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Dimensional Analysis of an object

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data
    The speed, v, of an object is given by the equation v=At^3 - Bt, where t refers to time. What are the dimensions of A and B?

    2. Relevant equations
    None, but I know that to find dimensions, we look at the units that make up the variable, like the area of a room is [L^2]. I understand the basic concept, but have trouble applying it.

    3. The attempt at a solution
    v = At^3 - Bt.
    I would think that since velocity is a measure of distance per time, we should get t to divide the distance. Could it be then that A is [L/T^4] and B= [L/T^2]?
    I can only figure out this way to get T as the denominator in a way that could get A and B added up to get [L/T] for the answer...

    Thanks for your help!
  2. jcsd
  3. Jan 28, 2007 #2
    Yes, this looks right because you know velocity should have dimensions of displacement/unit time. So you need a displacement in each of the coefficients, and also units of time that gives an inverse proportion in the coefficients.

    So, even though this can be done mentally, for other problems you would start with your original proportion and work towards making into the dimension you want; like so:

    [tex] A*t^3 = \frac{L}{t}[/tex]

    Now divide both sides by t^3.

    [tex] A = \frac{L}{t^4} [/tex]

    Or if you wanted to convert from km/h to m/s, you know that 1000m=1km and 60s=1m and 60m=1h, so:

    [tex]\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} = \frac{1}{60^2} \frac{km}{s} [/tex]

    So you see that the hours cancel and make minutes, then the minutes cancel and make seconds in the denominator. Now take care of the km units:

    [tex]\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} * \frac{1000m}{1km} = \frac{1000}{60^2} \frac{m}{s} [/tex]

    Just write out your proportions and see that the units cancel to get what you want.
    Last edited: Jan 28, 2007
  4. Jan 28, 2007 #3
    Thanks a lot!
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