Question 206289
an airplane flies between two cities 450 kilometers apart traveling with a wind
 of 45 mph going and against the same wind when returning. 
the trip out takes one less hour than the return trip. 
what is the speed of the plane in still air?
:
Let s = plane speed in still air
then
(s+45) = speed with the wind
and
(s-45) = speed against the wind
;
Write a time equation: Time = {{{dist/speed}}}
:
trip out + 1 hour = trip back
{{{450/((s+45))}}} + 1 = {{{450/((s-45))}}}
Multiply equation by (s+45)(s-45), results
:
450(s-45) + (s+45)(s-45)*1 = 450(s+45)
:
450s - 20250 + s^2 - 2025 = 450s + 20250
:
Arrange on the left
s^2 + 450s - 450s - 2025 - 20250 - 20250 = 0 
:
s^2 - 42525 = 0
;
s^2 = 42525
s = {{{sqrt(42525)}}}
s = 206.2 speed in still air
:
:
Check solution by finding the times of each trip
450/251.2 = 1.79 hrs
450/161.2 = 2.79 hrs
differs by 1 hr