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Chase problems!

  1. May 26, 2008 #1
    omg i have so much trouble doing these problems... can anyone help? and show process in which you took? thank you so much!

    a man starts 10m ahead of a finish line and races at a velocity of 10m/s. A women rides a bike from 0m (starting line) and accelerates 4m/s(square). when and where do they meet?
  2. jcsd
  3. May 26, 2008 #2


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    Welcome to PF!

    Hi napoodo! Welcome to PF! :smile:
    Isn't that typical of a man … starting at the finish line! :wink:

    I assume you mean that they started at the same time?

    Just write an equation in x and t for each of them:

    x = something t

    x = something else t​

    Then they meet when their x is the same, so you just write something t = something else t, which is an equation only in t, which you can solve! :smile:

    (if you're still having difficulty, show us what you've done so far, so that we know how to help)
  4. May 26, 2008 #3
    ok well i tried, and this is what i got. [Of woman]==> at=x=d/t <===[Of man] aka a=v/t and v = d/t and i put v is the same...

    4t = 10/t
    4t(square) = 10/4
    (square root) t (square) = (squareroot) 6
    t = 2.5s?
    would that time be right?
    or do you need the same equation to make it equal each other O_O because one side uses velocity, and time, and the other side uses acceleration and time

    shouldn't d(distance) be the same? = both
    Last edited: May 26, 2008
  5. May 26, 2008 #4


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    Hi napoodo! :smile:

    I'm sorry, I don't understand any of that. :confused:

    You really do need to write out equations beginning with "x ="

    What is the equation for the man? :smile:

    (going to be now :zzz:)
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