Comparing Runner Speeds in 1 km Races with Decimal Precision

[Beer w/Straw]

In 1 km races, runner 1 on track 1 (with time 2 min, 27.95
s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s).
However, length L2 of track 2 might be slightly greater than length
L1 of track 1. How large can L2 - L1 be for us still to conclude that
runner 1 is faster?

\frac{1000 m}{147.95 s}=\frac{L m}{148.15 s}

I get 1.35 the textbook gives an answer 1.4

I would have posted this in the homework section, but things have changed here and I haven't posted for years.

 

[jbriggs444]

The lap times are given to a precision of .01 seconds. The delta in lap time is .30 seconds. Two significant figures. One way to approach the problem (not the approach that you took) would have used the delta time and made this more obviously important. Two sig figs in the inputs, two sig figs in the outputs.

The distance that is computed by taking 1000m times 148.15 divided by 147.95 (the approach that you did take) results in an answer with 5 significant figures. To 5 significant figures, that result is 1001.4. Subtract 1000 and you get 1.4.

Remember that when adding or subtracting, it is the least precise decimal place that controls, not the number of significant figures.

[insert some handwaving here about the precision of the 1000 meter figure being unimportant as long as it is known to at least 2 significant figures]

 

[Beer w/Straw]

The way I did it got me 16 significant figures 1 001.351 808 043 258 I dropped the one thousandth and since the information was given to a precision of two decimals, I kept the last two.

Why is that a wrong answer?

:EDIT:

I tried posting this in the other subforum, but was having problems with the tex and and it wasn't user friendly.

 

[jbriggs444]

That's not the way significant figures work.

When you multiply or divide a set of numbers, the rule of thumb is that the number of significant figures in the result is taken from the number of significant figures in the inputs (whichever input has the least).

12.3 * 0.045678 = .562 (three digits)

When you add or subtract a set of numbers, the rule of thumb is that the decimal position of the last significant figure is taken from the decimal position of the last significant figure in the inputs (whichever input has it leftmost).

123.45 + 0.123 = 123.57 (the one hundredth's place).

In the case at hand you have both multiplication/division and you have addition/subtraction. So you do the operations in sequence. First the multiplication/division part which leaves you with five signficant figures. Then the addition/subtraction part. The last significant figure in the result is in the tenth's place. So the result of the addition/subtraction part is precise only to that digit.

You should keep all the computed digits in your intermediate results (good job) and keep track of which ones are significant so that you can round off at the end.

 

[Beer w/Straw]

...I guess the textbook wants me to use a less precise formula and I'm wrong because I'm doing it a different way.

 

[Khashishi]

No, that's not the right attitude to take here. jbriggs has given the right answer and it has nothing to do with the book being less precise. The book is as precise as the data allows.

 

[Beer w/Straw]

jbriggs444 said the precision was to two decimals .01 seconds. The answer is to .1 seconds.

What am I missing?I did change 1km to 1000m . Isn't it only 1 significant digit for the whole thing by the 1km?

 

[mfb]

The answer does not give seconds, it gives meters.
1.35 is not wrong, but the last 5 is not a true significant figure any more - it could equally be 6 or 4 (check that!) without a change in the problem statement, so we cannot trust this digit.

Another way to estimate the accurary: how far do they go in .01 seconds? It does not make sense to give an answer that is more precise than that.

 

[jbriggs444]

I did change 1km to 1000m . Isn't it only 1 significant digit for the whole thing by the 1km?

Following the rules of significant digits, it would be 1 significant digit, yes. But given the context, it is also a race track whose length is roughly the same as another race track to within something in the neighborhood of 1.4 meters! So we can justifiably consider it to be accurate to four significant figures.

 

[Beer w/Straw]

And the text answer, in more than one edition, is only two figures with one decimal. While it explicitly gives a precision to two decimals and ambiguously gives the the number of significant figures in the question, -but gives no hint that it is only two significant figures.

{Solve}\left[\frac{1000 m}{147.95 s}=\frac{L m}{148.15 s},L\right]

L\to 1001.35 Is the first answer I can get. (I only checked further to see if I should round the thousandth 1001.351808043258

I was thinking about faxing them (I can't find an email) and I think I will. People may think less of me, but it gives me something to do.

:EDIT:

My tex shows up in preview but not on post

 

[jbriggs444]

How many significant digits do you count in 1001.35? How many can be justified based on the input?

