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arlik [135]
4 years ago
11

a jetplane traveling at 500 miles per hour overtakes a propeller plane traveling at 200 miles per hour that had a 2 hour Head St

art. how far from starting point are the planes
Mathematics
1 answer:
labwork [276]4 years ago
7 0
Recall your d = rt, distance = rate * time.

bearing in mind that by the time the Jet matches the Propeller, the distance "d" miles is the same for each.

when that happens, if the Jet has travelled for "t" hours, we know the Propeller had 2 hours headstart, so by then the Propeller has been travelling "t + 2" hours.

\bf \begin{array}{lcccl}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Jet&d&500&t\\
Propeller&d&200&t+2
\end{array}
\\\\\\
\begin{cases}
d=500t\implies \cfrac{d}{500}=\boxed{t}\\\\
d=200(t+2)\\
-----------\\
d=200\left(\boxed{\frac{d}{500}}+2  \right)
\end{cases}

\bf d=200\left( \cfrac{d+1000}{500} \right)\implies \stackrel{\textit{multiplying both sides by the }\stackrel{LCD}{500}}{500d=200(d+1000)}
\\\\\\
500d=200d+200000\implies 300d=200000\implies d=\cfrac{200000}{300}
\\\\\\
d=\cfrac{2000}{3}\implies d=\stackrel{miles}{666\frac{2}{3}}\qquad \textit{or 3520000 feet}
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