Asked by Madasan
Suppose you have a set of data points for x and t. Using the below formula you need to graph a STRAIGHT LINE.
x = 1/2 a t^2
Now, I know I would graph x versus t, but do I graph:
-x versus t
-x versus the root of t
-x versus t^2
-or possibly x^2 versus t???
Soo confusing!
Please help!
Thank you so so so much!
x = 1/2 a t^2
Now, I know I would graph x versus t, but do I graph:
-x versus t
-x versus the root of t
-x versus t^2
-or possibly x^2 versus t???
Soo confusing!
Please help!
Thank you so so so much!
Answers
Answered by
Madasan
and please explain why so I understand how to do this!
thanks again!
thanks again!
Answered by
bobpursley
well, it makes no sense.
I assume you have a set of data points. If so, plot them. Is it a straight line, or a second degree curve? If it is a straight line, then use y=mt+b and find m, b from your data plot. If it is a parabola, use y=at^2+ b, and find a,b to match the points.
I assume you have a set of data points. If so, plot them. Is it a straight line, or a second degree curve? If it is a straight line, then use y=mt+b and find m, b from your data plot. If it is a parabola, use y=at^2+ b, and find a,b to match the points.
Answered by
Damon
Graph x versus t^2
Answered by
Damon
Here is the answer I gave to the same question yesterday:
============================
I would graph x versus t^2 but is that correct?
Thank you very much!
* Math/Calculus - Damon, Friday, July 2, 2010 at 8:36pm
Sure, that will work.
You could also use a sheet of graph paper with log scales.
log x = log [.5 a t^2] = log(.5 a) + 2 log t
============================
I would graph x versus t^2 but is that correct?
Thank you very much!
* Math/Calculus - Damon, Friday, July 2, 2010 at 8:36pm
Sure, that will work.
You could also use a sheet of graph paper with log scales.
log x = log [.5 a t^2] = log(.5 a) + 2 log t
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