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,
[tex]R=\text{velocity}=\dfrac{345\ m}{10\ s}\\\\=34.5\ m/s[/tex]
So, the airplane’s rate of change in elevation is 34.5 m/s.
