What kind of motion does a quadratic position time graph represent? [closed]

Question

When I make a function to represent an object's position at intervals of two seconds, and it is parabolic, for example $f(x)=\dfrac{5x^2}{2}$ with the following points:

$ \text{seconds} \ \ \text{metre} $

$ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 $

$ 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10 $

$ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 40 $

$ 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 90 $

What does this graph represent? I think it has to do with acceleration? I'm not exactly confident here, because it doesn't follow the standard form where you see acceleration written in explicitly. It isn't velocity... because velocity is linear. I've confused myself...

Answer 1

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a. $ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.

And once you know that the acceleration $a(t)$ doesn't actually depend on the time here, you can define the constant a = 5 to rewrite $s(t) = \frac{1}{2}at^2$ or with your variables $f(x) = \frac{1}{2}ax^2$ which should look familiar.

Answer 2

I think you meant to ask what it means that your object's position versus time function is second order.

So to answer in general terms, when your position versus time has a first order relationship $ f(x) = vx + c $ or $ s(t) = vt + c $ where c is the starting position at $t=0$ it means the object is moving at constant speed $v$ away from starting position $c$

When you get a second order equation $ f(x) = ax^2 + vx + c $ or $ s(t) = at^2 + vt + c $ it means that your object is moving at constant acceleration $a$ away from starting point $c$ where the initial speed was $v$.

So in your example the object is moving at constant acceleration of $2.5 \;\text{m/s}^2$ from an initial position of $0\;\text{m}$ and starting speed of $0\;\text{m/s}$

Note that this is only true when you're working with simplified cases where an object is moving in a straight line, when you start working with more dimentions equations get trickier.