How fast is the distance from the observer to the airplane increasing 30s later? An airplane flying horizontally at an altitude of 6km and at a speed of 960km/h passes directly above an observer on the ground.

Expert Answers

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First 960km/hr = 16km/min

The distance to the plane as a function of time after it passes overhead.

`s = sqrt(6^2+(16t)^2)=sqrt(36+256t^2)`

The rate of change of the distance is

 

`(ds)/(dt)=(d(sqrt(36+256t^2)))/(d(36+256t^2))*(d(36+256t^2))/(dt)=1/(2sqrt(9+256t^2))(256*2t)`

`(ds)/(dt)=(256t)/(sqrt(9+256t^2))`

So  30 seconds later.   30 seconds is 1/2 min.

So

`(ds)/(dt)=(256*1/2)/sqrt(36+256*1/4)=128/sqrt(36+64)`

`(ds)/(dt)=128/(10)=12.8"km/min"=768"km/hr"`

 

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