Chapter 1.1, Problem 55E

a

To determine

To find: the distance of sound wave that will travel in 0.05, 0.2, 1.4, and 5.9 seconds.

Answer to Problem 55E

  $\begin{array}{l}d\left(0.05\right)=14989622.9\text{ meters}\hfill \\ d\left(0.2\right)=59958491.6\text{ meters}\hfill \\ d\left(1.4\right)=419709441.2\text{ meters}\hfill \\ d\left(5.9\right)=1768775502.2\text{ meters}\hfill \end{array}$

Explanation of Solution

Given information:

The distance function, $d\left(t\right)$ in meters is as follows:

  $d\left(t\right)=299792458t$ … (1)

Time in seconds is given as follows:

  $\begin{array}{l}{t}_1=0.05\\ t_2=0.2\\ t_3=1.4\\ t_4=5.9\end{array}$

Calculation:

To find the distance of sound wave at $0.05\text{ s}$ , substitute $t_1=0.05$ in equation (1) and simply the equation as follows:

  $\begin{array}{c}d\left(t\right)=299792458t\\ d\left(0.05\right)=299792458\left(0.05\right)\\ =14989622.9\end{array}$

Thus, the distance of sound wave at $t_1$ is $14989622.9$ m.

Similarly, find the distance of sound wave at $0.2\text{ s}$ by substituting $t_2=0.2$ in equation (1) and simply the equation as follows:

  $\begin{array}{c}d\left(t\right)=299792458t\\ d\left(0.2\right)=299792458\left(0.2\right)\\ =59958491.6\end{array}$

Thus, the distance of sound wave at $t_2$ is $59958491.6$ m.

Similarly, find the distance of sound wave at $\text{1}\text{.4 s}$ by substituting $t_3=1.4$ in equation (1) and simply the equation as follows:

  $\begin{array}{c}d\left(t\right)=299792458t\\ d\left(1.4\right)=299792458\left(1.4\right)\\ =419709441.2\end{array}$

Thus, the distance of sound wave at $t_3$ is $419709441.2$ m.

Similarly, find the distance of sound wave at $\text{5}\text{.9 s}$ by substituting $t_4=5.9$ in equation (1) and simply the equation as follows:

  $\begin{array}{c}d\left(t\right)=299792458t\\ d\left(5.9\right)=299792458\left(5.9\right)\\ =1768775502.2\end{array}$

Thus, the distance of sound wave at $t_4$ is $1768775502.2$ m.

b

To determine

To find: the distance between transmitter and satellite, if the signal from satellite is received at a transmitter in 0.08 seconds.

Answer to Problem 55E

  $d\left(0.08\right)=23983396.64\text{ m}$

Explanation of Solution

Given information:The distance function, $d\left(t\right)$ in meters is as follows:

  $d\left(t\right)=299792458t$ … (1)

Time in seconds, $t=0.08\text{ s}$

Calculation:

To find the distance between transmitter and satellite, substitute $t=0.08$ in equation (1) and simply the equation as follows:

  $\begin{array}{c}d\left(t\right)=299792458t\\ d\left(0.08\right)=299792458\left(0.08\right)\\ =23983396.64\end{array}$

Thus, the transmitter is $23983396.64$ m far from the satellite.