Question 1026577: Please explain how to set up this problem as an algebraic equation. "Two cars started to move at the same time in the same direction, but one was moving twice as fast as the other. Six hours later, the cars were 204 miles apart. Find the speed of each car."
I'm not sure the "how" or "why" of the set up. Thank you for your help.
Ian
Found 3 solutions by josmiceli, stanbon, MathTherapy:
Answer by josmiceli(19441)   (Show Source):
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Please explain how to set up this problem as an algebraic equation.
"Two cars started to move at the same time in the same direction, but one was moving twice as fast as the other. Six hours later, the cars were 204 miles apart. Find the speed of each car."
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Faster car DATA:
time = 6 hrs ; rate = 2x ; distance = 6(2x) miles
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Slower car DATA:
time = 6 hrs ; rate = x ; distance = 6x miles
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Equation:
12x - 6x = 204 miles
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6x = 204
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x = 38 mph (slower car speed)
2x = 76 mph (faster car speed)
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Cheers,
Stan H.
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Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
Please explain how to set up this problem as an algebraic equation. "Two cars started to move at the same time in the same direction, but one was moving twice as fast as the other. Six hours later, the cars were 204 miles apart. Find the speed of each car."
I'm not sure the "how" or "why" of the set up. Thank you for your help.
Ian
Let the speed of the slower car be S
Then the speed of the faster car is: 2S
In 6 hours, the slower car covered 6(S), or 6S miles
In 6 hours, the faster car covered 6(2S), or 12S miles
Since they were 204 miles apart, and obviously travelling AWAY from each other, in the same direction,
the faster car would cover the distance the slower car has covered, plus 204 miles. This is illustrated as:
 . I assume you know what to do from here. This is as far as I’ll go since you want the SETUP.
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