逆风与顺风的飞机

总时间会变长。

去程速度 $v+w$，时间 $\frac{d}{v+w}$；返程速度 $v-w$，时间 $\frac{d}{v-w}$。

因此总时间：

$$T(w)=\frac{d}{v+w}+\frac{d}{v-w}=\frac{2dv}{v^2-w^2}.$$

当 $0<|w|<v$ 时，分母 $v^2-w^2$ 随 $|w|$ 增大而变小，所以 $T(w)$ 增大。直觉上：顺风带来的“节省时间”比逆风造成的“额外耗时”少。

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Original Explanation

The more the wind blows, the total time will be higher even though one of the journeys is faster.

Solution

Intuition

The time during which the plane's speed is boosted is shorter than the time during which it is slowed down, so the over all effect is of slow-down.

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Maths

Suppose the wind blows at a speed of $w$ from $A$ to $B$, and the plane's engine is capable of a steady speed of $s$ (without any wind). Ignore acceleration and de-acceleration.

Suppose the distance is $d$, the total time will be $2d/s$ without any wind.

Speed from $A$ to $B$ is $(s + w)$ and $B$ to $A$ is $(s - w)$

$t = \dfrac{d}{s+w} + \dfrac{d}{s-w}$

$= d ( \dfrac{2 s}{s^2 - w^2})$

$= \dfrac{2d}{s - (w^2/s) } > \dfrac{2d}{s}$

For $0 < w < s$, the denominator in the first term is less than $s$, and therefore the total time is greater than the time without the wind.

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Footnotes

1. Consider the edge case, where the plane's speed is the same as the wind. In one direction, the plane's speed is doubled, while in the other, it becomes zero! (i.e. it takes an infinite amount of time to finish the journey!)