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-   -   Quadratic speed distance time (https://www.askmehelpdesk.com/showthread.php?t=265806)

  • Oct 2, 2008, 01:12 PM
    duffman11
    Quadratic speed distance time
    A woman begins jogging at 3pm, running due north at a 6 minute/mile pace. Later she reverses direction and runs due south at a 7 minute/mile pace. If she returns to her starting point at 3:45, find the total number of miles run.

    I usually make a chart for Speed, Distance, Time problems such as these.
    S D T
    A 6min/mile x x/6min/mile t=d/s


    B 7min/mile x x/7min/mile

    Let x be the distance travelled by the woman.
    I figure X is = to both distance A and B since she runs the same distance there and back.
    The speeds are given and to find time I divide distance by speed. Here's what I get for an equation:

    6x+7x=45

    I've tried it several ways so far and I just can't seem to get the right equation. Where am I going wrong!
  • Oct 2, 2008, 06:04 PM
    galactus
    This problem is worded wacky. Be careful of units. It is in minutes per mile instead of miles per minute.

    d=rt. If it takes 6 minutes to run a mile, that is 10 mph.

    If it takes 7 minutes to run a mile on the way back, then that is about 8.57 mph.

    45 minutes is 3/4 hours.

    6t=d on the way up.

    The distances are the same up and back.

    So, the distance back is 7(3/4-t)
  • Oct 30, 2010, 11:31 AM
    mygod123456789
    ERISON VALDEZ ANSWER

    MINUTES=MT START RUNNING AT 3:00PM NORTH
    NORTH=N 3:06=1mlN
    SOUTH=S 3:12=2mlN
    MILES=ml 3:18=3mlN
    3:24=4mlN
    REVERSE TO BEGIN SOUTH
    3:31=1mlS
    3:38=2mlS
    3:45=3mlS
    TOTAL MILES RAN IN 45 MINUTES=
    4mlN+3mlS=7ml in 45 minutes
  • Oct 30, 2010, 11:34 AM
    mygod123456789
    Comment on mygod123456789's post
    Nicw neatly understanding

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