Question 126797
A weather plane flies due north into a 40 mph wind until it reaches the North Pole, at which point it returns to its original position along the same route. The average velocity for the round trip was 396 mph, even though the plane was flying at its maximum velocity the entire time. What is the maximum velocity of the plane (in mph) when there is no wind?
:
Let s = speed of the plane in still air
:
(s-40) = speed against the wind
(s+40) = speed with the wind
:
Let distance one way = 500
Round trip = 1000 mi
then using the given average speed:
Time = {{{1000/396}}} =  2.525 hrs for the round trip
------------------------------------------------------
Sorry punched the wrong button, to continue:
:
Write a time equation: Time = {{{dist/speed}}}
{{{500/((s+40))}}} + {{{500/((s-40))}}} = 2.525 hrs
Multiply equation by (s+40)(s-40) to get rid of the denominators
:
500(s-40) + 500(s-40) = 2.525(s-40)(s+40)
:
500s - 20000 + 500s - 20000 = 2.525(s^2 - 1600)
:
1000s = 2.525s^2 - 4040
:
s^2 - 1000s - 4040 = 0; a quadratic equation
:
Using the quadratic formula, I came up with: s = 400 as the positive solution
:
400 mph is the max speed of the aircraft in still air.
:
WE can check that by finding the time and the average speed
:
{{{500/440}}} + {{{500/360}}} = 
1.136 + 1.389 = 2.525 hrs
{{{1000/2.535}}} = 396 mph av speed

I don't think that using a different distance will effect the solution.
Try it using a 1 way distance of 1000 mi (2000 mi round trip)