Is this correct? (I made up this example) On a short car trip the mileage is obs
ID: 1707176 • Letter: I
Question
Is this correct? (I made up this example)
On a short car trip the mileage is observed and recorded at intervals of 5.0min+6sec (0.1min). The limit error for the odometer readings is + 0.1kms because that's the smallest unit the gauge registers. I'm trying to determine the limit error for the calculated average velocity in km/hr.
(tm)min+0.1
(th)hrs+0.0016(??)
(d)kms+0.1
0.0
0.0
0.0
5.0
0.0833
7.6
10.0
0.1666
15.1
15.0
0.25
22.4
20.0
0.333
30.2
25.0
0.4166
37.6
30.0
0.5
44.9
35.0
0.5833
52.4
40.0
0.666
59.8
45.0
0.75
67.4
Question: (Vaverage) = 67.4km/.75h, or 90km/hr+????
Am I calculating this limit error correctly? With tm = +0.9? Do I add up error to take into acount all of the intervals? What about tm? Do I calculate t using minutes and then multiply t by the 1/60h factor? If I am correct in these steps, then the limit error here looks like this...
There is no calcuable limit to how crazy I'm going with this stuff! PLEASE help! Sincerely - J.C.
(tm)min+0.1
(th)hrs+0.0016(??)
(d)kms+0.1
0.0
0.0
0.0
5.0
0.0833
7.6
10.0
0.1666
15.1
15.0
0.25
22.4
20.0
0.333
30.2
25.0
0.4166
37.6
30.0
0.5
44.9
35.0
0.5833
52.4
40.0
0.666
59.8
45.0
0.75
67.4
Explanation / Answer
at each point you solve for the error. it would be .1min/5m * (actual time so far in minutes) + .1km/1km * (actual distance so far) solve for the average velocity now in km/m, convert it to km/hr and then throw in the +- percent. i.e. you calculated 50/60 km/m, which is 50 km/hr and the error in time and distance was .1 you would then have 50 km/hr +- 5km/hr thats how i learned it in my classes...
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