Chapter 5, Problem 9PEB

You see condensed steam expelled from a ship’s whistle 2.50 s before you hear the sound. If the air temperature is 20.0°C, how many meters are you from the ship?

To determine

The distance of the ship from the person when the condensed steam expelled from the ship’s whistle is seen $2.50\text{ s}$ before its sound is heard.

Answer to Problem 9PEB

Solution:

The distance of the ship from the person is $858\text{ m}$.

Explanation of Solution

Given data:

The air temperature is $20.0°\text{ C}$. Also, the steam from the ship is seen $2.50\text{ s}$ before its sound is heard.

Formula used:

Write the expression for the velocity of a sound wave at the given temperature.

$v_{T\left(P\right)}=v_0+\left(\frac{0.600\text{m/s}}{°\text{ C}}\right)\left(T_P\right)$ …… (1)

Here, $v_{T\left(P\right)}$ and $v_0$ represent the velocity of the sound wave at present temperature and $0°\text{ C}$ respectively, $T_P$ represents the present temperature.

Write the expression for the time.

$t=\frac{d}{v}$

Here, $d$ and $v$ are the distance and the velocity respectively.

Explanation:

When the event occurs, it emits light and sound waves. The sound wave and light wave travelsby the speed of sound and light in air respectively. Take the case of travel of sound from the ship to the person. Calculate the velocity of sound in air.

Substitute $331\text{ m/s}$ for $v_0$, and $20.0°\text{ C}$ for $T_P$ in the equation (1).

$\begin{array}{c}{v}_{T\left(P\right)}=\left(331\text{ m/s}\right)+\left(\frac{0.600\text{ m/s}}{°\text{ C}}\right)\left(20.0°\text{ C}\right)\\ =331\text{ m/s}+12\text{ m/s}\\ =343\text{ m/s}\end{array}$

Consider the distance between the person and the sip as $d$. Then, calculate the time required by the sound to reach the person from the point of source or the ship.

$t_{\text{sound}}=\frac{d}{v_{T\left(P\right)}}$

Substitute $343\text{ m/s}$ for $v_{T\left(P\right)}$.

$t_{\text{sound}}=\frac{d}{343\text{ m/s}}$

Now, consider the case of travel of light from the ship to the person. The velocity of light in air is $3\times {10}^8\text{ m/s}$. Calculate the time taken by the light to reach the person.

$t_{\text{light}}=\frac{d}{v_{\text{light}}}$

Substitute $3\times {10}^8\text{ m/s}$ for $v_{\text{light}}$.

$t_{\text{light}}=\frac{d}{3\times {10}^8\text{ m/s}}$

The time taken by the sound is $2.50\text{ s}$ more than the light. Thus, write the expression for their time difference.

$t_{\text{sound}}-t_{\text{light}}=2.50\text{ s}$

Substitute $\frac{d}{343\text{ m/s}}$ for $t_{\text{sound}}$, and $\frac{d}{3\times {10}^8\text{ m/s}}$ for $t_{\text{light}}$.

$\begin{array}{c}\frac{d}{343\text{ m/s}}-\frac{d}{3\times {10}^8\text{ m/s}}=2.50\text{ s}\\ \frac{\left(3\times {10}^8\text{ m/s}\right)d-\left(343\text{ m/s}\right)d}{\left(343\text{ m/s}\right)\left(3\times {10}^8\text{ m/s}\right)}=2.50\text{ s}\\ \left(\text{299999657 m/s}\right)d=2.50\text{ s}\left(343\text{ m/s}\right)\left(3\times {10}^8\text{ m/s}\right)\end{array}$

Further, solve.

$\begin{array}{c}d=\frac{2.50\text{ s}\left(343\text{ m/s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\text{299999657 m/s}}\\ =857.5\text{ m}\\ \approx 858\text{ m}\end{array}$

Conclusion:

The ship is $858\text{ m}$ away from the person.