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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