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Listening post are at A, B, and C. Point A is 2000ft north of point B, and point C is 2000ft east of B. The sound of a gun reaches A and B simultaneously one second after it reaches C. Show that the coordinates of a gun's position are approximately (860,1000), where the x - axis passes through B and C and the origin is midway between B and C. Assume that sound travels 1100ft/sec
The coordinates of the points A, B and C are: (-1000, 2000), (-1000,0), (1000,0).Quote:
Originally Posted by ^_^Engineer_Adam^_^
Let the gun's position be (x,y), the we are effecticly told that (as the
distances of the gun from A and from B are equal):
(-1000-x)^2 + (2000-y)^2 = (-1000-x)^2 + y^2.
Which fixes y=1000.
Also we are told that as the sound reaches B 1 second after it reaches C:
sqrt[(-1000-x)^2 + y^2] = sqrt[(1000-x)^2 + y^2] + 1100
This is a curve with a single branch and so if y=1000, has a single root,
and this can be found numericaly to be at x=858.0.
(you can substitute the suggested x=860 y=1000 into the above equation
and show that the two sides are approximatly equal)
RonL
HI,Quote:
Originally Posted by ^_^Engineer_Adam^_^
CaptainBlack types twice as fast as I can do, so I don't reply to your problem.
If you like to get some further information see here: LORAN - Wikipedia, the free encyclopedia
EB
Hello, ^_^Engineer_Adam^_^!
I have no idea why they placed the origin that way . . . very inconsiderate!
Quote:
Listening post are at A, B, and C.
Point A is 2000ft north of point B, and point C is 2000ft east of B.
The sound of a gun reaches A and B simultaneously one second after it reaches C.
Show that the coordinates of a gun's position are approximately (1860,1000),
where the origin is at B. .Assume that sound travels 1100ft/sec
Code:A *
|*
| *
| *
| * d+1100
| *
2000 | *
| * G
| *(x,y)
| * * d+1100
| *d *
| * *
* - - - - - - - - - - - *
B 2000 C
The gun is at G(x,y).
Let GB = d. .Then GA = GC = d + 1100
Since G is equidistant from A and C, it lies on the 45°-line: x = y
The distance from G(x,x) to B(0,0) is: .√(2x²) .= .d
The distance from G(x,x) to C(2000,0) is: .√[(x - 2000)² + x²] .= .d + 1100
And we have this system of equations:
. . . . . . . . . . . 2x² .= .d²
. . (x - 2000)² + x² .= .(d + 1100)²
Good luck!