Galileo's Acceleration Hypothesis

1. Feb 29, 2012

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!

2. Feb 29, 2012

tiny-tim

Welcome to PF!

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

have you done calculus yet?

3. Feb 29, 2012

FMAgent

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

4. Feb 29, 2012

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 + $\frac{1}{2}$at2?

5. Feb 29, 2012

FMAgent

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

6. Feb 29, 2012

SHISHKABOB

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

7. Feb 29, 2012

FMAgent

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

8. Feb 29, 2012

SHISHKABOB

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?

9. Feb 29, 2012

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?

10. Feb 29, 2012

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?

11. Feb 29, 2012

FMAgent

d= 1/2at^2 + 0

12. Feb 29, 2012

SHISHKABOB

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?

13. Feb 29, 2012

FMAgent

yes thats the relationship I was looking for.

14. Feb 29, 2012