Respuesta :
recall your d = rt, distance = rate * time
so... let's say, the plane's speed in still air is "p", and the speed of the wind is "w".
when the plane is travelling with the wind over those 1000 miles, is really not going "p" fast, is going " p + w " fast, since it's going with the wind.
now, when the plane is travelling against the wind, is not going "p" fast either, is going " p - w " fast, since the wind is subtracting speed from it.
bearing in mind he cover the 1000 miles as well as the 880 miles in 2 hrs each way.
[tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{with the wind}&1000&p+w&2\\ \textit{against the wind}&880&p-w&2 \end{array} \\\\\\ \begin{cases} 1000=2(p+w)\\ \qquad 500=p+w\\ \qquad 500-p=\boxed{w}\\ 880=2(p-w)\\ ----------\\ \frac{880}{2}=p-w\\ 440=p-\left( \boxed{ 500-p }\right) \end{cases} \\\\\\ 940=2p\implies \cfrac{940}{2}=p\implies 470=p[/tex]
what's the speed of the wind? well, 500 - p = w.
so... let's say, the plane's speed in still air is "p", and the speed of the wind is "w".
when the plane is travelling with the wind over those 1000 miles, is really not going "p" fast, is going " p + w " fast, since it's going with the wind.
now, when the plane is travelling against the wind, is not going "p" fast either, is going " p - w " fast, since the wind is subtracting speed from it.
bearing in mind he cover the 1000 miles as well as the 880 miles in 2 hrs each way.
[tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{with the wind}&1000&p+w&2\\ \textit{against the wind}&880&p-w&2 \end{array} \\\\\\ \begin{cases} 1000=2(p+w)\\ \qquad 500=p+w\\ \qquad 500-p=\boxed{w}\\ 880=2(p-w)\\ ----------\\ \frac{880}{2}=p-w\\ 440=p-\left( \boxed{ 500-p }\right) \end{cases} \\\\\\ 940=2p\implies \cfrac{940}{2}=p\implies 470=p[/tex]
what's the speed of the wind? well, 500 - p = w.
