The pilot of a Coast Guard patrol aircraft on a search mission had just spotted a disabled fishing trawler and decided to go in for a closer look. Flying in a straight line at a constant altitude of 1000 ft and at a steady speed of 232 ft/sec, the aircraft passed directly over the trawler. How fast was the aircraft receding from the trawler when it was 1600 ft from the trawler?

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Manetho Manetho · Answer 1 · 2019-12-23T08:07:14+01:00

Answer:

[tex]\frac{dD}{dt}= 181\frac{168}{1600}[/tex]

Step-by-step explanation:

h= 1000 ft

dx/dt= 232 ft/sec

D= 1600 ft

First, we have to find a distance x ( shown in the figure)

applying pythagorus theorem

D^2= h^2+x^2................1

x^2= D^2-h^2

= 1600^2-1000^2

x^2= 1560000

x=1248.99 m

Now we can find out how fast the aircraft is receding from the trawler

( notice that the height is not changing)

so differentiating equation 1 w.r.t t we get

[tex]\frac{d}{dt} (D^2)= \frac{d}{dt}(h^2+x^2)[/tex]

2DD'= 2hh'+2xx'

[tex]D\frac{dD}{dt}= h\frac{dh}{dt}+x\frac{dx}{dt}[/tex]

now putting values we get

[tex]1600\frac{dD}{dt}= 1000\times0+232\times1249[/tex]

[tex]\frac{dD}{dt}= 181\frac{168}{1600}[/tex]