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Kinematics Problem, seemingly missing information

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data

    In reaching her destination, a backpacker walks with an average velocity of 1.36 m/s, due west. This average velocity results, because she hikes for 5.18 km with an average velocity of 2.05 m/s due west, turns around, and hikes with an average velocity of 0.699 m/s due east. How far east did she walk (in kilometers)?

    2. Relevant equations

    avg. v = distance/time

    3. The attempt at a solution

    the closest i was able to get to a solution was

    1.36 = (5180 meters + distance of second part of walk)/(2526.829268 + time of second part of walk)

    as far as i can tell, there doesn't seem to be enough information for me to figure out the question.
    Last edited: Aug 31, 2008
  2. jcsd
  3. Aug 31, 2008 #2


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    Homework Helper

    Rewrigh the above step as
    1.36 = (5180 meters + distance of second part of walk)/(2526.829268 +
    distance of second part of walk/0.699) and solve the equation for distance of the second part of walk.
  4. Sep 1, 2008 #3


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    Homework Helper

    You need to remember that velocity is a vector, and that Avg velocity in this case is total displacement from the start - not total distance - divided by the total time.

    The eastward walk after the westward should be treated as minus not plus, even though its contribution to total time will be plus.

    [tex]1.36 = \frac{5180 - Dist}{\frac{5180}{2.05} + \frac{Dist}{.699}}[/tex]
    Last edited: Sep 1, 2008
  5. Sep 1, 2008 #4
    thanks for the help guys
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