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Expert-verified Found in: Page 11 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # A marathon runner completes a 42.188 km course in 20h, 30 min and 12 s . There is an uncertainty of 25 m in the distance travelled and uncertainty of 1 s in the elapsed time. (a) Calculate the percent uncertainty in the distance. (b) Calculate the uncertainty in the elapsed time. (c) What is the average speed in meters per second? (d) What is the uncertainty in the average speed?

1. The percentage uncertainty is the distance is 0.0592% .
2. The uncertainty in the elapsed time is 0.1%.
3. The average speed in meters per second is ${\mathrm{v}}_{\mathrm{avg}}=4.6813\mathrm{m}/\mathrm{s}$.
4. The uncertainty in the average speed is 534.03% .
See the step by step solution

## Step 1: Defining concept of uncertainty and percentage uncertainty

The uncertainty of the estimated value is the interval around that value, so that any recurrence of the measure will produce a new result within this interval.

The percentage uncertainty is equal to the total uncertainty divided by the scale, 100% times.

Given Data:

Consider the given data as below.

Distance, $\mathrm{x}=42.188\mathrm{km}$

Time taken, $\mathrm{t}=2\mathrm{hours},30\mathrm{min},12\mathrm{sec}=9012\mathrm{sec}$

Uncertainty in distance, $\mathrm{\delta x}=25\mathrm{m}$

Uncertainty in time, $\mathrm{\delta t}=1\mathrm{s}$

## Step 2: (a) Explanation ofthe percentage uncertainty in the distance:

The percentage uncertainty in the distance is

$%\delta x=\frac{\delta x}{x}×100%\phantom{\rule{0ex}{0ex}}=\frac{25}{42.188}×100%\phantom{\rule{0ex}{0ex}}=0.0592%$

## Step 3: (b) Calculating the uncertainty in the elapsed time

The percentage uncertainty in the elapsed time is,

$%\mathrm{\delta t}=\frac{\mathrm{\delta t}}{\mathrm{t}}×100%\phantom{\rule{0ex}{0ex}}=\frac{1}{9012}×100%\phantom{\rule{0ex}{0ex}}=0.1%$

## Step 4: (c) Calculating the average speed in meters per second

Average speed of the runner is-,

${\mathrm{v}}_{\mathrm{avg}}=\frac{\mathrm{x}}{\mathrm{t}}\phantom{\rule{0ex}{0ex}}=\frac{42.188\mathrm{km}}{9012\mathrm{s}}\phantom{\rule{0ex}{0ex}}=\frac{42.188×{10}^{3}\mathrm{m}}{9012\mathrm{s}}\phantom{\rule{0ex}{0ex}}=4.6813\mathrm{m}/\mathrm{s}$

## Step 5: (d) Calculating the uncertainty in the average speed

Uncertainty of the speed is-,

$\mathrm{\delta v}=\frac{\mathrm{\delta x}}{\mathrm{\delta t}}\phantom{\rule{0ex}{0ex}}=\frac{25\mathrm{km}}{1\mathrm{s}}\phantom{\rule{0ex}{0ex}}=\frac{25×{10}^{3}\mathrm{m}}{1\mathrm{s}}\phantom{\rule{0ex}{0ex}}=25000\mathrm{m}/\mathrm{s}$

Therefore, the uncertainty in average speed is-,

$%\mathrm{\delta v}=\frac{\mathrm{\delta v}}{\mathrm{v}}×100\phantom{\rule{0ex}{0ex}}=\frac{25}{4.6813}×100\phantom{\rule{0ex}{0ex}}=534.03%$ ### Want to see more solutions like these? 