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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?

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Hi can u help me in a question
merlynthewhizz
The total elevation that the scuba driver does is from [tex]-29 \frac{7}{10} [/tex] to [tex]-28 \frac{9}{10} [/tex]

The ascending distance = [tex]-28 \frac{9}{10}--29 \frac{7}{10} [/tex]
The ascending distance = [tex] \frac{4}{5} feet[/tex]

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 = [tex] \frac{4}{5} [/tex] ÷ 6.375 = 0.125 feet/minute

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