Lets say someone is flying at constant speed from place $X$ to place $Y$ and back. Going to X takes 5 hours (with the wind) and coming back from X to Y takes 6 hours. Lets assume that the wind is at a constant speed, then how long would it take for a piece of paper being propelled by the wind alone to travel from $X$ to $Y$.
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Heh, this problem is cute! Suppose our (constant) speed is $v$, the (constant) speed of the wind is $w$, the time it takes for us to get from $Y$ to $X$ is $t_1$ and going back takes $t_2$. If the distance between $X$ and $Y$ is $d$, then because the speeds are all constant, we know $$ (v+w)t_1 = d = (v-w)t_2,$$ where the plus vs. the minus sign is because the faster direction ($Y$ to $X$) is where we're travelling in the direction of the wind. Putting the times in, we get $$ 5v + 5w = 6v - 6w,$$ and so $w = v/11$. Now, the distance the piece of paper has to travel is that same as the distance we travel, so we have $$ t_1(v+w) = 5(v+w) = d = wt,$$ where $t$ is the amount of time the piece of paper takes to travel. Solving for $t$, we get $$ t = \frac{5(v + v/11)}{v/11} = \frac{5(12v/11)}{v/11} = 5\cdot 12 = 60.$$ So the piece of paper takes 60 hours to get from $Y$ to $X$. |
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Let $\,v=\,$ someone's constant speed, and let $\,x=\,$ wind's constant speed from X to Y . If $\,d=\,$ distance between X and Y, we get (using the basic formula $\,v=d/t\,$) $$t=5=\frac{d}{v+x}\,\,\,,\,\,\,6=\frac{d}{v-x}\Longrightarrow 5v+5x=6v-6x\Longrightarrow v=11x\Longrightarrow$$ $$\Longrightarrow5=\frac{d}{12x}\Longrightarrow x=\frac{d}{60}$$ Thus, a paper fying from X to Y without dragging and gliding will make the distance in $$t=\frac{d}{x}=60\,\,\text{ hours}$$ |
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