homework and exercises - 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?

Answer

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.