- #1

SchroedingersLion

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I came across the following brainteaser:

A plane travels from airport A to airport B and then returns to A from B. There is no wind, both trips follow a straight line, and the plane flies at constant engine speed.

Suppose now that a constant wind is blowing from A to B. Will the required time for the round trip change or will it remain the same?

If you want to think about it first, don't scroll down.

Solution and my question:The given solution is that the round trip with wind takes longer. While the wind decreases the plane's speed on the way to A by the same amount as it increases its speed on the way to B, the retardation on the way to A lasts a greater amount of time than the boost on the way to B.

As an intuitive explanation, one is supposed to imagine the situation in which the wind speed is equal to the engine speed. Then the plane would be able to reach B twice as fast, but would not move from its place on the way back.

I try to write down some basic mechanics to show this, but I embarassingly failed.

I take "engine speed" to mean that the engine runs at a certain power ##P_0##. Let the distance be given by ##s## and the time taken for the trip ##t_0 = t_{A,0}+t_{B,0}=2t_{A,0}##. The power is related to the velocity by ##v_0=P_0 / F_0## with the force that needs to be overcome ##F_0##, so that the needed time can be expressed as $$t_0=2t_{A,0}=2 \frac{s_A}{v_0}=2\frac{s}{P_0}F_{0}.$$

Now, let the wind blow from A to B. Can't I just model it as a constant force ##F_w## applied in that direction?

Then, the force from A to B with wind is ##F_{A,w}=F_0 - F_w##, and the force from B to A ##F_{B,w}=F_0 + F_w##.

Then, applying the formula above, the required time will be

$$t_w = t_{A,w} + t_{B,w} = \frac {s} {P_0} (F_{A,w} + F_{B,w}) = \frac {s} {P_0} *2F_0 = t_0. $$

Where is my error? Needless to say, it is pretty frustrating to study physics for years, obtain an MSc., only to fail on petty brainteasers that are not even meant to require a physics education