Expert: Josh - 8/24/2009

Question
A pilot flies her airplane to Columbus, Ohio and then back. The airplane can go a speed of 139.66 mph in still air, but there is a tail wind on the way to Columbus and a headwind on the way back. If Columbus is 180.00 miles away, and her average speed for the trip is 1.310E+02 mph, find the speed of the wind.  

Answer
Imagine flying from A to B to A.

A.................B (Columbus)

Velocity v1 for outbound journey consists of two components: A ---------->--->           B
The larger arrow is the jet speed (139.66 mph). The smaller arrow due to the tail wind assists the plane, it also points in the direction that it is going. Let this speed component be X.

On the return flight, velocity v2 consists of two components: A           ---><----------  B  
We assume the wind has the same magnitude, but now it points in the opposite direction. So, it slows down the plane.

In summary, v1=139.66+X, v2=139.66-X.  ....[*]

Note: the average speed for the trip is NOT a simple average of v1 and v2.
This is because the return flight takes longer. The average is really the total distance covered (from A to B, then from B to A) divided by the flight time.

Given the average speed for the trip is V=131.0 mph, and the total distance D=180*2 miles, the total time T=D/V. I'll let you work this out on a calculator.

Now, the value of T is known. T = t1 + t2 consists of two components, t1 is the time it takes to cover 180 miles from A to B, t2 is the time it takes to cover the same distance from B to A.

Since t1=180/v1, t2=180/v2 [time = distance/|velocity| ],
using the expressions in [*], we have
T=180/v1 + 180/v2
=180(v1+v2)/(v1*v2)
=180(139.66+X + 139.66-X)/(139.66+X)(139.66-X)
=(180*2*139.66)/(139.66*139.66-X^2)

Using the value of T you obtained earlier,
2.748091603 = 50277.6/(19504.92-X^2)
(19504.92-X^2) = 50277.6/2.748091603
X = sqrt(19504.92 - (50277.6/2.748091603))
 = 34.777 mph

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We can now verify the following:

Outbound velocity, v1 = 139.66+X = 174.437 mph (approximately)
Inbound velocity,  v2 = 139.66-X = 104.882 mph (approximately)

t1 = 180/v1 = 1.031889446 hr
t2 = 180/v2 = 1.716202818 hr

Total time, T = t1+t2 = 2.748092264 hr

Average speed, V = 360/T , equals to 131 mph as claimed.