Question

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.80 s apart. How far away did the impact occur? See Table 12-1.

Short answer

The distance between the person and impact occur is $309.82\mathrm{m}$.

Step-by-step solution

Understanding the concept of distance travelled by a moving system

In this problem, the distance travelled by the moving system is determined by the product of the speed of the moving system and the time of the journey.

Given data

The time difference between the sound travel in air and concrete is $$\mathrm{\Delta }t=0.80\mathrm{s}$$.

Evaluating the total distance travelled by the sound by using the relation among speed, distance, and time

The standard value for the speed of sound in air and concrete from Table-12.1 are $v_{\mathrm{a}\mathrm{i}\mathrm{r}}=343\mathrm{m}/\mathrm{s}$ and $v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}=3000\mathrm{m}/\mathrm{s}$ respectively.

The total distance travelled by the sound in air and concrete is equal.

The time taken by the sound to travel in air is calculated below:

$\begin{array}{r}c{d}_{\mathrm{a}\mathrm{i}\mathrm{r}}=d_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}\\ v_{\mathrm{a}\mathrm{i}\mathrm{r}}{t}_{\mathrm{a}\mathrm{i}\mathrm{r}}=v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}{t}_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}\\ v_{\mathrm{a}\mathrm{i}\mathrm{r}}{t}_{\mathrm{a}\mathrm{i}\mathrm{r}}=v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}\left(t_{\mathrm{a}\mathrm{i}\mathrm{r}}-\mathrm{\Delta }t\right)\\ t_{\mathrm{a}\mathrm{i}\mathrm{r}}=\frac{v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}}{v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}-v_{\mathrm{a}\mathrm{i}\mathrm{r}}}\left(\mathrm{\Delta }t\right)\end{array}$

Here, $d_{\mathrm{a}\mathrm{i}\mathrm{r}}$ and $d_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}$ are the distance travelled by the sound in air and concrete, $t_{\mathrm{a}\mathrm{i}\mathrm{r}}$ and $t_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}$ are the time taken by the sound in air and concrete.

The total distance travelled by the sound is calculated below:

$\begin{array}{r}cd=v_{\mathrm{a}\mathrm{i}\mathrm{r}}{t}_{\mathrm{a}\mathrm{i}\mathrm{r}}\\ =v_{\mathrm{a}\mathrm{i}\mathrm{r}}\left[\frac{v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}}{v_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{t}\mathrm{e}}-v_{\mathrm{a}\mathrm{i}\mathrm{r}}}\left(\mathrm{\Delta }t\right)\right]\end{array}$

Substitute the values in the above equation.

$\begin{array}{r}c=343\mathrm{m}/\mathrm{s}\left[\frac{3000\mathrm{m}/\mathrm{s}}{3000\mathrm{m}/\mathrm{s}-343\mathrm{m}/\mathrm{s}}\left(0.80\mathrm{s}\right)\right]\\ =309.82\mathrm{m}\end{array}$

Hence, the distance between the person and impact occur is $309.82\mathrm{m}$.