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Please help - average velocity question

  1. Sep 12, 2007 #1


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    1. The problem statement, all variables and given/known data

    In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.68m/s, due west, turns around, and hikes with an average velocity of 0.447 m/s, due east. How far east did she walk?

    2. Relevant equations

    3. The attempt at a solution
    v1=2.68m/s west
    d1=6440m west

    v2=0.477m/s east

    At this point I notice I have two variables? t2? and d2? And I do not where to continue from here.

    please help me thanks in advance
    Last edited: Sep 12, 2007
  2. jcsd
  3. Sep 12, 2007 #2
    Well having two variable means you need two equations to solve this, right?

    Your equation for average velocity turns out to be (- is west, + is east)

    -1.34\frac{m}{s} = \frac{displacement}{t_1 + t_2}

    Where [tex]t_1, t_2[/tex] are your independent time it takes to travel west/east from given information.

    And displacement = [tex]d_1-d_2[/tex]

    You've managed to find [tex]t_1[/tex], which is 2388.0 sec. and [tex]d_1[/tex] is given to you. Plug that into the equation above.

    You have another equation for the distance travelled east, which is

    d_2 = 0.447 * t_2

    Now you have two equations and two unknowns! Rest is up to your math.
  4. Sep 12, 2007 #3


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    Ahh im going to have to use a system of equations i see..
    thanks a lot..ill post back once i solve it
  5. Sep 12, 2007 #4
    Unless if you know Calculus :D There's more of a dynamics-approach of solving this.
  6. Sep 12, 2007 #5


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    awesome thanks so much i got the answer - this place will probably be another home for me during this physics semester :) haha thanks again man
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