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astrosander правка от 5 авг. 2026 г., 06:43

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			### Statement

			$1.1.5^*.$ 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?

			![ For problem 1.1.5 \|582x162, 39%](../../img/1.1.5/statement.png)

			### Solution

			Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:

			$$
			t_A = t_0 + \\frac{L + x}{c}, \\tag{1}
			$$

			where $x = BO$. Similar to point $B$

			$$
			t_B = t_0 + \\frac{x}{c}, \\tag{2}
			$$

			For point $C$

			$$
			t_C = t_0 + \\frac{L - x}{c}, \\tag{3}
			$$

			Let's subtract the second equation from the first equation:

			$$
			t_A – t_B = \\frac{L}{c}. \\tag{4}
			$$

			And from the first equation we subtract the third equation

			$$
			t_A – t_C = \\frac{2x}{c}. \\tag{5}
			$$

			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:

			$$
			AO = L + x = L + \\frac{t_A – t_C}{2}\\frac{L}{t_A – t_B}
			$$

			After the transformation

			$$
			AO = \\frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \\cdot L. \\tag{6}
			$$

			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

			$$
			t_0 = t_B - \\frac{1}{2} \\cdot (t_A – t_C)
			$$

			#### Answer

		@@ -81,11 +81,3 @@Answer
			$$
			AO = \\frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \\cdot L, \\; t_0 = t_B - \\frac{1}{2} \\cdot (t_A – t_C)
			$$
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	### Statement		### Statement

	$1.1.5^*.$ 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?

	![ For problem 1.1.5 \|582x162, 39%](../../img/1.1.5/statement.png)		![ For problem 1.1.5 \|582x162, 39%](../../img/1.1.5/statement.png)

	### Solution		### 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}, \\tag{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}, \\tag{2}		t_B = t_0 + \\frac{x}{c}, \\tag{2}
	$$		$$

	For point $C$		For point $C$

	$$		$$
	t_C = t_0 + \\frac{L - x}{c}, \\tag{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}. \\tag{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}. \\tag{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}
	$$		$$

	After the transformation		After the transformation

	$$		$$
	AO = \\frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \\cdot L. \\tag{6}		AO = \\frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \\cdot L. \\tag{6}
	$$		$$

	To determine the moment of time at which the explosion occurred, we substitute in the expression		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}		t_A = t_0 + \\frac{L + x}{c}, \\; c = \\frac{L}{t_A – t_B}
	$$		$$

	$$		$$
	\\frac{t_A – t_C}{2} = \\frac{x}{c}		\\frac{t_A – t_C}{2} = \\frac{x}{c}
	$$		$$

	after transformations		after transformations

	$$		$$
	t_0 = t_B - \\frac{1}{2} \\cdot (t_A – t_C)		t_0 = t_B - \\frac{1}{2} \\cdot (t_A – t_C)
	$$		$$

	#### Answer		#### Answer

	@@ -81,11 +81,3 @@Answer
	$$		$$
	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)		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)
	$$		$$

	test