Question

An airplane flying at constant air speed from Indianapolis to St. Louis (assume they are directly east-west of each other) in calm weather (no wind of any kind) would log the same flying time for both legs of the trip. Suppose the same trip is taken when there is wind from the west at a constant speed all day long and into the next. How would the total (round trip) time in windy weather compare with the total time in calm weather

Answer 1

Answer:

Explanation:

Suppose the distance between the two cities is D and the velocity in calm weather is V . The total time taken in two way travel is given by

Total distance / velocity

= 2 D / V

Suppose velocity of wind is v . Then in one way the velocity of airplane will become V + v and in the return trip its velocity will be V - v

Total time taken

= $\frac{D}{V+v} +\frac{D}{V-v}$

= $\frac{2DV}{V^2-v^2}$

= $\frac{2V^2D}{V(V^2-v^2)}$

= $\frac{2D}{V(1 - \frac{v^2}{V^2}) }$

= The denominator contains a factor

$1-\frac{v^2}{V^2}$

which is less than one so time calculated will be more than

2D / V

Hence in the second case time taken will be more .