Question 1163051: Help please
A submarine passes 30 m beneath a boat while traveling at 8.89 m/s on a course of N34°21'W. The boat continues on a heading of N55°39'E at a speed of 13.3 m/s.
Round answers to 3 significant digits.
(a) How far do radio signals between the two craft have to travel 3 s later?
m
(b) How fast are the two craft moving apart at this time?
m/s
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Since one vessel is on a course east of north and the other is on a course west of north, simply adding the course values gives the difference in their courses. Since the sum is 90 degrees, the horizontal distance is easily calculated by the Pythagorean theorem:
^2\ +\ (3\,\times\,13.3)^2})
But the distance radio waves would have to travel is dependent on the height above the waterline for the radio antenna for the boat and the actual depth of the radio antenna for the submarine. Further, since radio waves are refracted when changing media from atmosphere to water, and the index of refraction for air is temperature dependent and the index of refraction for water is temperature and salinity dependent, there is insufficient information to answer your question as posed.
However, making the highly irrational assumption that the boat and submarine are point sources of radio frequency energy and discounting refraction, an additional application of Pythagoras with one leg of the triangle the results of the above calculation and the other leg measuring 30 meters, will provide the desired answer.
John
My calculator said it, I believe it, that settles it
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