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Homework Help: Dimensional analysis of this equation

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data
    In the equation below, x and x0 are distances in meters, t is the time in seconds, and v0 is a speed in meters per second. Find the units of a.
    x-x0 = v0(t) + (1/2)a(t^2)
    Find the dimensions (or units) of v0 that will make this equation dimensionally correct.


    3. The attempt at a solution
    So I found that a is acceleration, or m/s^2, but the last question confuses me- possibly I don't understand the wording? When I went through the equation and subbed in the units, I got:

    m-m = (m/s)(s) + (1/2)(m/s^2)(s^2) which comes down to
    m= m + (1/2)m = (some number)m

    So what exactly is the problem with the units or dimensions or whatnot? Or is the point for us to realize there isn't one? Or perhaps I found a's units wrong.

    Thanks in advance for any help.
     
  2. jcsd
  3. Aug 30, 2010 #2

    diazona

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    Re: units

    The last question is "Find the dimensions (or units) of v0 that will make this equation dimensionally correct." They just want you to find the units of v0.

    Why would you think you might have found the units of a wrong? What else could they possibly be?
     
  4. Aug 30, 2010 #3
    Re: units

    But didn't they give us the units of v0? As in meters per second?

    Sorry if this is a stupid question, but I don't understand why they're asking for what I think they gave us.
     
  5. Aug 30, 2010 #4

    diazona

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    Re: units

    Oh yeah, they do... I'm not sure why they do that either. Maybe whoever wrote the question just wasn't paying attention.
     
  6. Aug 30, 2010 #5

    D H

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    Staff Emeritus
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    Re: units

    Yes, they gave you units for v0, but are they the right units?
     
  7. Aug 30, 2010 #6
    Re: units

    Uh, I think so? Otherwise the first term wouldn't be in meters, which would be a problem. Wouldn't it? Ugh.
     
  8. Aug 30, 2010 #7

    diazona

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    Re: units

    Yes, that would be a problem, if the first term (or any term) didn't have the same units as all the other terms.
     
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