Airplane in the headwind and tailwind
An airplane makes daily round-trip flights between two airports. On a normal day, the flight takes each way. Typically, the plane is assumed to fly at equal and constant speed both ways.
One day severe winds in one direction make the plane go 20% faster than usual on the outbound flight, and 20% slower than usual on the return flight.
How long is the total time (in minutes) of the outbound and return flights on that windy day? Submit your answer to 2 decimal places.
The answer is 406.25.
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1 solution
Assume d is the distance between the airports, and v is the normal daily speed on the windy day:
Time outbound= d / ( v +0.2 v)= 0.8333 d/v
Time return= d / (v-0.2 v)= 1.25 d/v
Total time on the windy day: 0.8333+1.25=2.08333 d/v
In the above, d/v is the normal one way flight time, which is 195 minutes
Total time= 2.083333 d/v=2.08333X195=406.25 minutes
A Note: Looking at this problem quickly, one may assume the 20% faster one way and 20% slower the another way would make up for each other, and the total flight time is the same as normal, but the above calculation shows the actual time is about 4% slower. Now, as to why this is counter-intuitive: We tend to think time-distance-speed problems are linear. However, the speed variable in t=d/v is in denominator. Therefore a linear change in speed variable gives a non-linear result in time.