A scientist uses a submarine to study ocean life.
She begins at sea level, which is an elevation of 0 feet.
She descends for 67 seconds at a speed of 1.7 feet per second.
She then travels directly up for 25 seconds at a speed of 0.6 feet per second.
After this 92-second period, how much time, in seconds, will it take for the scientist to travel back to sea level at 3.6 feet per second? If necessary, round your answer to the nearest tenth of a second.
The scientist descends for 67 seconds at a speed of 1.7 feet per second, covering a total distance of 67 * 1.7 = <<67*1.7=113.9>>113.9 feet.
She then ascends for 25 seconds at a speed of 0.6 feet per second, covering a total distance of 25 * 0.6 = <<25*0.6=15>>15 feet.
Therefore, the depth of the submarine after the 92-second period is 113.9 - 15 = <<113.9-15=98.9>>98.9 feet.
To get back to sea level, the scientist needs to cover a distance of 98.9 feet.
At a speed of 3.6 feet per second, it will take her 98.9 / 3.6 = <<98.9/3.6=27.47222222222222>>27.5 seconds to travel back to sea level.
Rounding this to the nearest tenth of a second gives 27.5 seconds. Answer: \boxed{27.5}.