Help in constructing a (x-xi)/t vs t graph

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To construct a "(x-xi)/t vs t" graph, it's important to clarify that this represents instantaneous velocity versus time, not average velocity. The discussion emphasizes calculating Δx/Δt, where Δx is the change in position and Δt is the change in time between consecutive measurements. The third column in the provided data represents average velocity, which does not directly relate to acceleration as instantaneous velocity does. The graph should yield a straight line, allowing for the determination of acceleration from its slope. Understanding these distinctions is crucial for accurately completing the assignment.
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Help in constructing a "(x-xi)/t vs t" graph

Homework Statement


I am asked to construct a "(x-xi)/t vs t" graph. from the graph determine th eobject's acceleration. state the equation of the graph and explain how i determined the acceleration.



Homework Equations





The Attempt at a Solution


i was first given 11 times and positions and i had to fill in a position/time column of a table then construct a position vs. time graph first. i don't understand which variables i have to include in the x and xi parts of the second graph
 
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Welcome to PF. This sounds like a good old tickertape experiment!

So you have a table with columns for time, position x, Δx.
I'm putting Δx instead of x-xi in case I am misunderstanding you. My Δx means the difference between the distance at one time and the distance at the previous time (one row up in the table). You then add another column for velocity = Δx/Δt, where Δt is the difference in times between the current row and the previous row.

Finally, you graph the velocity column vs the time column. I trust you remember the operation you perform on the velocity vs time graph to find the acceleration. The graph should be approximately a straight line, so you can find its y-intercept (b) and slope (m) and write its equation y = mx+b.
 


thank you for the quick response.
actually the table that i have does not have a column for deltaX, the third column is for position/time (m/s) so doesn't that column already give me the velocity?
but are you saying for me to construct a (x-xi)/t VS. t graph i have to take the second time, subtract by the first, divide by time itself and have that on the X column of my graph and have the original time as the y-axis for my graph?
but i don't think that's what you suggested to do, you said my velocity column is deltax/deltat where deltat is the difference in times between current and previous rows?
 


I am asked to construct a "(x-xi)/t vs t" graph.
Sorry, I didn't read that carefully the first time round.
It seems to me it should say Δx/Δt vs t or (x-xi)/(t-ti) vs t, which is instantaneous velocity vs time.
It doesn't make sense to do Δx/t because this is not velocity.
I am not sure I understand what the x in (x-xi) is. Is it the distance value in the row below row i?

the third column is for position/time (m/s) so doesn't that column already give me the velocity?
Position/time is an average velocity. It's graph vs time is not related to the acceleration in the direct way that instantaneous velocity is.

The experiment I am thinking of is written up here: http://www.utm.edu/staff/cerkal/Phy201Lab2.html. Scroll half way down the document to see the table for accelerated motion.
There are other ways to do it and perhaps I don't understand what you are supposed to do. Maybe you should post your question again so others will see it unanswered.
 
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