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I have some data for an object moving under constant acceleration.

Velocity $(m/s)$ $[0.84,1.58,2.32,3.06,3.80,4.54,5.28]$

Time $(s)$ $[0,1.5,3.0,4.5,6.0,7.5,9.0]$

The usual approach would be to graph velocity (y-axis) against time (x-axis).

But I am experimenting with plotting velocity (y-axis) against the inverse of time (x-axis).

The reason I want to do this is that the instantaneous slope will give a direct value for displacement.

$v=\frac{d}{t}=d(\frac{1}{t})$

$y=mx$

I have plotted my data, $v$ against $\frac{1}{t}$ and now I want to obtain an equation which describes this graph.

I am thinking it will have the form of $y=\frac{a}{x}$ but I can't model my data that way. Is there another form with another constant or two in there? Like $\frac{a}{x^b}+c$ or something like that? What is the general form of this data such that I can model it on a graph?

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    $\begingroup$ What did you do with the first data point where $t=0$ ? $\endgroup$ Sep 15 '18 at 14:40
  • $\begingroup$ @ClaudeLeibovici Well just for the purpose of plotting I just used a large value to deal with the infinity. $\endgroup$
    – Kantura
    Sep 16 '18 at 7:58
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I cannot figure out what exactly is the difficulty you have.

If one can draw $\quad\begin{cases} x=t \\ y=v=ax+b \end{cases}\quad$ it is obvious that one can draw : $$\quad\begin{cases} X=\frac{1}{t} \\ y=v=\frac{a}{X}+b \end{cases}\quad$$

with $\quad\begin{cases} a\simeq 0.493333 \\b\simeq 0.840\end{cases}\quad$ from numerical regression calculus.

enter image description here

Of course, the point at $t=0$ cannot be drawn on the hyperbolic curve because $X=\infty$. In fact it gives the horizontal asymptote $y=b$.

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  • $\begingroup$ That answered my question fully, thanks. $\endgroup$
    – Kantura
    Sep 18 '18 at 11:54

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