Help finding velocity using data table

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To find the velocity at specific time intervals using the provided data points, the formula v = change in position/change in time is applied. For calculating the velocity at 0.25 seconds, the appropriate method is to use the data points (0,0) and (0.25,0.29), resulting in v = (0.29-0)/(0.25-0). For the interval between 0.25 seconds and 0.50 seconds, the calculation should use (0.25,0.29) and (0.50,1.15), yielding v = (1.15-0.29)/(0.50-0.25). Exact velocities cannot be determined from limited data points, but interpolating between them provides reasonable approximations.
terasnap
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Homework Statement



Using the data found, in this chart, create a time-velocity data table.

Time(s),Position(m)[down]: (0,0), (0.25,0.29), (0.50,1.15)


Homework Equations



v= change in position/change in time

The Attempt at a Solution



I'm just wondering which sets of data points I should be using to calculate the velocity at 0.25 seconds: for example v=(0.25-0)/(0.29-0) is different from v=(0.50-0.25)/(1.15-0.29).
 
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Since the graph is linear between endpoints, select the points and compute the slope of the line, using the velocity equation you indicated and the endpoints of the line. Do the same for the second line, and you obtain the velocities.
 
NasuSama said:
Since the graph is linear between endpoints, select the points and compute the slope of the line, using the velocity equation you indicated and the endpoints of the line. Do the same for the second line, and you obtain the velocities.

Sorry, but I don't really understand how to find the exact velocity at 0.25 seconds, since your method calculates the average velocity between 0s-0.25s.
 
terasnap said:

Homework Statement



Using the data found, in this chart, create a time-velocity data table.

Time(s),Position(m)[down]: (0,0), (0.25,0.29), (0.50,1.15)


Homework Equations



v= change in position/change in time

The Attempt at a Solution



I'm just wondering which sets of data points I should be using to calculate the velocity at 0.25 seconds: for example v=(0.25-0)/(0.29-0) is different from v=(0.50-0.25)/(1.15-0.29).
You would use v=(0.29-0)/(0.25-0) to get your best approximation to the velocity at 0.125, and you would use v=(1.15-0.29)/(0.5-0.25) to get your best approximation to the velocity at 0.375.
 
terasnap said:
Sorry, but I don't really understand how to find the exact velocity at 0.25 seconds, since your method calculates the average velocity between 0s-0.25s.
Hi terasnap, welcome to Physics Forums.

There is no way to calculate the "exact" velocity at any moment, when all you have are a few data points. It is reasonable to assume that speed changes smoothly, so the data points you have should give their best approximation to the speed at those times midway between your data values. Calculate these values of velocity, and you can then use these figures to interpolate intermediate values of velocity.

Chestermiller has shown you the way to begin.
 
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