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Percentage uncertainty

  1. Feb 4, 2008 #1
    [SOLVED] Percentage uncertainty

    If I measure the time of something to one decimal place, lets say I get 0.6 seconds. Would my absolute uncertainty for this be 0.05 seconds? I have read that it should be 0.01 seconds but that does not make sense to me. Surely my figures form that result could range from 0.55 to 0.65 so that is 0.05 seconds either way?

    My actual question here is measuring the percentage uncertainty. If I measure a distance of 0.150 metres, with an absolute uncertaintly of 0.001 metres and a time of 0.6 with an absolut euncertainty of 0.05, what would be the percentage uncertaintly of the result of the average speed.

    I am measuring time to a tenth of a second, and distance to the nearest millimetre.


    Here is my attempt.

    [tex]speed=\frac{distance}{time}[/tex]

    % uncertainty for the time is [tex]\frac{0.05}{0.6}[/tex]

    % uncertainty for the distance is [tex]\frac{0.001}{0.150}[/tex]

    Then to calculate the percentage uncertainty of the speed calculated I would plug those figures into the formula?


    Any help would be great!!!!
     
    Last edited: Feb 4, 2008
  2. jcsd
  3. Feb 4, 2008 #2
    I can't imagine it is particularily hard, I'm just having trouble with wether what I am doing is correct, it's more a question of confidence in my own ability. I do not know if what I have done is correct.
     
  4. Feb 4, 2008 #3
    I'm going to have to guess, i've done my research on the web and in my book and can't find an answer.
     
  5. Feb 4, 2008 #4

    Hootenanny

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    You are correct.
    Note that you have calculated the fractional uncertainty in your quantities, not the percentage. In general, when performing multiplication or division with quantities that have an associated error you must add them in quadrature. So for your example above let [itex]s,d,t[/itex] be speed, distance & time and [itex]\Delta s, \Delta d, \Delta t[/itex] be their absolute uncertainties respectively. Then;

    [tex]\frac{\Delta S}{S} = \sqrt{\left(\frac{\Delta d}{d}\right)^2 + \left(\frac{\Delta t}{t}\right)^2}[/tex]

    I'm sorry that I don't have time for a more detailed explanation right now since it's late. I'll check back in tomorrow and try and add some more.
     
  6. Feb 4, 2008 #5
    I don't think I can put in words how grateful I am for your help, this is very much last minute stuff as I only got one night to do this work. Thank you.
     
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