Question about Significant Figures

[Mr Davis 97]

I have a question about sig figs and how they relate to measuring. Say I am finding the velocity of a small object. From an experiment, I gather the data that the object moved .0765 meters in .07 seconds. As one can tell, my watch is less precise than my fancy ruler. Using these data, I calculate the velocity my dividing .0765 m by .07 s. My question is, how many significant figures will the resulting calculation have? I'm sure that it will be 1 significant figure since the time measurement only has one. However, if this is true, would it mean that if I measured the time interval to, say, 1 second, the velocity calculation would have 2 significant figures? Is it true that the longer I measure the time interval, the more precise it will be?

 

[Simon Bridge]

If the time interval was 1.07s then that is 2sig fig, yes, even though the measurement is no more accurate.
sig fig is very rough. If this worries you, then use error propagation instead.
Generally, the longer you measure the better the data... if your stopwatch is accurate to 100th of a second, then it may be 100% wrong if you timed only 0.01s or so, but 1% wrong if you timed 1s.
Per your example, you will find you have more confidence measuring longer times and, thus, slower speeds.

 

[HallsofIvy]

I have a question about sig figs and how they relate to measuring. Say I am finding the velocity of a small object. From an experiment, I gather the data that the object moved .0765 meters in .07 seconds. As one can tell, my watch is less precise than my fancy ruler. Using these data, I calculate the velocity my dividing .0765 m by .07 s. My question is, how many significant figures will the resulting calculation have? I'm sure that it will be 1 significant figure since the time measurement only has one. However, if this is true, would it mean that if I measured the time interval to, say, 1 second, the velocity calculation would have 2 significant figures? Is it true that the longer I measure the time interval, the more precise it will be?

Writing ".07" is a little ambiguous- it is not clear whether that is to have two significant figures or one. Better would be $7.65 \times 10^{-2}$ showing it has three significant figures and either $7 \times 10^{-2}$ or $7.0 \times 10^{-2}$ depending upon how many significant figures you intend. An operation Involving significant figures is no more accurate than the least accurate number.

 

[Simon Bridge]

http://en.wikipedia.org/wiki/Significant_figures#Identifying_significant_figures