I know that slope of x vs t2 curve gives a2. How do I prove it?
This is what I did :
x=ut+at22
x=at22 (for u=0)
dxdt2=a2
But! Here I assumed that initial velocity is zero. But what if it is non-zero? I know that slope would still be a2, but how do I prove it?
Answer
Taking what you ask at face value, we have
ddt2x=ddt2(12at2+ut)=ddt2(12at2+u(t2)1/2)=12a+12u(t2)−1/2=12(a+ut)
So the slope asyptotically approaches a/2 as you get far enough away from t=0 that the initial velocity is ignorable. But the slope never is actually equal to a/2. It's illustrative for you to open your favorite plotting program and compare the well-behaved parabolae of x=12at2+ut+x0 as a function of t to the very different shape they assume as a function of t2.
If you're actually fitting data, I advise polynomial regression in x versus t, in which a/2, u, and x0 enter as first-order parameters. What you're doing is linear regression on x versus t2, which is a superior analytical technique if your computational tools are a straightedge and a pencil. Even spreadsheet programs support polynomial regression these days.
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