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# A freight train and an express train leave towns..

## Tags: MATH

16:12 Uhr, 23.05.2023

Tag,
guys, is this correct?. Vielen Dank.

A freight train and an express train leave towns 390km apart, traveling toward one another. The freight train travels 30km per hour slower than the express train. They pass each other 3 hours later. What are the speeds of the freight and express train respectively?.

Freight data
distance $=x\phantom{\rule{0.12em}{0ex}}$ kms apart from the other train
time $=3$ hrs
Speed $=\frac{d\phantom{\rule{0.12em}{0ex}}}{t\phantom{\rule{0.12em}{0ex}}}=\frac{x\phantom{\rule{0.12em}{0ex}}}{3}$ kmph

Express data
distance $=\left(390-x\phantom{\rule{0.12em}{0ex}}\right)$ kms apart from the other train

time $=3$ hrs (the time it takes for them to pass each other)

Speed= $\frac{d\phantom{\rule{0.12em}{0ex}}}{t\phantom{\rule{0.12em}{0ex}}}=\frac{390-x\phantom{\rule{0.12em}{0ex}}}{3}$

then,
Speed of express - Speed of freight $=30$ kmph

$\frac{390-x\phantom{\rule{0.12em}{0ex}}}{3}-\frac{x\phantom{\rule{0.12em}{0ex}}}{3}=30$ mph

$390-x\phantom{\rule{0.12em}{0ex}}-x\phantom{\rule{0.12em}{0ex}}=90$

$-2x\phantom{\rule{0.12em}{0ex}}=-300$

$x\phantom{\rule{0.12em}{0ex}}=150$ km this is the distance that the freight train has traveled
$S\phantom{\rule{0.12em}{0ex}}=\frac{d\phantom{\rule{0.12em}{0ex}}}{t\phantom{\rule{0.12em}{0ex}}}=\frac{x\phantom{\rule{0.12em}{0ex}}}{3}=\frac{150}{3}=50$ kmph (rate of speed of the freight train)

then to find distance for the express,
$390-150=240\left($ this is the distance that the express train has traveled

now to find the rate of the speed for the express
$S\phantom{\rule{0.12em}{0ex}}=\frac{d\phantom{\rule{0.12em}{0ex}}}{t\phantom{\rule{0.12em}{0ex}}}=\frac{240}{3}=80$ kmph

so the freight train was going at a rate of $50$ kmph
the express train was going at $80$ kmph.

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

16:36 Uhr, 23.05.2023

Zum Vergleich meine Rechnung mit Einsetzen:

${v\phantom{\rule{0.12em}{0ex}}}_{1}=$ Gschwindigkeit expresstrain, ${v\phantom{\rule{0.12em}{0ex}}}_{2}=$ Geschwindigkeit freight train

${v\phantom{\rule{0.12em}{0ex}}}_{1}\cdot 3+{v\phantom{\rule{0.12em}{0ex}}}_{2}\cdot 3=390$

${v\phantom{\rule{0.12em}{0ex}}}_{2}={v\phantom{\rule{0.12em}{0ex}}}_{1}-30$

${v\phantom{\rule{0.12em}{0ex}}}_{1}\cdot 3+\left({v\phantom{\rule{0.12em}{0ex}}}_{1}-30\right)\cdot 3=390$

${v\phantom{\rule{0.12em}{0ex}}}_{1}=80\to {v\phantom{\rule{0.12em}{0ex}}}_{2}=50$ jeweils in $\frac{k\phantom{\rule{0.12em}{0ex}}m\phantom{\rule{0.12em}{0ex}}}{h\phantom{\rule{0.12em}{0ex}}}$

Ich denke, das ist leicht nachvollziehbar und vlt. etwas einfacher.

18:50 Uhr, 23.05.2023

dasa gefallt mir!. Megacool! Vielen Dnake.