Galileo's Acceleration Hypothesis

FMAgent

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!

tiny-tim

Hi FMAgent! Welcome to PF!
How you prove that depends on how much you know …

have you done calculus yet?

FMAgent

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

SHISHKABOB

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 + [itex]\frac{1}{2}[/itex]at²?

FMAgent

Yes, so should I explain that d=vit*1/2at^2 is similar to y=mx+b by comparing the variables?

SHISHKABOB

well, y = mx + b is a linear equation, where x = v₀t + [itex]\frac{1}{2}[/itex]at² is a quadratic equation

FMAgent

But my initial velocity is 0 therefor removing the variable x time completly correct? or at least making it zero.

SHISHKABOB

if we rewrite x = v₀t + [itex]\frac{1}{2}[/itex]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 + [itex]\frac{1}{2}[/itex]ax²

so what happens if b (which is the initial velocity) is zero?

FMAgent

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?

SHISHKABOB

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?

FMAgent

d= 1/2at^2 + 0

SHISHKABOB

right, so we have two equations

d = [itex]\frac{1}{2}[/itex]at²
y = [itex]\frac{1}{2}[/itex]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?

FMAgent

yes thats the relationship I was looking for.

SHISHKABOB

glad I could help :)