Galileo's Acceleration Hypothesis

1. The problem statement, all variables and given/known data
We rolled a ball down a ramp and timed the time it took to cover a certain distance (15cm, 30cm, 45cm, 60cm, 75cm, 90cm, 105cm, 120cm.) we input the data into a graph (x= time (s^2), y= distance (m)) my partner and I understand that the slope is the 1/2 of the acceleration, but we don't understand how to explain/prove the slope is 1/2 of a.

any help would be greatly appreciated!

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Hi FMAgent! Welcome to PF!
 Quote by FMAgent … my partner and I understand that the slope is the 1/2 of the acceleration, but we don't understand how to explain/prove the slope is 1/2 of a.
How you prove that depends on how much you know
have you done calculus yet?

 No, I haven't done calculus, this is my first actual physics course in highschool.

Galileo's Acceleration Hypothesis

then you're going to want to apply what you have been taught about motion with constant acceleration

for example: do you know the equation x = v0t + $\frac{1}{2}$at2?

 Yes, so should I explain that d=vit*1/2at^2 is similar to y=mx+b by comparing the variables?
 well, y = mx + b is a linear equation, where x = v0t + $\frac{1}{2}$at2 is a quadratic equation

 Quote by SHISHKABOB well, y = mx + b is a linear equation, where x = v0t + $\frac{1}{2}$at2 is a quadratic equation
But my initial velocity is 0 therefor removing the variable x time completly correct? or at least making it zero.

 if we rewrite x = v0t + $\frac{1}{2}$at2 with x as the variable and y as the independent variable, and then the two constants v0 and a as b and a respectively, we get y = bx + $\frac{1}{2}$ax2 so what happens if b (which is the initial velocity) is zero?
 well b is zero, because my Vi is zero, and zero times anything is zero. so if I made distance the y, t^2 the x, then wouldn't 1/2a be my slope? vit being the b but not really nessisary because its zero?
 it's not so much that it's not necessary, it's just that it's zero. And well, what does it look like if we write it like that?
 d= 1/2at^2 + 0
 right, so we have two equations d = $\frac{1}{2}$at2 y = $\frac{1}{2}$ax where y = d, and t2 = x the second equation looks like your graph, right? And you know that the first equation is a fundamental equation of constant acceleration, right? so can you see the relationship?
 yes thats the relationship I was looking for.