Interpolate value in velocity vs time table

AI Thread Summary
The discussion centers on finding the velocity of a 4kg mass at 3.8 seconds based on a provided velocity vs. time table. Initial attempts to calculate the velocity using the average of surrounding points yielded an incorrect result. The correct approach involves understanding the non-linear nature of the data, leading to a more accurate interpolation method. Suggestions were made to explore Bessel's and LaGrange's interpolation formulas for better results. Ultimately, the expected velocity at 3.8 seconds is approximately 27.1 m/s, highlighting the importance of precise interpolation techniques.
rasen58
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Homework Statement


Table gives the rectilinear motion of a 4kg mass over a 5 s time interval.
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When the time was 3.8 seconds, the velocity of the mass was approximately what

Homework Equations

The Attempt at a Solution


I first just tried to find the average of the velocities at 3.5 and 4.0 seconds, so (23+30)/2 = 26.5 m/s.
But that was wrong as the answer should be 27.1 m/s.

Then, I also saw that it wasn't linear, so I tried modeling the data to an exponential function of form A= Pe^(rt).
And tried plugging in two points to find the constants and then plugged in 3.8 for t, but that didn't really work and gave me different values depending on which two initial points I used.

How should this problem be done?
 
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rasen58 said:
I first just tried to find the average of the velocities at 3.5 and 4.0 seconds, so (23+30)/2 = 26.5 m/s.
That gives you an approximation of the velocity at 3.75 s.

From the table, how much does the velocity increase in the 0.5 s between 3.5 s and 4.0 s? So approximately how much does it increase in the 0.3 s between 3.5 s and 3.8s?
 
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MrAnchovy said:
That gives you an approximation of the velocity at 3.75 s.

From the table, how much does the velocity increase in the 0.5 s between 3.5 s and 4.0 s? So approximately how much does it increase in the 0.3 s between 3.5 s and 3.8s?
Oh, I see now.
So the velocity increased by 7 m/s over .5 s. So to find how much it increased every 0.1 s, I would do 7/5 = 1.4 m/s.
To find the v at 3.8, that is .3 s past 3.5 so 23 + 3 * 1.4 = 27.2 m/s, which is pretty much the answer.
Thanks!
 
Hmmm, if the answer they are looking for is 27.1 m/s I think they may be expecting you to plot the points on graph paper and sketch a curve fitting them.
 
But if you're sketching it by hand, then you probably wouldn't know if it's exactly 27.1
 
Try it, I think you'll find that it is closer to 27.1 than 27.0 or 27.2.
 
rasen58, have you covered any other interpolation methods besides a linear fit? There are several methods of tabular interpolation that effectively take into account the shape of the curve around the point of interest by considering surrounding data points.

If you are interested you might do a bit of research on "Bessel's interpolation formula" and "LaGrange's interpolation formula". Both are fairly straightforward to apply.
 
gneill said:
rasen58, have you covered any other interpolation methods besides a linear fit? There are several methods of tabular interpolation that effectively take into account the shape of the curve around the point of interest by considering surrounding data points.

If you are interested you might do a bit of research on "Bessel's interpolation formula" and "LaGrange's interpolation formula". Both are fairly straightforward to apply.
Thanks I'll look into it.
 
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