Question

A scuba diver dove from the surface of the ocean to an elevation of -59 9/10 feet at a rate of -18.4 feet per minute. After spending 12 minutes at that elevation, the diver ascended to an elevation of -18 9/10 feet. The total time for the dive so far was 19 1/8 minutes. What was the rate of change in the diver's elevation during the ascent? Round your answer to the nearest hundredth.
The rate of change of the diver's elevation was _______ ft/min.


Please include all of the work required to complete the mathematical problem. I would really appreciate it! <3

Answer 1

Answer:

The rate of change of the diver's elevation was 10 53/89 ft./min

Step-by-step explanation:

The given parameters are;

The initial elevation of the diver = 0 feet (Ocean surface level)

The depth to which the diver dove = -59 9/10 feet

The rate at which the diver descended = -18.4 feet/minute

The duration the diver spends at the elevation, t₁ = 12 minutes

The depth to which the diver ascended to = -18 9/10 feet

The total time the diver takes so far in the ocean, $t_{(tot)}$ = 19 1/8 minutes

Therefore, we have;

The time it takes the diver to descend to -59 9/10 feet t₂ = -59 9/10/(-18.4) = 3 47/184 minutes

The duration of the divers ascent = $t_{(tot)}$ - t₁ - t₂ = 19 1/8 - 12 - 3 47/184 = 3 20/23 minutes

The rate of change of the divers elevation in ascent = Distance of elevation/Time

The rate of elevation of the divers elevation in ascent = (-18 9/10 - (-59 9/10))/(3 20/23) = 41/(3 20/23) = 943/89 feet/minute = 10 53/89 feet/minute

The rate of change of the diver's elevation = 10 53/89 ft./min ≈ 10.6 ft./min.