Galileo's Acceleration Hypothesis

[FMAgent]

Homework Statement

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]

Welcome to PF!

Hi FMAgent! Welcome to PF!

… 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? ​

 

[FMAgent]

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

 

[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?

 

[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 + $\frac{1}{2}$at² is a quadratic equation

 

[FMAgent]

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 completely correct? or at least making it zero.

 

[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?

 

[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 = $\frac{1}{2}$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?

 

[FMAgent]

yes that's the relationship I was looking for.

 

[SHISHKABOB]

glad I could help :)