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Math Help - Hard word problem, equation with 2 variables?

  1. #1
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    Post Hard word problem, equation with 2 variables?

    the problem says: cyclists lance and alberto each average 30km/h for the first hour of a 100 km race. At the end of the first hour lance crashes and loses 12 mins. alberto finishes the remainder of the race at an average of 25 km/h and lance averages 27.5 km/h after starting up again. How long after the START of the race will lance catch up?

    I have been trying and everyone i know cant get it. i have an answer but i have nothing to base it off of to see if it is right. I would include my work but it is all over and trying to put it into words in a forum would be hard.

    Thanks
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  2. #2
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    alberto finishes the remainder of the race at an average of 25 km/h
    Lets convert everything into m/s to make it easier. 25km/h = 6.9444... or 125/18 meters per second. He travels for x amount of seconds.


    lance averages 27.5 km/h after starting up again
    27.5km/h = 7.638888.... or 275/36 meters per second. He trevels for x amount of seconds to catch up, minus 12 minutes (720 seconds) of travel time.

    This is only a single variable question.

    Therefore: \dfrac{125x}{18} = \dfrac{275x}{36} - 720

    Solve for x, which is the time travelled in seconds after lance crashes. Then add on the extra hour (3600 seconds) and you will have your total time after the start of when lance will catch up.
    Last edited by Educated; September 18th 2010 at 10:55 PM. Reason: Time in seconds
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