Slope of $x$-$t²$ curve (displacement vs time squared) I know that slope of $x$ vs $t^2$ curve gives $\frac{a}{2}$. How do I prove it? 
This is what I did : 
$x = ut + \frac{at^2}{2}$ 
$x = \frac{at^2}{2}$    (for $u=0$) 
$\frac{dx}{dt^2} = \frac{a}{2}$ 
But! Here I assumed that initial velocity is zero. But what if it is non-zero? I know that slope would still be $\frac{a}{2}$, but how do I prove it? 
 A: Taking what you ask at face value, we have
\begin{align}
\frac{\mathrm d}{\mathrm dt^2}
x
&=
\frac{\mathrm d}{\mathrm dt^2}
\left(
\frac12 at^2 + ut
\right)
\\
&=
\frac{\mathrm d}{\mathrm dt^2}
\left(
\frac12 at^2 + u\left(t^2\right)^{1/2}
\right)
\\ &=
\frac 12 a + \frac12 u \left(t^2\right)^{-1/2}
\\&= \frac12 \left(a + \frac ut\right)
\end{align}
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 $x=\frac12 at^2 + ut + x_0$ as a function of $t$ to the very different shape they assume as a function of $t^2$.
If you're actually fitting data, I advise polynomial regression in $x$ versus $t$, in which $a/2$, $u$, and $x_0$ enter as first-order parameters.
What you're doing is linear regression on $x$ versus $t^2$, which is a superior analytical technique if your computational tools are a straightedge and a pencil.
Even spreadsheet programs support polynomial regression these days.
A: If $u=0$ then a graph of $s$ against $t$ will be a straight line through the origin  of gradient $\frac a 2$.
You are comparing $s=\frac 12a \, t^2 +0 $ with the general equation of a straight line $y=mx+c$.  
If there is an initial velocity then to linerise the graph write the equation as $\frac s t = \frac 12a\,t + u$ and compare it with $y=mx+c$.  
Thus a graph of $\frac s t$ against $t$ should be a straight line of gradient $\frac 12 a$ and intercept on the $\frac st$ axis of $u$.
