At a certain instant an aircraft flying due east at 240 miles per hour passes directly over a car traveling due southeast at 60 miles per hour on a straight, level road. If the aircraft is flying at an altitude of .5mile, how fast is the distance between the aircraft and the car increasing 36 seconds after the aircraft passes directly over the car?

Answer 1

at time t=0,
let the plane be at (0,0,.5)
let the car be at (0,0,0)
let the +x axis be due east, so +y is due north

at time t hours,

plane is at (240t,0,.5)
car is at (60/√2 t,-60/√2 t,0)

the distance d between car and plane is

d^2 = (240t - 60/√2 t)^2 + (60/√2 t)^2 + .5^2

36 seconds = 0.01 hours, so

d^2 = (2.4-.6/√2)^2 + (.6/√2)^2 + .5^2
d = 2.0817

2d dd/dt = 7200(17-4√2)t
2(2.0817) dd/dt = 7200(17-4√2)(.01)
dd/dt = 196.16 mph