Question

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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!

Answer 1

Answer:

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

Step-by-step 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,

$R=\text{velocity}=\dfrac{345\ m}{10\ s}\\\\=34.5\ m/s$

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