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!
and please explain why so I understand how to do this!
thanks again!
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.
Graph x versus t^2
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
To graph a straight line using the formula x = (1/2)at^2, you need to plot x on the vertical axis (y-axis) and t on the horizontal axis (x-axis).
Because the equation is given as x = (1/2)at^2, it implies that the relationship between x and t is quadratic, not linear. Therefore, the graph of this equation will not be a straight line.
The options you provided (-x versus t, -x versus the root of t, -x versus t^2, and x^2 versus t) do not accurately represent the equation x = (1/2)at^2. None of these options would yield a straight line graph.
To graph a quadratic equation, like the one given, you need to have multiple data points for different values of t in order to plot them on the graph. Then, you can fit a curve through these points that represents the overall trend.
If you specifically want to graph a straight line, you would need a linear equation of the form y = mx + c, where m is the slope and c is the y-intercept. The equation x = (1/2)at^2 does not have this form, so it cannot be graphed as a straight line.