91Ó°ÊÓ

Speed of the sound, $v=343\text{″¾}/\text{s}$

We can equate the time difference between the flash seen and the thunder heard to the time taken by the sound waves since the speed of light is much greater than that of the velocity of sound. We can write this time in terms of distance and velocity by inserting the value of the velocity of sound waves to find the integer to divide the number of seconds by the distance in kilometers.

Formula:

The speed of the wave,

$\mathit{S}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{=}\frac{\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$ …(¾±)

Time difference between the flash seen and thunder heard is given as:

$T=t_{sound}−t_{light}$

Speed of light is much greater than that of the velocity of the sound. So, we neglect the term. So, using equation (i), we get the time of the traveling sound as:

$\begin{array}{l}{t}_{sound}=T\\ =\frac{d}{V_{sound}}\end{array}$ $t_{sound}=\frac{d}{343\text{″¾}/\text{s}}$ …(²¹)

We want the distance in kilometers; we convert the velocity to kilometers per second.

$\begin{array}{c}343\text{″¾}/\text{s}=343\text{″¾}/\text{s}×\frac{1\text{ k³¾}}{1000\text{″¾}}\\ =0.343\text{ k³¾}/\text{s}\end{array}$

Substituting the value in km/s in equation (a), we get

Fromtheabove equation we can say that, the distance of the sound travel is given as:

$\begin{array}{c}d=(0.343\text{ k³¾}/\text{s})×t\\ =\frac{t}{\left(\frac{1}{0.343\text{ k³¾}/\text{s}}\right)}\end{array}$

So, the integer we will divide the number of seconds will be $\frac{1}{0.343}\text{ k³¾}/\text{s}≈\frac{1}{0.3}\text{ k³¾}/\text{s}$.