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Galileo's Acceleration Hypothesis

  1. Feb 29, 2012 #1
    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. jcsd
  3. Feb 29, 2012 #2

    tiny-tim

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    Welcome to PF!

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

    have you done calculus yet? :wink:
     
  4. Feb 29, 2012 #3
    No, I haven't done calculus, this is my first actual physics course in highschool.
     
  5. Feb 29, 2012 #4
    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?
     
  6. Feb 29, 2012 #5
    Yes, so should I explain that d=vit*1/2at^2 is similar to y=mx+b by comparing the variables?
     
  7. Feb 29, 2012 #6
    well, y = mx + b is a linear equation, where x = v0t + [itex]\frac{1}{2}[/itex]at2 is a quadratic equation
     
  8. Feb 29, 2012 #7
    But my initial velocity is 0 therefor removing the variable x time completly correct? or at least making it zero.
     
  9. Feb 29, 2012 #8
    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?
     
  10. Feb 29, 2012 #9
    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?
     
  11. Feb 29, 2012 #10
    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?
     
  12. Feb 29, 2012 #11
    d= 1/2at^2 + 0
     
  13. Feb 29, 2012 #12
    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?
     
  14. Feb 29, 2012 #13
    yes thats the relationship I was looking for. :smile:
     
  15. Feb 29, 2012 #14
    glad I could help :)
     
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