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Why only 2 sig. fig.?

  1. May 31, 2013 #1
    I have the following info for a question:

    x0 = 380m
    a = -9.80m/s^2
    t = ?
    v0 = 0

    I use the formula x = x0 +v0t + (1/2)at^2

    When I solve for 't' I get 8.80631

    I figure I would round to 3 sig. figs. and have t = 8.81s, but my book is telling me t = 8.8 which is only 2 sig. figs.

    Why are they reporting only 2 sig. figs. for this question?
  2. jcsd
  3. May 31, 2013 #2


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    Staff: Mentor

    I agree 8.81s looks better. Apparently authors of the book were not treating significant figures too religiously. Please remember they are only an approximated way of dealing with precision, and not a very good one.
  4. May 31, 2013 #3
    Here is the full question in case I missed something:

    Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380m high).

    I assume a = -9.80m^2. The only other thing I see is that the height of the building can be seen as only 2 sig. figs. since the last digit is '0', but I would normally take that as 3 sig. figs. since it is not stated as 'approximate'.

    Any input?
  5. May 31, 2013 #4


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    Staff: Mentor

    You haven't missed anything.
  6. May 31, 2013 #5
    In questions involving the acceleration due to gravity at the Earth's surface the key word "Estimate" usually means "assume a = 10 ms-2". So an estimate would be the square root of 76, or 8.7.

    8.81 would be more appropriate if the question said "Calculate..."
  7. May 31, 2013 #6
    There is an ambiguity when dealing with significant figures and different people/textbooks use different conventions.

    Often 380m indicates two significant figures.
    Including a decimal point after the last digit (380.m) would indicate three.

    Using scientific notation removes the ambiguity
    3.8 is 2 sig figs
    3.80 is 3
    3.8000 is 5

    Many of the introductory physics books I've seen use this convention.
  8. May 31, 2013 #7


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