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Question

A conservation research team has tagged several humpback whales with radar monitors to assist in pinpointing their location in three dimensions. The radar shows that one of the whales has suddenly plummeted straight down and is considerably behind the others. This has the team very concerned. The research team is 700 meters east and 500 meters south of their research station, at a depth of 100 meters in their submarine. The stray humpback is approximately 200 meters west, 200 meters south of the research station, at a depth of 600 meters below the ocean's surface. Assuming the submarine averages about 800 meters per minute and the whale doesn't move, how fast will the team reach the whale?

Answer 1

the displacement (900^2+500^2)^1/2=1,060,000^1/2 =>1000 m.approximately and so time taken =1000/800 min=1.234 min
approximately

Answer 2

1.20 minutes.

First triangulation:
The ship and whale are 900 meters apart. One is 300 meters further south than the other, making a triagle which sets the linear distance along the water of 948.68 meters. (Figure this by using standard right angle formulas Square root of (900*900)+(300+300) ..)

Second triangulation.
One is 500 meters further down than the other and they are as mentioned 948.68 meters apart. Second triangle figuring puts the angular difference at 961.77 meters. (Same formulas as above)

Divide 961.77 / 800 meters per min, gives an answer of 1.20 minutes.