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Homework Help: How to determine significant figures for quadratic?

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s2.

    For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus?

    2. Relevant equations

    x = ∫vdt

    v = ∫adt

    Quadratic formula

    3. The attempt at a solution


    x2=40.0 + 2*0.170*t2

    The person and bus meet when x1=x2, so to solve for t, I rewrite as a quadratic equal to zero and use the formula:

    5.0t = 40.0 + 0.0850*t2 → 0.0850*t2 - 5.0t + 40.0 = 0

    t = [itex]\frac{5.0 \pm \sqrt{(5.0)^{2} - 4(0.0850)(40.0)}}{2(0.0850)}[/itex]

    So here is where my question actually comes in. How do I solve this using the appropriate number of significant digits? My book says the answers are 9.55 and 49.3, which you get if you calculate the exact answer, and then use 3 significant digits, but why 3 in this case?

    My book gives significant digit rules for addition/subtraction and multiplication/division, but does not say anything about equations using both types of operations. I have thoroughly searched online resources and actually find that most sites out there use slightly different methods, and several I referenced actually contradicted themselves or miscalculated their own examples based on their own rules. It doesn't seem like this should be that confusing.

    If I follow my interpretation of my book's rules, here's what I get:

    1. Because 5.0 has two significant digits, 5.0*5.0=25
    2. Because 4 is an exact number in the formula it is ignored when determining significant digits, and 4*0.0850*40.0 = 13.6, since both 0.0850 and 40.0 have three significant digits.
    3. Because 2 is an exact number in the formula it is ignored when determining significant digits, and 2*0.0850 = 0.170, since 0.0850 has three significant digits.

    t = [itex]\frac{5.0 \pm \sqrt{25-13.6}}{0.170}[/itex]

    4. Because 25 is the most uncertain numbers, 25 - 13.6 = 11

    t = [itex]\frac{5.0 \pm \sqrt{11}}{0.170}[/itex]

    5. Because 11 has two significant digits, sqrt(11) = 3.3

    t = [itex]\frac{5.0 \pm 3.3}{0.170}[/itex]

    6. Because both 5.0 and 3.3 have the same uncertainty, the numerator becomes 8.3 and 1.7.
    7. Because both 8.3 and 1.7 have two significant digits, the two of them divided by the denominator would result in two significant digit answers:

    t = 49 and t = 10

    Now, I know that I'm compounding the rounding, which is leading my answer away from more accurate values, but isn't that the point?

    MANY, many, many sites have told me to perform the equation as a whole first, and then round the final answer, but none of them explain how to determine the number of significant digits to use when doing it that way... I'm a bit disappointed that my book doesn't address how to handle significant figures for equations containing both adding/subtraction and multiplication/division operations. It would appear that if all the variables had the same number of significant digits, then the answer would too, but in this case, one of the variables has 2, and the others have 3.

    Thanks for any help you all can provide! It is greatly appreciated!
  2. jcsd
  3. Apr 24, 2012 #2


    Staff: Mentor

  4. Apr 24, 2012 #3


    User Avatar

    Staff: Mentor

    Those MANY, many, many sites have the right idea :smile: How many significant figures is enough? As many as it takes :smile: But seriously, the easiest thing to do is to keep all intermediate results in machine registers (calculator memories, spreadsheet cells, that sort of thing) and round only for presentation purposes. If you must work by hand and record intermediate results ("Paper Memory"), then get into the habit of recording six decimal places -- it requires no thought and will hold enough accuracy for most lengthy calculations that result in only 1 to 3 significant figures at the end.

    If you want to be more formal about it, recognize that significant figures in the given data imply a certain precision in those values that can be expressed as uncertainties in those values. So a number like 5.2 implies 5.2 +/- 0.05 (so the uncertainty is a "half digit" in the last digit of the given value). You can then perform all calculations using error propagation, which I guarantee will eventually drive you batty and slow your progress to a crawl.

    Instead, use full precision for all intermediate values and learn to use the partial derivative method of finding the error in the result of calculations that are more involved than an addition or multiplication or two.
    Last edited: Apr 24, 2012
  5. Apr 24, 2012 #4
    Thanks a lot gneill! The part about using the partial derivative method really helps, and I bet that will be what I use further down the road; however, at the present, I'm concerned about coming to different answers than my textbook. For the time being, it is instructing me to use the significant figures in the variables to determine how many to use in the quiz answers. I need to understand why I should have came to the conclusion of 9.55 (three significant figures) without using error propagation or partial derivatives (as those techniques have not been introduced), because if I used two or four, I could very well lose points. Are there not significant figure rules for equations involving both addition/subtraction and multiplication/division?

    BTW, as an aside, I'm studying this material on my own in advance of going to school for physics, so I don't have an instructor I can ask yet to see what's acceptable vs. not acceptable when it comes to significant digits.

    Thanks again!
  6. Apr 24, 2012 #5


    User Avatar

    Staff: Mentor

    Without doing a lot of analysis of a particular problem, the general rule of thumb is to give results with as many significant figures as the least number of significant figures in the input values. In this case we note that the speed, 5.0 m/s, has two significant figures so that should be an acceptable number. Yet the problem solution made a point of giving a result with three significant figures for the time (and they specifically stated that they were doing so). The reason may be that later result values are sensitively dependent on the intermediate value of this time, and they'd either have to give you the extra precision here or give you another value with more figures to use as an intermediate value for further calculations.

    I note that a distance value calculation depends upon the square of the time. This will tend to "magnify" any error in t in that calculation. Try plugging in "9.5s", "9.6s", and "9.55s" for time values and see if the significant figures for the distance calculation are affected.
  7. Apr 24, 2012 #6
    For both distance functions, the three values of t you listed result in the same distance number when significant figures are considered: 48.

    Since I began trying to figure this out, I've wondered if I was doing something wrong, or if the book was wrong, or if it doesn't use its own significant figure rules in its answer. I think it's starting to look like one of the later two explanations is the case.

    In addition, jedishrfu's link has a section about the disagreements and imperfection of using significant figures, and it states that they are more of an introductory method that will eventually give way to more reliable techniques such as the error propagation or partial derivative methods you mentioned. So perhaps I shouldn't worry about it much for now and just move on. I just didn't want to move forward without having a firm grasp of the material I've already read, but I think in this case, worst case scenario, I can just ask my instructor what he wants after I start the actual class.

    Thanks for all of your help! You too jedishrfu!
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