #11
vela
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Not so sure now... please see my edit to post #2.
A quadratic curve fits the data quite well, or it could be that the OP was expecting constant acceleration based on his model of the system from which the data came. Either way, an initial assumption of constant acceleration is reasonable.

The problem with calculating derivatives from the data is that differentiation amplifies the noise in the data. With so few data points, it's difficult to separate the signal from the noise, especially in the case of the acceleration.
 
  • #12
haruspex
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With so few data points,
I think there are enough if we can assume constant acceleration and use two parameter regression analysis. Otherwise, yes, it is almost meaningless to derive an acceleration independently from each consecutive triple.
 
  • #13
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I would dump the data into Excel and ask it for a parabolic trendline. Looks like this data fits an equation of the form s = ut + 1/2 a t^2.
 
  • #14
BvU
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Second neil. And once you convince yourself the figure is a parabola you could easily also plot ##\sqrt{\rm position}## and fit a straight line through the origin ot obtain a fairly good value for the acceleration.....

The error discussion could still be quite interesting: maybe a 2 or 3 parameter fit (the parabola yes or no through the origin) is more sensible than a 1 parameter fit (straight line through the origin). And taking the square root messes with the weigths of the individual observations.

Most important thing to do is first make an accurate plot and look at it carefully. Draw in estimated horizontal and vertical error bars, etc.
 
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