 

[Beer w/Straw]

Don't you keep more significant digits during the process and then round off at the end?

1.4 is two, 1.35 is three are they not?

How many significant digits is "2 min 27.95 s"?

 

[jbriggs444]

How many significant digits do you count in 1001.35? How many can be justified based on the input?

 

[Beer w/Straw]

Isn't the significant digits a Red Herring in this case and the decimal precision is what matters?

If I do it differently, and fudge the math by rounding immediately, I could end up with answers of 0 or 1.5. However, the same approach without fudging the math gives me 1.35

 

[dean barry]

Constant speed is assumed (no acceleration at the start)
You can calculate the track length that gives runner two the identical velocity.
Start by calculating runner 1 velocity :
= 1000 / 147.95 = 6.7590420 m/s
Now calculate the track length required by runner 2 traveling at an identical speed :
= 6.7590420 * 148.15
= 1001.3518 metres

 

[jbriggs444]

Isn't the significant digits a Red Herring in this case and the decimal precision is what matters?

It is not clear what you mean by "decimal precision". Possibly you mean that since there are two places to the right of the decimal point in the inputs, there should be two decimal places to the right of the decimal point in the outputs. If so, that's flatly wrong.

Take 1234.5 times 0.0000054321. The "decimal precision" is at the 0.1 level. Rounding based on this gives, a result of 0.0. Using significant digits, the correctly rounded result is 0.0067059.

A proper error analysis would examine how uncertainty in the inputs propagates to uncertainty in the computed result. For instance, one might write down the complete formula, take its partial derivative with respect to each of the inputs and multiply each partial by the corresponding uncertainty. Keeping track of significant digits is a short cut rule of thumb approach that can produce roughly the same computed uncertainty without as much work.

 

[Beer w/Straw]

And the number of significant figures for the question are one, four, or five. To two decimal places.

The text gives an answer of only two figures to one decimal place.

Is the text telling me I shouldn't bother computation and round immediately to have only one or two decimal positions? Or should I compute a certain amount of figures and then round at the end to something that gives less precision than the input?

How many significant figures should the answer to this yield 1000/147.95?

Should I round to two decimal places?

 

[jbriggs444]

You have not answered my leading questions. I will answer yours. Consider returning the favor.

The number of significant figures in "1000" is ambiguous. It can reasonably be taken as having one significant figure. It can reasonably be taken as having four. Without any further context one would reasonably say that 1000/147.95 has one significant figure.

However... In post #16 above I mentioned an approach using partial derivatives. Write down the complete formula for the result you will report in terms of the inputs you have:

$r$ is the computed result -- the number of meters longer track 2 needs to be than track 1
$l$ is length of the first track
$t_1$ is the first runner's time
$t_2$ is the second runner's time

$r = \frac{l t_2}{t_1} - l$

The partial derivative of r with respect to l is $\frac{t_2}{t_1} - 1$ which is about 0.0014. An error of 1 meter in l results in an error of 1 millimeter in r. In other words, the number of significant digits in 1000 is not that signficant. In post #2 above I mentioned a handwave. This is what makes that handwave rigorous.

No. You should not round to two decimal places. You should pay attention to significant figures. Treat $l$ as having unlimited precision. $\frac{l t_2}{t_1}$ has five significant figures because $t_1$ and $t_2$ have five significant figures and the formula consists entirely of multiplications and divisions. That result is 1001.3518... The rightmost significant digit is the in the 0.1's place.

Now again treat $l$ as having unlimited precision. As above, the $\frac{l t_2}{t_1}$ has its rightmost significant digit in the 0.1's place. $\frac{l t_2}{t_1} - l$ is a formula involving addition and or subtraction of two terms, one of which is only precise to the 0.1's place. So that's where the result has its rightmost significant digit.

So the result is 1001.3518 - 1000 = 1.3518 rounded in the 0.1's digit to 1.4

 

[Beer w/Straw]

You mean I should round before completing the entire calculation. Concerning myself more with the method rather than an answer.

 

[jbriggs444]

No. You should not round before completing the calculation. But you should keep track of the digit position that you would have rounded to. That way you know the precision of the intermediate results.

 

[Beer w/Straw]

I'll see if that is consistent but I can't verify it right now as I am busy elsewhere.

Thanks.