An evaluation of a distance outdoors is done by measuring the time t between sending and receiving the echo of a sound signal returned from a very massive rock. The time obtained is:
t = 1 min. + 4.332 sec., with an error = ±0.001 sec. |
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v = 344 m/sec with an error of ±5 m/sec |
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The distance d between the the source and the rock is $d=\frac{vt}{2}$ .
The time should be converted to the units of seconds before the calculation:
The time is given in 5 significat figures and the speed in 3. The relative error of the speed is much larger, therefore the error of the result is dominated by that of the speed. The speed is given to 3 significant figures and we can take 3 - for the final result, or 4 at most. We obtain:
Since the relative error of the speed is dominant, we'll use it for the final result of the distance.
The correct final result of part 4 is worth 10 points. Even a result like
If the final result is not achieved then: