$1.1.5^*.$ Three microphones located on the same straight line at points $A$, $B$, $C$ recorded successively at the moments $t_A > t_B > t_C$ the sound of an explosion that occurred at point $O$, which lies on the segment $AC$. Find the length of the segment $AO$ if $AB = BC = L$. At what point in time did the explosion occur?
Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:
where $x = BO$.
Similar to point $B$
$$t_B = t_0 + \frac{x}{c}, \,(2)$$
For point $C$
$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$
Let's subtract the second equation from the first equation:
$$t_A – t_B = \frac{L}{c}. \,(4)$$
And from the first equation we subtract the third equation
$$t_A – t_C = \frac{2x}{c}. \,(5)$$
From equation $(4)$ let's express $c = \frac{L}{t_A – t_B}$, а из $(5)$ $x = \frac{t_A – t_C}{2} \cdot c$. Then the required distance
$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
After the transformation
$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$
To determine the moment of time at which the explosion occurred, we substitute in the expression
$$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$ $$\frac{t_A – t_C}{2} = \frac{x}{c}$$
after transformations
$$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$