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?

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

2DD'= 2hh'+2xx'

now putting values we get

correct answer:

i^31 = -i

y + 11 = (1/4)(x + 16)

step-by-step explanation: