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 x between the boat and the station emitting the signals.

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.

Problem diagram

As the sound wave that spreads through the water has greater speed it arrives before to the boat. If t is the wave propagation time in the water, the wave propagation time in the air is t+Δt (it will be the sum of the time t of propagation in the water with the delay Δtthat it has to be slower).

The boat is at a distance x from a sound station, the sound velocity in the air is 341 m/s, and the sound velocity in the water is 1504 m/s, there is a Delta t interval of 5 seconds between the arrival of the signals.

figure 1

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 S_0ar = S_0ag = 0, the position of the boat is x, the final position where the waves should arrive S_ar = S_ag = x.

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 Least Common Multiple (LCM) between 1 and 1504 is 1504

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

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

Close

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