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Solved Problem on Kinematics
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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 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 tt (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 S0ar = S0ag = 0, the position of the boat is x, the final position where the waves should arrive Sar = Sag = 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
\[ \frac{1504x-341x}{1504}=1705\\ 1163x=1705 \times 1504\\ 1163x=2564320\\ x=\frac{2564320}{1163} \]
\[ \bbox[#FFCCCC,10px] {x\simeq 2205\;\text{m}} \]

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