Question

A scuba diver dove down from a surface of the ocean to a elevation of -29 7/10 feet at a rate of -18.8 feet per minute. after spending 12.75 minutes at the elevation, the diver ascended to a elevation o -28 9/10 feet. the total time for the dive so far was 19 1/8 minutes. what was the rate of change in the divers elevation during the ascent?

Answer 1

The total elevation that the scuba driver does is from $-29 \frac{7}{10}$ to $-28 \frac{9}{10}$

The ascending distance = $-28 \frac{9}{10}--29 \frac{7}{10}$
The ascending distance = $\frac{4}{5} feet$

The time (t) taken during the ascend = 19.125 - 12.75 = 6.375 minutes

The average change in speed is given by the ascending distance divided by time taken during the ascend = $\frac{4}{5}$ ÷ 6.375 = 0.125 feet/minute

Answer 2

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