Two friends A and B are standing a distance x apart in an open...

Let $v$ be the velocity of sound in still air and $u$ be the velocity of wind.

In the first case, the wind blows from $A$ to $B$. The velocity of sound relative to the ground is $v + u$.

The time taken for sound to travel distance $x$ is $t_1 = \dfrac{x}{v + u}$.

$v + u = \dfrac{x}{t_1}$

In the second case, $A$ and $B$ swap positions. The sound now travels from the new position of $A$ to the new position of $B$, which is opposite to the direction of the wind.

The velocity of sound relative to the ground is $v - u$.

The time taken for sound to travel distance $x$ is $t_2 = \dfrac{x}{v - u}$.

$v - u = \dfrac{x}{t_2}$

Adding the two equations:

$(v + u) + (v - u) = \dfrac{x}{t_1} + \dfrac{x}{t_2}$

$2v = x \left( \dfrac{1}{t_1} + \dfrac{1}{t_2} \right)<br><br>$v = $\dfrac{x}{2} \left( \dfrac{1}{t_1} + \dfrac{1}{t_2} \right)<br><br>Subtracting the second equation from the first:<br><br>$(v + u) - (v - u) = $\dfrac{x}{t_1} - \dfrac{x}{t_2}<br><br>$2u = x \left( \dfrac{1}{t_1} - \dfrac{1}{t_2} \right)$<br><br>$u = \dfrac{x}{2} \left( \dfrac{1}{t_1} - \dfrac{1}{t_2} \right)