Question

TWO PLANES LEAVE SIMULTANEOUSLY FROM AIRPORT, ONE FLYING DUE NORTH AND THE OTHER DUE EAST. THE NORTHBOUND PLANE IS FLYING 50 MPH FASTER THAN THE EASTBOUND PLANE. AFTER 3 HRS, THE PLANES ARE 2440 MILES APART. FIND THE SPEED OF EACH PLANE.

Answer 1

Two questions, how far have you gotten and do you have the correct answer to check your work?

Answer 2

no answer

Answer 3

Try this: x = Speed of northbound plane; y = Speed of eastbound plane.
x=y+50 9x^2+9y^2=5953600
So, 18y^2+100y+2500=5953600-->9y^2+50y-5952350=0
Solve, and you get y=572.22 mph, x=622.22 mph.

Answer 4

@albertedison98 That's actually incorrect I believe, it should be y = x+50 since the northbound plane is the one flying faster.
@gcgmpng Do you understand where he got these equations?

Answer 5

Sorry, that's wrong. You'd get 18y^2+900y+22500=5953600, which yields
y=549.57 mph, x=599.57 mp

Answer 6

yes

Answer 7

@lasttccasey I think y=x+50 would be right, but it doesn't matter due to symmetry.

Answer 8

Well from mathematics it does matter, the answers would be flipped if you had it the other way around. But as long as @gcgmpng understand it, its all good.

Answer 9

THANK YOU :)