# Finding acceleration from graphs (s vs t^2)

• Nov 11th 2007, 05:47 PM
SportfreundeKeaneKent
Finding acceleration from graphs (s vs t^2)
This is a bit of a physics question but I think you guys would know it, it's for a lab, anyways:

If I’m to graph a plot of s (displacement) vs t^2 (time squared) then what would be the independent and dependent variable, t or s? I’m supposed to supposed to use this graph to find a slope with error bars and use the appropriate kinematics equation [which I think is s(t) = 1/2 at^2 (the vi*t is 0)]. The slope I will attain from the t^2 versus s graph, what will it represent? And how could I possibly plug it back into the equation to find acceleration? Thanks.

Also when I graph my results, should it be a positive or negative slope if the distance and time are decreasing every time? Because if it's a negative slope than that would mean that the x values for time squared would decrease going from 9.826 to 7.132

This is the table I'll use
Average Displacement (cm)---------->Average Time squared (t²) s
Distance 1 152.90 ± 0.05 cm--------->9.686
Distance 2 143.66 ± 0.05 cm--------->9.321
Distance 3 133.30 ± 0.04 cm--------->8.588
Distance 4 123.56 ± 0.02 cm--------->7.896
Distance 5 113.76 ± 0.02 cm--------->7.132
• Nov 11th 2007, 09:51 PM
topsquark
Quote:

Originally Posted by SportfreundeKeaneKent
This is a bit of a physics question but I think you guys would know it, it's for a lab, anyways:

If I’m to graph a plot of s (displacement) vs t^2 (time squared) then what would be the independent and dependent variable, t or s? I’m supposed to supposed to use this graph to find a slope with error bars and use the appropriate kinematics equation [which I think is s(t) = 1/2 at^2 (the vi*t is 0)]. The slope I will attain from the t^2 versus s graph, what will it represent? And how could I possibly plug it back into the equation to find acceleration? Thanks.

Also when I graph my results, should it be a positive or negative slope if the distance and time are decreasing every time? Because if it's a negative slope than that would mean that the x values for time squared would decrease going from 9.826 to 7.132

This is the table I'll use
Average Displacement (cm)---------->Average Time squared (t²) s
Distance 1 152.90 ± 0.05 cm--------->9.686
Distance 2 143.66 ± 0.05 cm--------->9.321
Distance 3 133.30 ± 0.04 cm--------->8.588
Distance 4 123.56 ± 0.02 cm--------->7.896
Distance 5 113.76 ± 0.02 cm--------->7.132

s will be the dependent variable and $t^2$ will be the independent variable.

So the usual form for a line graph is $y = mx + b$

In your case y = s and $x = t^2$. So your equation should be of the form
$s = mt^2 + b$

and you correctly predicted that the correct Physics equation is
$s = \frac{1}{2}at^2$

So when you graph (ie. do the regression analysis) the slope m will be equal to $\frac{1}{2}a$ and your intercept b should be equal to (ie. close to) 0.

-Dan
• Nov 12th 2007, 02:03 PM
SportfreundeKeaneKent
I was wondering if it would still work if I switched my numbers around so instead of going from 9.896 s, 9.321 s---> 7.132 s, I go from 7.132 s to 9.896 s on the graph for the x axis. So basically, I'd be switching this table going from bottom to top:

152.9---------9.686
143.66--------9.321
133.3---------8.588
123.56--------7.896
113.76--------7.132
• Nov 12th 2007, 08:37 PM
topsquark
Quote:

Originally Posted by SportfreundeKeaneKent
I was wondering if it would still work if I switched my numbers around so instead of going from 9.896 s, 9.321 s---> 7.132 s, I go from 7.132 s to 9.896 s on the graph for the x axis. So basically, I'd be switching this table going from bottom to top:

152.9---------9.686
143.66--------9.321
133.3---------8.588
123.56--------7.896
113.76--------7.132

The order in which you graph the points is immaterial. The graph still looks the same.

-Dan