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S = ut + 1/2 at^2

  1. Feb 22, 2006 #1
    hi there, i need a little help:

    s = ut + 1/2 at^2

    0.3 = 0 x 0.02 + 0.5 x a x 0.02^2

    0.3= 0.5 x a x 0.02^2


    Thats wher i got a bit confused, what do i do to get a the subject, and what happens to the t^2, bearing in ind that the graph i have to plot at the end involves t^2, and not t

    Thanks, dragon2309
     
  2. jcsd
  3. Feb 22, 2006 #2

    Hootenanny

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    What are you plotting [itex] t^2 [/itex] against what?
     
  4. Feb 22, 2006 #3
    im plotting t^2 against s, which is displacement (or distance)
     
  5. Feb 22, 2006 #4

    Hootenanny

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    And is [itex] t^2 [/itex] on the [itex] y [/itex] axis? If that is the case you need to re-arrange until you get something like [itex] t^2 = k.s + c [/itex].
     
  6. Feb 22, 2006 #5
    You are plotting [itex]t^2[/itex] against s? But in your equation above, the only unknown is a. Do you have a set of values for t and s and have to determine the acceleration from them?
     
  7. Feb 22, 2006 #6
    Yes i do, i have a set of values for s and t, im getting really confused, i just dont know what im supposed to be doing now, and tryig to think about it just threw up more questions.
     
  8. Feb 22, 2006 #7

    Hootenanny

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    I'll help you through it, ignore the numbers for the moment. Like I said before you trying to get something that looks like [itex] t^2 = k.s + c [/itex]. Start with [itex]s = ut + \frac{1}{2} a t^2 [/itex] and see how you can manipulate it. It would be easier however to plot [itex] t [/itex] against [itex] s [/itex].
     
    Last edited: Feb 22, 2006
  9. Feb 22, 2006 #8

    Hootenanny

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    HINT: You can cancel the [itex] ut [/itex] because [itex] u =0 [/itex] [itex] \Rightarrow s = \frac{1}{2} a t^2 [/itex]. Nevermind, it seems [itex] t^2 [/itex] is easier to plot. oops
     
    Last edited: Feb 22, 2006
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