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Solved Problem on Kinematics

During a fog, a navigator receives two signals simultaneously sent by a station on the coast, one through the air and the other through the water. Between the receptions of the two sounds, a time interval t=5 seconds elapses. Under the conditions of the experiment, the speed of sound in the air is 341 m/s and 1504 m/s in the water. Find the distance

Problem data

- speed of sound in air:
*v*_{ar}= 341 m/s; - speed of sound in water:
*v*_{ag}= 1504 m/s; - the time interval between receptions: Δ
*t*= 5 s.

As the sound wave that spreads through the water has greater speed it arrives before to the boat. If

We chose a reference frame oriented to the right with origin in the sound emission position. The initial position of the sound waves will be

Solution

The equation of displacement as a function of time is given by

\[ \bbox[#99CCFF,10px]
{S=S_{0}+\mathit{vt}} \tag{I}
\]

Using the expression (I) for the wave propagating through water, we have
\[
S_{\text{w}}=S_{0\text{w}}+v_{\text{w}}t
\]

substituting the problem data
\[
\begin{align}
x=0+1504t\\
x=1504t \quad \tag{II}
\end{align}
\]

using the expression (I) for the wave propagating through the air, we have
\[
S_{\text{a}}=S_{0\text{a}}+v_{\text{a}}t
\]

substituting the problem data
\[
\begin{align}
x=0+341\;(\;t+\Delta t\;)\\
x=341\;(\;t+5\;) \quad \tag{III}
\end{align}
\]

From the expression (II) we have the time t
\[
\begin{align}
t=\frac{x}{1504} \tag{IV}
\end{align}
\]

substituting the expression (IV) in (III), we have
\[
x=341\;\left(\;\frac{x}{1504}+5\;\right)\\
x=\frac{341}{1504}x+341 \times 5\\
x=\frac{341}{1504}x+1705\\
x-\frac{341}{1504}x=1705
\]

the
\[
\frac{1504x-341x}{1504}=1705\\
1163x=1705 \times 1504\\
1163x=2564320\\
x=\frac{2564320}{1163}
\]

\[ \bbox[#FFCCCC,10px]
{x\simeq 2205\;\text{m}}
\]

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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .