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zhenek [66]
3 years ago
10

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

Mathematics
1 answer:
Makovka662 [10]3 years ago
3 0

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.

Learn more about ascending rate and descending rate here:

brainly.com/question/1477877?referrer=searchResults

brainly.com/question/1072871?referrer=searchResults

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