SOLUTION: a submarine traveling at 9.0 mph is descending at an angle of depression of 6 degrees. How many minutes, to the nearest tenth, does it take the submarine to reach a depth of 80 fee

Algebra ->  Trigonometry-basics -> SOLUTION: a submarine traveling at 9.0 mph is descending at an angle of depression of 6 degrees. How many minutes, to the nearest tenth, does it take the submarine to reach a depth of 80 fee      Log On


   

Question 815433: a submarine traveling at 9.0 mph is descending at an angle of depression of 6 degrees. How many minutes, to the nearest tenth, does it take the submarine to reach a depth of 80 feet?
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
we need the vertical velocity of the sub:
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user vector addition:
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the sub's descent velocity vector (d) is inclined 6 degrees from the horizontal and has magnitude 9 mph.
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sin A = opp/hyp
opp = magnitude of vertical velocity vector mag(v)
hyp = magnitude of descent velocity vector mag(d)
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sin(6 degrees) = mag(v) / mag(d)
sin(6 degrees) = mag(v) / 9
sin(6 degrees) * 9 = mag(v)
mag(v) = 0.9407561 mph
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s = d / t
t = d / s
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time to depth:
t = 80 feet / (0.9407561 miles/hour * 5280 feet/mile * 1/60 hour/minute)
t = 0.9663406 minutes
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Answer:
the sub will reach 80 feet of depth in 0.966 minutes or roughly 1.0 minute
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