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I'm a little confused as to when to use significant figures for my physics class. For example, I'm asked to find the average speed of a race car that travels around a circular track with a radius of $500~\mathrm{m}$ in $50~\mathrm{s}$.

Would I need to apply the rules of significant figures to this step of the problem? $$ C = 2\pi (1000~\mathrm{m}) = 6283.19 $$

Or do I just need to apply significant figures at this step? $$ \text{Average speed} = \frac{6283.19~\mathrm{m}}{50~\mathrm{s}} = 125.664~\mathrm{m}/\mathrm{s} $$

Should I round $125.664~\mathrm{m}/\mathrm{s}$ to $130~\mathrm{m}/\mathrm{s}$ since the number with the least amount of significant figures is two?

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    $\begingroup$ mathematical-physics probably isn't the right tag for this. But I don't know what is, mathematics isn't appropriate either. $\endgroup$
    – tpg2114
    Commented Feb 2, 2013 at 0:23
  • $\begingroup$ @tpg2114 I agree, I couldn't find an appropriate tag for this question. $\endgroup$
    – Scott
    Commented Feb 2, 2013 at 0:25
  • $\begingroup$ Related: physics.stackexchange.com/q/32078/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jun 18, 2013 at 18:06

4 Answers 4

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You should always find an answer that is a formula, and then only apply significant figures once you get to the one final step of substituting your numbers back into the problem in place of variables.

Avoid multiple intermediate steps of substituting numbers at all costs. Not only will this save your pencil a lot of work, but it will also cause your answer to be more accurate, as rounding errors can pile up, even when using a calculator.

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  • $\begingroup$ So the correct answer would just be 130 m/s then, right? $\endgroup$
    – Scott
    Commented Feb 2, 2013 at 0:23
  • $\begingroup$ @Scotty Yes, if the 500 and 50 given to you each had at least 2 sig figs. Some conventions are "50" has one sig fig (so your answer should round to 100) and "50." (note the decimal) is more precise. You should check with your instructor/textbook for what the conventions are. $\endgroup$
    – user10851
    Commented Feb 2, 2013 at 0:37
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    $\begingroup$ This is good advice, but sometimes it makes sense to, e.g., write down an intermediate result so that you can check whether it's reasonable, because it's of some interest on its own, or because you're doing calculations in your head or on paper without your calculator handy. (Or for those of us old enough to remember slide rules ...) In this situation, you need to keep enough extra sig figs to avoid accumulating significant rounding errors. If there are 10 arithmetic operations, then we expect that roughly 1 extra sig fig is needed. If 100 operations, 2 extra sig figs, etc. $\endgroup$
    – user4552
    Commented Jun 19, 2013 at 0:50
  • $\begingroup$ The other thing to point out to the OP is the reason for sig fig rules. The reason is to avoid miscommunicating to other people the precision of your result, e.g., in the CIA fact book that gives the population of Nigera to 8 sig figs. If intermediate steps aren't being communicated to anyone else, then their sig figs aren't an issue. $\endgroup$
    – user4552
    Commented Jun 19, 2013 at 0:52
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Keep precision all the way through to the number you report and then truncate accordingly at the end.

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    $\begingroup$ No, this is silly. What if your intermediate result is $\sqrt{2}$? Are you going to write down infinitely many sig figs, or keep infinitely many sig figs in the memory of your calculator? If you're going to record intermediate numerical results, then it makes sense to keep enough extra sig figs (typically 1 or 2) to keep from accumulating significant rounding errors. $\endgroup$
    – user4552
    Commented Jun 19, 2013 at 0:46
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I have been taught and continue to teach that you should record (and use) the answers to your intermediate calculations (and/or conversions) with one extra significant digit beyond the number of significant digits that will be in your final answer.

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  • $\begingroup$ I don't think the question is about intermediate calculations, but about how to express the final result. Right? $\endgroup$ Commented Oct 4, 2018 at 23:53
  • $\begingroup$ What is the reason that you think this rule is correct? Just stating the rule without a reason isn't very helpful. This rule is not correct in some cases. For example, if you have a complicated calculation with hundreds of intermediate numerical steps, then one extra sig fig might be insufficient. $\endgroup$
    – user4552
    Commented Oct 5, 2018 at 1:27
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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/02%3A_Measurement_and_Problem_Solving/2.04%3A_Significant_Figures_in_Calculations

Just revisited this in chemistry this year. You actually have to apply the correct number of significant figures based on the rules of operations (addition/subtraction or multiplication/division) as you perform each calculation in a problem with multiple calculations. The purpose of using significant figures is not to get accurate results or results closest to what a calculator would get. The point of using significant figures is not to mislead the reader or person following your work in thinking that you used exact numbers when in fact you did not. So the more steps you have in a problem, the further your answer will be from the answer a calculator would derive. If you apply the correct number of significant figures in every step, you are accurately accounting for what you actually MEASURED (this is why they are applied to measured numbers, not exact numbers). And if you’re sitting somewhere computing the problem, you did not measure anything. We also do not just chose to use an extra significant figure or two but have to stick to the rules of operation. ie- least amount of significant figures for multiplication and division and lowest decimal place for addition and subtraction.

Hope the link above with consistent explanations and videos helps.

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  • $\begingroup$ So above, the correct work to show would be 6283 (4 sig fig)/ 50s $\endgroup$
    – Kim
    Commented Mar 31, 2021 at 9:27
  • $\begingroup$ So above, the correct work to show would be 6283 (4 sig fig) / 50s (2 sig fig)= 125.66, which becomes 130 ( 2 sig fig). This also turns out to be the same answer even if you don’t use the appropriate significant figures in the first step but that won’t always be the case so it’s best to practice the right way so that you always get the questions right on an exam, if your professor uses the answers from the textbook and doesn’t edit them or formulate problems without using the correct rules. $\endgroup$
    – Kim
    Commented Mar 31, 2021 at 9:33

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