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Q24PE

Expert-verifiedFound in: Page 11

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?**

- The percentage uncertainty is the distance is 0.0592% .
- The uncertainty in the elapsed time is 0.1%.
- The average speed in meters per second is ${\mathrm{v}}_{\mathrm{avg}}=4.6813\mathrm{m}/\mathrm{s}$.
- The uncertainty in the average speed is 534.03% .

**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, $\mathrm{\delta t}=1\mathrm{s}$

The percentage uncertainty in the distance is

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

The percentage uncertainty in the elapsed time is,

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

Average speed of the runner is-,

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

Uncertainty of the speed is-,

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

Therefore, the uncertainty in __average__ speed is-,

$\%\mathrm{\delta v}=\frac{\mathrm{\delta v}}{\mathrm{v}}\times 100\phantom{\rule{0ex}{0ex}}=\frac{25}{4.6813}\times 100\phantom{\rule{0ex}{0ex}}=534.03\%$

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