Error propagation in calculations

AI Thread Summary
The discussion revolves around error propagation in calculations related to converting time from minutes to hours and its impact on average velocity. The user calculates the conversion and associated errors, confirming that the error remains consistent across different time values despite the percentage error decreasing as time increases. It is clarified that the fixed error in minutes leads to a proportional error in hours, maintaining the same percentage error. The user also explores how to calculate average velocity and its associated error using the rise/run method from a graph. The final average velocity is determined to be 104.5 km/hr with an error of ±1 km/hr.
jakerue
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This is an issue I am running into at the beginning of my physics course.

Homework Statement



Given distance and time in minutes, calculate the time in hours (part of a larger average velocity question) and graph over 170minutes. Include error bars in the graph

Homework Equations



So if tm = 10.0 +/- 0.1min and I use min -> hr conversion as 1hr/60min = 0.0167

th = 10.0min * 0.0167 hrs/min = 0.167 min

The Attempt at a Solution



The error in min is +/- 0.1 min. Now if I am using 0.0167hrs/min as an exact constant factor I should use z=k*x and \Deltaz= k \Deltax

So I will use hours = 0.0167 * minutes and my \Deltahours = 0.0167 * 0.1min making the \Deltahours = 0.00167hours.

By sig fig this value is 0.002 hours correct?

At 10 minutes th = 0.167 +/- 0.002hrs

at 120 minutes th = 2.00 hrs +/- 0.002 hrs

I think I am right but I want to make sure my answer is correct and that this error holds true for all values of minutes from 0-170min.

Thanks for any help, most appreciated.

JakeRue
 
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Looks good, JakeRue. Certainly 10±0.1 minutes = 0.167±0.00167 hr
and it makes sense to round the .00167 to .002.
Very likely your 2 hours is ±.002 as well, though you haven't given any information from which that can be determined. The ±0.1 minute was given in the question for the 10 minute time. Was it also given in the question for the 2 hours? In reality it would depend on how the time was measured. Always a good idea to post the question exactly as given so we can check your interpretation of it.
 
Thanks for the reply Delphi51.

The only errors given in the question are with the minutes (+0.1 min) and with the distance (+0.1 km). The question asks to convert minutes to hours and then graph distance over time.

I then use the km/h to calculate average velocity and best fit line.

I am asking about the 2 hours because the minutes is given in the form of 10.0min and then 120.0 min (3 sig fig, then 4 sig fig). If I calculate the error using only the conversion coefficient of 0.00167 * \Deltatm then the error would stay the same no matter if the sig fig is 3 digits or 4 digits - correct?

What is tripping me up with the error propagation is that with minutes having a + of 6 seconds, or 10% error, the hour then has +0.00167 which is an error of less than 10% of the hour (I could be wrong here).

I don't understand how can I get less error when I multiply by a unit conversion coefficient (1/60)?
 
The error is fixed at 0.1 minute for both 10 minutes and 120 minutes, so certainly the % error will be 12 times less for the 120 minutes. But the size of the error bars will be the same - 0.1 minute for all times.

Changing the units did not change the % error. You had .1 minute per 10 minutes, which is 1%. It converted to .00167 hr per .167 hr, which is also 1%.

How will you calculate the average velocity? You could draw a line of best fit and its slope would be the average velocity. Or just use the last point (total distance and total time).
 
OK I can see the error being at 1% for both minutes and hours. So converting 120.0min +0.1min using the 0.0167 hrs/min conversion coefficient will give me 2.000 hours +0.002hrs.

I was doing velocity by rise/run from two points (does it matter which two I choose since it won't be the actual value?) and getting y-intercept from that.

When I get my slope I think I must multiply the error of both the time in hours (0.1km) and distance in km (0.002) to get a final average velocity.

(209.0 - 17.5km)/(2.00 - 0.167hrs) = 191.5km/1.833 hrs = 104.5km/hr and the error is

∆average velocity =[velocity] [(\Deltatime/time + \Deltadistance/distance)]
∆average velocity = 104.5 [(0.0167/1.833hrs) + (0.1/191.5km)] = 104.5(0.00911+0.000522) = 104.5*0.009632

∆average velocity = 1.006544 then sig fig brings it to 1km/hr

So final average velocity = 104.5km/hr +1 km/hr
 
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