11. Long distance radio navigation for aircraft and ships uses synchronized puls
ID: 3014823 • Letter: 1
Question
11. Long distance radio navigation for aircraft and ships uses synchronized pulses transmitted by widely separated transmitting stations. These pulses travel at the speed of light (186,000 miles per second). The difference in the times of arrival of these pulses at an aircraft or ship is constant on a hyperbola having the transmitting stations as foci. Assume that two stations, 300 miles apart, are positioned on a rectangular coordinate system at coordinates
(-150, 0) and (150, 0), and that a ship is traveling on a hyperbolic path with coordinates (x, 75), as shown in the figure below.
(a) Find the x-coordinate of the position of the ship if the time difference between the pulses from the transmitting stations is 1000 microseconds (0.001 second). (Round your answer to one decimal place.)
(b) Determine the distance between the ship and station 1 when the ship reaches the shore. (Round your answer to one decimal place.)
(c) The captain of the ship wants to enter a bay located between the two stations. The bay is 29 miles from station 1. What should the time difference be between the pulses? (Round your answer to five decimal places.)
(d) The ship is 60 miles offshore when the time difference in part (c) is obtained. What is the position of the ship? (Round your answers to one decimal place.)
(x, y) =
Explanation / Answer
A hyperbola is defined as the set of all points in the planes the difference of whose distances from two fixed points (the foci) is a positive constant:
r=|r1-r2|=k
In this problem, two transmitter stations are placed on the foci of the hyperbola along which the ship travels.
Since the speed of light is constant, a constant time difference of arrival of the signal corresponds to a constant difference of distances:
r=c t=186000*0.001=186 miles=k.
Let's introduce a coordinate system such that the foci are at (±150,0). The equation of a hyperbola is
(x/a)² - (y/b)² = 1
The two vertices are separated by 2a=k=186 miles, so that a=93 miles.
To find b, we use the well-known relationship for a hyperbola,
c²=a²+b², where c is the position of one of the foci, in this case c=150. Therefore,
b²=150²-93², b=118 miles.
Therefore, this hyperbola has an equation
(x/93)² - (y/118)² =1.
To answer part a), let y=75, then
(x/93)² = 1+(75/118)²,
which gives us
x=±110 miles.
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