An airplane travels 3605 km against the wind in five hours and 4605 km with the wind in the same amount of time. What is the rate of the pla

Question

An airplane travels 3605 km against the wind in five hours and 4605 km with the wind in the same amount of time. What is the rate of the plane in still air and what is the rate of the wind?

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Charlie 10 mins 2021-09-14T23:11:29+00:00 1 Answer 0

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    2021-09-14T23:13:12+00:00

    recall your d = rt, distance = rate * time.

    p = speed of the plane

    w = speed of the wind

    let’s keep in mind that, when the plane is going with the wind, is not really going “p” fast is actually going “p + w” since the wind is adding speed to it, likewise, when the plane is going against the wind, the plane is going “p – w” fast, since the wind is subtracting speed from it.

    \bf \begin{array}{lcccl} &\stackrel{km s}{distance}&\stackrel{kph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{against the wind}&3605&p-w&5\\ \textit{with the wind}&4605&p+w&5 \end{array}~\hfill \begin{cases} 3605=(p-w)5\\ \frac{3605}{5}=p-w\\ 721=p-w\\ 721+w=\boxed{p}\\ \cline{1-1} 4605=(p+w)5 \end{cases} \\\\\\ 4605=(p+w)5\implies \cfrac{4605}{5}=p+w\implies \stackrel{\textit{substituting in the 2nd equation}}{921=\left( \boxed{721+w} \right)+w}

    \bf 921=721+2w\implies 200=2w\implies \cfrac{200}{2}=w\implies \blacktriangleright 100=w \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{p=721+w\implies }p=721+100\implies \blacktriangleright p=821 \blacktriangleleft

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