Question about sig figs and rounding: How to handle when there are 2 parts to the problem?

[Arnav]

Hey I just started taking a physics course and I am a bit confused about sig figs and rounding.

My question is that if there are two parts to a physics problem and the latter part requires an answer from the first part, then do you use the sig fig/ rounded value or the calculated value.

For example, you're solving for distance and you know initial velocity, final velocity and time. You use the formula
a = (vf - vi)/t

And then you use the formula
d = vi*t + 1/2 * a * t^2

When you substitute a (acceleration) in the second formula, would you use the sig fig/rounded value or the value that the calculator gives.

 

[pixel]

Best approach is to use the not-rounded-off acceleration value from the first equation that's in the calculator when evaluating the second equation. This is the same as just substituting the first equation into the second one and using just vi, vf and t to calculate d.

It's not too bad with just two equations, but if you had a series of equations, each of which carried over a result from the previous one and they were rounded off, the errors would compound. It all depends on how accurate the final result has to be.

 

[fresh_42]

In principle errors during a longer calculation add up, so it's always best to round at the end of a calculation. However, if your initial data are e.g. in hours and kilometers, it won't make much sense to carry seconds and meters through a calculation. A result can't be more accurate as the initial data. In my opinion it is good to be as precise as possible from the start and adjust only the final result. This might get you into trouble if the expected solution (at school) differs from your result, but hopefully not decisive. If in doubt, ask the teacher. In a research environment you need to keep track of the error margins in any case. It should be (an important) part of the experimental setup.

 

[Agent Smith]

I was never at a level in science where sig figs mattered. I understood it to mean the number of digits a measurement was 100% certain about. So $28$ cm implies that both the $2$ and $8$ are digits the ruler is certain of having measured accurately, hence $2$ sig figs.

 

[Steve4Physics]

I was never at a level in science where sig figs mattered. I understood it to mean the number of digits a measurement was 100% certain about. So $28$ cm implies that both the $2$ and $8$ are digits the ruler is certain of having measured accurately, hence $2$ sig figs.

You can think in terms of ranges.
27cm means the length is somewhere between 26.5 cm and 27.4999... cm.
28cm means the length is somewhere between 27.5 cm and 28.4999... cm.
29cm means the length is somewhere between 28.5 cm and 29.4999... cm.
etc.

Edit. Very old significant figures joke:
Man to museum guide: How old is that dinosaur skeleton?
Guide: 280,000,003 years old.
Man: How do you know to the nearest year?
Guide: I was here when they installed it 3 years ago. And they told me it was 280,000,000 years old.

 

[Agent Smith]

@Steve4Physics , gracias for the quick lesson on sig figs. So the last digit has been rounded, didn't know that, A measurement of $28$ cm means sure, $20$, but the $8$ could be anything between $7.5$ and $8.5$.

There are other questions on sig figs. One was about how many sig figs in $3000$ km (km is a unit that I'm attaching). Common sense informs that the distance can't be exactly $3000$ km and even if it is no measuring instrument can achieve this kind of $100$% accuracy. So the actual distance must be within a range e.g. $2999.5$ to $3000.5$ or $2500$ to $3500$ and so on. The number of sig figs then would depend on which place value was rounded. In the first example we have $4$ sig figs (all 4 digits before the decimal point) i.e. the nearest km. and in the second case we have $1$ sig fig, to the nearest $1000$ km. Is this correct?

 

[Steve4Physics]

So the last digit has been rounded, didn't know that, A measurement of $28$ cm means sure, $20$, but the $8$ could be anything between $7.5$ and $8.5$.

Yes. If we call the length in cm "$L$" and if you are familiar with mathematical notation, we could write $27.5 \le L \lt 28.5$.

There are other questions on sig figs. One was about how many sig figs in $3000$ km (km is a unit that I'm attaching). Common sense informs that the distance can't be exactly $3000$ km and even if it is no measuring instrument can achieve this kind of $100$% accuracy. So the actual distance must be within a range e.g. $2999.5$ to $3000.5$ or $2500$ to $3500$ and so on. The number of sig figs then would depend on which place value was rounded. In the first example we have $4$ sig figs (all 4 digits before the decimal point) i.e. the nearest km. and in the second case we have $1$ sig fig, to the nearest $1000$ km. Is this correct?

You are describing an ambiguous situation. (And maybe "nearest $1000$ km" should be "nearest $500$ km".)

$3000$ km has only 1 sig. fig. because, for values with no decimal point, all trailing (and any leading) zeroes are ignored when counting sig. figs. For example $5070200$ has 5 sig. figs.

For $3000$ km, the implied range is $2500$ km to $3499. \dot 9$ km. But for convenience we usually say $2500$ km to $3500$ km.

Of course, you might be in a situation where you actually know the distance more accurately than the nearest $500$ km. For example, suppose you know the distance to an accuracy of plus or minus $50$ km. In this case you need to supply extra information if you want to convey information about the accuracy.

You could use 'standard form' (sometimes called scientific notation) and write the value as $3.0 \times 10^3$ km. Since $3.0$ has a decimal point, it has 2 sig. figs. and its implied range is $2.95$ to $3.04\dot 9$.

Or, even clearer, you could be explicit and write the value as $(3000 \pm 50)$ km. This is a good way to convey accuracy information as it is unamiguous.

 

[Agent Smith]

3000 km has only 1 sig. fig. because, for values with no decimal point, all trailing (and any leading) zeroes are ignored when counting sig. figs. For example 5070200 has 5 sig. figs.

How do I report a distance L (in m) such that $5070199.5 < L < 5070200.5$ if the precision is to the nearest meter. As $5070200$ meters or $5070200.0$ meters?

Does that mean to say the distance is $5070200$ meters (with $5$ sig figs), it actually means that $50700150 < \text{Distance (5070200)} < 5070250$

 

[Steve4Physics]

How do I report a distance L (in m) such that $5070199.5 < L < 5070200.5$ if the precision is to the nearest meter. As $5070200$ meters or $5070200.0$ meters?

Does that mean to say the distance is $5070200$ meters (with $5$ sig figs), it actually means that $50700150 < \text{Distance (5070200)} < 5070250$

You have a typo': '$50700150$' should be '$5070150$'. Andd the correct spelling is 'metre', not 'meter'.

Using $L = 5070200$ m implies that $5070150 ~\text m \le L \lt 5070250 ~\text m$.

If $L = 5070200$ m but you actually know it with a precision of $\pm 0.5$ m and you want to convey this information, use $(5.0702000 \pm 0.0000005) \times 10^6$ m or $(5070200.0 \pm 0.5)$ m.

 

[Frabjous]

Andd the correct spelling is 'metre', not 'meter'.

British vs American English

 

[Vanadium 50]

Lets be clear - "sig figs" are not the same as a rigorous error analysis. They are there to keep people from spouting obvious nonsense ("The distance from Chicago to San Francisco is 1846.00000001 miles")

 

[gmax137]

Andd the correct spelling is 'metre', not 'meter'.

British vs American English

I didn't know the British spelling is "Andd."

 

[jbriggs444]

I didn't know the British spelling is "Andd."

Andd was the navigator. He was all right. Buddy went to pieces. It was awful how he came unglued...

 

[gmax137]

Andd was the navigator. He was all right. Buddy went to pieces. It was awful how he came unglued...

wow, that's going back... wayyy back...