]> Exercise 2

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 1: Real Numbers; page 4

Presentation, Exercise 2


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.    
On the other hand the speed of sound was badly evaluated in view of unknown atmospheric conditions:
v = 344 m/sec with an error of ±5 m/sec    
What is the distance between the source of the sound signal and the rock, in most significant figures? What is the error in the distance?

Parts 1

The distance d between the the source and the rock is   d= vt 2 .

Part 2

The time should be converted to the units of seconds before the calculation:

t=64.332sec   ±0.001sec

Part 3

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:

d=11.07× 10 3 m=11.07km .

Part 4

Since the relative error of the speed is dominant, we'll use it for the final result of the distance.

The relative error of the speed is = 5 344 .
For obtaining the error on the distance we have to multiply this relative error to the calculated distance. Finally we obtain:
d=11.07km   ±0.16km .


The correct final result of part 4 is worth 10 points. Even a result like

d=11.1km   ±0.2km is acceptable.

If the final result is not achieved then:

Part 1 is worth 2 points.
Part 2 is worth 1 point.
Part 3 is worth 4 points.
Part 4 is worth 3 points.