Solved Problem on One-dimensional Motion

FISICAEXE

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, an interval of time, 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. Determine the distance x between the boat and the station emitting the signals.

Problem data:

Speed of sound in air: v_a = 341 m/s;
Speed of sound in water: v_w = 1504 m/s;
Time interval between receptions: Δt = 5 s.

Problem diagram:

As the sound wave that spreads through the water has greater speed, it arrives before 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).

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_0a = S_0w = 0, the position of the boat is x, the final position where the waves should arrive S_a = S_w = x.

Solution:

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

\[ \begin{gather} \bbox[background:#99CCFF; padding:10px; border-radius:10px;] {S=S_0+v t} \tag{I} \end{gather} \]

Using the equation (I) for the wave propagating through water

\[ \begin{gather} S_w=S_{0w}+v_{w}t \\[5pt] x=0+\left(1504\;\mathrm{\small{\frac{m}{s}}}\right)t \\[5pt] x=1504t \tag{II} \end{gather} \]

Using the equation (I) for the wave propagating through the air

\[ \begin{gather} S_a=S_{0ar}+v_at \\[5pt] x=0+\left(341\;\mathrm{\small{\frac{m}{s}}}\right)(t+\Delta t) \\[5pt] x=\left(341\;\mathrm{\small{\frac{m}{s}}}\right)(t+5;\mathrm s) \tag{III} \end{gather} \]

From the equation (II), we have the time t

\[ \begin{gather} t=\frac{x}{1504\;\mathrm{\frac{m}{s}}} \tag{IV} \end{gather} \]

substituting the expression (IV) into equation (III)

\[ \begin{gather} x=\left(341\;\mathrm{\small{\frac{m}{s}}}\right)\left(\frac{x}{1504\;\mathrm{\frac{m}{s}}}+5\;\mathrm s\right) \\[5pt] x=\frac{341\;\mathrm{\frac{\cancel m}{\cancel s}}}{1504\;\mathrm{\frac{\cancel m}{\cancel s}}}x+\left(341\;\mathrm{\small{\frac{m}{\cancel s}}}\right)(5\;\mathrm{\cancel s}) \\[5pt] x=\frac{341}{1504}x+1705\;\mathrm m \\[5pt] x-\frac{341}{1504}x=1705\;\mathrm m \end{gather} \]

multiplying both sides of the equation by 1504

\[ \begin{gather} \qquad\qquad\qquad x-\frac{341}{1504}x=1705\;\mathrm m \qquad (\times 1504) \\[5pt] 1504 x-\cancel{1504}\times\frac{341}{\cancel{1504}}x=1705\;\mathrm m\times 1504 \\[5pt] 1163x=2564320\;\mathrm m \\[5pt] x=\frac{2564320\;\mathrm m}{1163} \end{gather} \]

\[ \begin{gather} \bbox[background:#FFCCCC; padding:10px; border-radius:10px;] {x\approx 2205\;\mathrm m} \end{gather} \]