Question

A scuba diver dove from the surface of the ocean to an elevation of -79 9/10 feet at a rate of -21.3 per minute. After spending 10.5 minutes at that elevation, the diver ascended to an elevation of -8 9/10 feet. The total time for the dive so far was 17 1/8 minutes. What was the rate of change in the diver's elevation ascent? Round you answer to the nearest hundredth.

Answer 1

Answer:

The rate of change of the divers elevation is approximately 24.71 feet/minute

Step-by-step explanation:

The given information are;

The depth to which the scuba diver dove =$-79\frac{9}{10} \ feet$

The rate at which he dove = -21.3 feet/minute

The time which he spent at that elevation = 10.5 minutes

The elevation the diver then ascended to = $-8\frac{9}{10} \ feet$

The total time for the dive = $17\frac{1}{8} \ minutes$

Therefore, the time, $t_d$, with which the scuba diver descended to $-79\frac{9}{10} \ feet$ is given as follows;

$t_d = \dfrac{Distance}{Speed} = \dfrac{-79\frac{9}{10} }{-21.3} \approx 3.75 \ minutes$

The time, $t_e$, it took the scuba diver to elevate to $-8\frac{9}{10} \ feet$ is given as follows;

$t_e$ = $17\frac{1}{8} \ minutes$ - (3.75 + 10.5) minutes ≈ 2.8738 minutes

The rate of change of the divers elevation = (Final elevation - Initial elevation)/(Time taken)

∴ The rate of change of the divers elevation = ($-8\frac{9}{10} \ feet$ - ($-79\frac{9}{10} \ feet$ ))/(2.87 minutes) = 71/2.8738 ≈ 24.71 feet/minute to the nearest hundredth

The rate of change of the divers elevation ≈ 24.71 feet/minute.