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An airplane descended 345 m in 10 seconds. What was the airplane’s rate of change in elevation? Show your work and don’t forget the unit!

User Ferid
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1 Answer

3 votes

Answer:

The airplane’s rate of change in elevation is 34.5 m/s.

Explanation:

Given that,

An airplane descended 345 m in 10 seconds.

We need to find the airplane’s rate of change in elevation. Let the rate is R. Its rate is equal to the distance divided by time. So,


R=\text{velocity}=(345\ m)/(10\ s)\\\\=34.5\ m/s

So, the airplane’s rate of change in elevation is 34.5 m/s.

User BergQuester
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