$1.1.5^*.$ 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?
### Solution
@@ -8,23 +8,23 @@Solution
Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:
−
$$t_A = t_0 + \frac{L + x}{c}, \, (1)$$
+
$$t_A = t_0 + \frac{L + x}{c}, \tag{1}$$
where $x = BO$. Similar to point $B$
−
$$t_B = t_0 + \frac{x}{c}, \,(2)$$
+
$$t_B = t_0 + \frac{x}{c}, \tag{2}$$
For point $C$
−
$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$
+
$$t_C = t_0 + \frac{L - x}{c}, \tag{3}$$
Let's subtract the second equation from the first equation:
−
$$t_A – t_B = \frac{L}{c}. \,(4)$$
+
$$t_A – t_B = \frac{L}{c}. \tag{4}$$
And from the first equation we subtract the third equation
−
$$t_A – t_C = \frac{2x}{c}. \,(5)$$
+
$$t_A – t_C = \frac{2x}{c}. \tag{5}$$
From equation $(4)$ let's express
$$c = \frac{L}{t_A – t_B},$$
and from $(5)$
$$x = \frac{t_A – t_C}{2}\cdot c.$$
Then the required distance:
$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
$1.1.5^*.$ 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?
$1.1.5^*.$ 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?
### Solution
### Solution
@@ -8,23 +8,23 @@Solution
Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:
Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:
$$t_A = t_0 + \frac{L + x}{c}, \, (1)$$
$$t_A = t_0 + \frac{L + x}{c}, \tag{1}$$
where $x = BO$. Similar to point $B$
where $x = BO$. Similar to point $B$
$$t_B = t_0 + \frac{x}{c}, \,(2)$$
$$t_B = t_0 + \frac{x}{c}, \tag{2}$$
For point $C$
For point $C$
$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$
$$t_C = t_0 + \frac{L - x}{c}, \tag{3}$$
Let's subtract the second equation from the first equation:
Let's subtract the second equation from the first equation:
$$t_A – t_B = \frac{L}{c}. \,(4)$$
$$t_A – t_B = \frac{L}{c}. \tag{4}$$
And from the first equation we subtract the third equation
And from the first equation we subtract the third equation
$$t_A – t_C = \frac{2x}{c}. \,(5)$$
$$t_A – t_C = \frac{2x}{c}. \tag{5}$$
From equation $(4)$ let's express
From equation $(4)$ let's express
$$c = \frac{L}{t_A – t_B},$$
$$c = \frac{L}{t_A – t_B},$$
and from $(5)$
and from $(5)$
$$x = \frac{t_A – t_C}{2}\cdot c.$$
$$x = \frac{t_A – t_C}{2}\cdot c.$$
Then the required distance:
Then the required distance:
$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
