Distance or Displacement-time graph equation explanation requested

[tsuria]

Homework Statement

Plotted a distance vs time graph for acceleration on an inclined plane (constant acceleration) and got graph equation of y = 10.242x^2 + 25.996x - 10.315
So this means that d = 1/2at^2 + vt - y intercept
Meaning 1/2 acceleration is 10.242 and initial velocity is 25.996 right?
Then I plotted distance vs time squared and got y=18.448x+8.3244 so this means d=1/2at^2 + y intercept right? So the 1/2 acceleration here is 18.448.

Question is why are the two 1/2 accelerations of different values? How do I explain it?

Homework Equations

The Attempt at a Solution

 

[berkeman]

Homework Statement

Plotted a distance vs time graph for acceleration on an inclined plane (constant acceleration) and got graph equation of y = 10.242x^2 + 25.996x - 10.315
So this means that d = 1/2at^2 + vt - y intercept
Meaning 1/2 acceleration is 10.242 and initial velocity is 25.996 right?
Then I plotted distance vs time squared and got y=18.448x+8.3244 so this means d=1/2at^2 + y intercept right? So the 1/2 acceleration here is 18.448.

Question is why are the two 1/2 accelerations of different values? How do I explain it?

Homework Equations

The Attempt at a Solution

Welcome to the PF.

Could you show the full problem statement and any figures that go along with it? That would help a lot in trying to answer your questions. Thanks.

 

[haruspex]

Then I plotted distance vs time squared

So call this new "x" coordinate z=t², while your original x coordinate was t.
What happens if you rewrite your quadratic equation for y in terms of z instead of t?

 

[kuruman]

I concur with the suggestion by @berkeman for future reference, however the particular question you ask can be answered immediately. Presumably you used a fitting program to extract coefficients for a second degree polynomial using two different methods, one assuming a linear term and one assuming no linear term. The fitting algorithm did its best in each case to find the best fit to your data. The algorithm is dumb and just did what you asked it to do, find the best fit to the data given a model. If you force the linear term be zero, then you should not expect it to find the same quadratic coefficient (acceleration) and maintain goodness of fit. Something has to give. It's up to you to interpret the results you got in terms of measurable quantities and a reasonable physical model. So ask yourself, in terms of what you observed in the lab which set of values is more physically meaningful? One glaring difference is not the acceleration, but the position at t = 0 (or x = 0). One model says it's -10.315 (centimeters I presume) and the other +8.32 cm. The difference is huge. You know what you did, which initial position (if any) is correct?

If you wish additional help on this, please post your data and the problem statement as @berkeman suggested.