Question

A scuba diver descends in the water at a rate of 23 1/2

Answer 1

The scuba diver's position relative to sea level after the 4.6 minutes is 5.1 ft.

Given:

A scuba diver descends in the water at a rate of 23 1/2  feet per minute for 2.6 minutes.

A scuba diver ascends at a rate of 28 feet per minute for 2 minutes after seeing a big fish.

To find :

The scuba diver's position relative to sea level after the 4.6 minutes.

Solution:

Rate of a scuba diver descending in the water for 2.6 minutes = $R _1$

$R_1= 23\frac{1}{2}ft/min=\frac{47}{2} ft/min$

Distance covered by a scuba diver in 2.6 minutes = $d_1$

$d_1=R_1\times 2.6min=\frac{47}{2} ft/min\times 2.6 min=61.1 ft$

Rate of a scuba diver ascending for 2 minutes = $R _2$

$R_2= 28ft/min$

Distance covered by a scuba diver in 2 minutes = $d_2$

$d_2=R_2\times 2min=28 ft/min\times 2 =56 ft$

The position of scuba diver after 4.6 minutes relative to sea level:

$d_1-d_2=61.1 ft-56 ft=5.1ft$

The scuba diver's position relative to sea level after the 4.6 minutes is 5.1ft.

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