# Distance directly proportional to time squared

I am quite confused with the last bit, "acceleration is uniform".

I have the graph with x-axis being distance and y-axis as time squared.

I'll talk about how time squared is directly proportional to distance but how do I link that with "acceleration is uniform"

how do I know if it is uniform, do I have to do any calculations? or does the line from the graph represents that? or is it assumed that it is already constant during the experiment?

We were only told to square the time and make a graph for it with distance and btw the title is Galileo's hypothesis experiment • Aren't you familiar with the formula $s=ut+\frac12 at^2$? – sammy gerbil Mar 25 '18 at 0:28
• sammy gerbil Modern teaching rejects lecture and exposition. Students are expected to discover all of mathematics and physics through group cooperative activities. Yes. It is true, and explains a lot. – C. Towne Springer Mar 25 '18 at 3:16

You should be familiar with the formula $s=ut+\frac12 at^2$ which is one of the SUVAT equations for uniform motion which relate distance $s$, initial and final velocities $u$ and $v$, acceleration $a$ and time $t$.
If $s=0$ at $t=0$ then $u=0$. If $a$ is constant then $s=\frac12 at^2$ so plotting $s$ vs $t^2$ should give you a straight line of the form $y=mx$ going through the origin. The slope is $m=\frac12 a$.
Possibly there is uncertainty in your data points, so that they are spread randomly around the line $y=mx$.
More likely is that $s \ne 0$ at $t=0$. Then the relationship between $s$ and $t^2$ is more complicated. The easiest option is probably to write the equation as
$s=\frac12 a(t^2+2\frac{u}{a}t+\frac{u^2}{a^2})-\frac{u^2}{2a}=\frac12 a(t+\frac{u}{a})^2-\frac{u^2}{2a}$
which is of the straight-line form $y=mx+c$ if $x=(t+\frac{u}{a})^2$.
To obtain a straight line you need to add or subtract an amount $t_0$ to your values of $t$ and adjust this value until the $R^2$ value of your trend line is as close as possible to 1. This is quite easy to do in a spreadsheet. The amount $t_0$ which maximises $R^2$ gives you a value for $t_0=\frac{u}{a}$. As a check, the y-intercept on your graph should be $c=-\frac{u^2}{2a}=-mt_0^2$ where $m=\frac12 a$ is the slope of your graph.