Source: High school physics (Chinese)
Problem
An airplane flies a round trip from point A north to point B and then back to A. The airplane's speed relative to the air is $v$. The round-trip time in still air (when air speed relative to ground, $u$, is 0) is $t_0$. Now, consider a wind blowing from south to north with speed $u$ (where $u eq 0$).
The round-trip time with wind, $t$, is derived as follows: The time for the northbound trip with a tailwind is $t_{A \to B} = L/(v+u)$. The time for the southbound trip with a headwind is $t_{B \to A} = L/(v-u)$. The total time is $t = t_{A \to B} + t_{B \to A} = \frac{L}{v+u} + \frac{L}{v-u} = \frac{2Lv}{v^2-u^2}$. Using the still-air time $t_0 = 2L/v$, we substitute $2L = vt_0$ to get:
$$t = \frac{(vt_0)v}{v^2 - u^2} = \frac{v^2 t_0}{v^2 - u^2}$$Dividing the numerator and denominator by $v^2$ yields the final proven formula:
$$t = \frac{t_0}{1 - u^2/v^2}$$Let $L$ be the one-way distance between points A and B. The airplane's speed relative to the air is $v$, and the wind speed relative to the ground is $u$.
1. Time in Still Air ($t_0$) In still air ($u=0$), the airplane's speed relative to the ground is $v$ for both the northbound and southbound trips. The total time for the round trip is:
$$t_0 = t_{A \to B} + t_{B \to A} = \frac{L}{v} + \frac{L}{v} = \frac{2L}{v}$$2. Time with Wind ($t$) The wind blows from south to north. We use the concept of relative velocity, where the velocity of the plane relative to the ground ($v_{pg}$) is the vector sum of the velocity of the plane relative to the air ($v_{pa}$) and the velocity of the air relative to the ground ($v_{ag}$).
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Trip from A to B (Northbound): The airplane flies with a tailwind. Its speed relative to the ground is the sum of its airspeed and the wind speed.
$$v_{A \to B} = v + u$$The time for this leg is:
$$t_{A \to B} = \frac{L}{v+u}$$ -
Trip from B to A (Southbound): The airplane flies with a headwind. Its speed relative to the ground is the difference between its airspeed and the wind speed.
$$v_{B \to A} = v - u$$The time for this leg is:
$$t_{B \to A} = \frac{L}{v-u}$$
The total round-trip time with wind, $t$, is the sum of the times for each leg:
$$t = t_{A \to B} + t_{B \to A} = \frac{L}{v+u} + \frac{L}{v-u}$$3. Derivation To combine the terms, we find a common denominator:
$$t = L \left( \frac{(v-u) + (v+u)}{(v+u)(v-u)} \right) = L \left( \frac{2v}{v^2 - u^2} \right)$$ $$t = \frac{2Lv}{v^2 - u^2}$$From the still-air case, we have $t_0 = 2L/v$, which can be rearranged to $2L = vt_0$. Substituting this into the expression for $t$:
$$t = \frac{(vt_0)v}{v^2 - u^2} = \frac{v^2 t_0}{v^2 - u^2}$$Finally, we divide the numerator and the denominator by $v^2$ to arrive at the desired form:
$$t = \frac{t_0}{1 - u^2/v^2}$$This completes the proof. Note that since $u^2/v^2 > 0$ (for $u eq 0$), the denominator is less than 1, which implies $t > t_0$. A round trip always takes longer with a headwind/tailwind than in still air.