Chapter 7.2, Problem 45E

(a)

To determine

To find: The approximate air temperature after $t=2.9\text{ s}$ .

Answer to Problem 45E

  ${23}^{\circ }C$

Explanation of Solution

Given:The time t (in seconds) it takes for sound to travel 1 km can be modeled by

  $t=\frac{1000}{0.6T+331}$

Where T is the air temperature (in degree Celsius).

First draw graph of $t=\frac{1000}{0.6T+331}$ using graphing utility.

  

In graph clearly seen the air temperature after hear the thunder 2.9 seconds was 23 degree Celsius.

(b)

To determine

To find: The average rate of change of time as the air temperature change $0^{\circ }C$ to ${10}^{\circ }C$ .

Answer to Problem 45E

  $3\text{ seconds}$

Explanation of Solution

Given: The time t (in seconds) it takes for sound to travel 1 km can be modeled by

  $t=\frac{1000}{0.6T+331}$

Where T is the air temperature (in degree Celsius).

Formula Used: $f_{avg}=\frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(t\right)dt}$

The given function of time, $t=\frac{1000}{0.6T+331}$

The average of time as air temperature change $0^{\circ }C$ to ${10}^{\circ }C$ .

Using formula, $t_{avg}=\frac{1}{10-0}{\displaystyle {\int }_0^{10}\frac{1000}{0.6T+331}dT}$

  $\begin{array}{l}{t}_{avg}=\frac{1000}{10}\cdot \frac{1}{0.6}{\ln \left|0.6T+331\right||}_0^{10}\\ t_{avg}=\frac{1000}{6}\left(\ln 337-\ln 331\right)\\ t_{avg}\approx 3\end{array}$

Hence, the average time is 3 seconds.