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A difficult 1D kinematics problem dealing with displacement and velocity

  1. Sep 9, 2009 #1
    In reaching her destination, a backpacker walks with an average velocity of 1.31 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.54 m/s, due west, turns around, and hikes with an average velocity of 0.475 m/s, due east. How far east did she walk?


    Is there not enough information? I feel like they should have given me another variable or something.


    I figured out that it took her 2535.433 seconds to hike for the 6.55km stretch. But I'm really in the dark on this one, I appologize. I've looked all over the internet to try and find a problem simular to this one but I had no luck.
     
  2. jcsd
  3. Sep 9, 2009 #2

    Hootenanny

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    Welcome to Physics Forums.

    There is sufficient information to solve the problem.

    HINT: What is the definition of average velocity?
     
  4. Sep 9, 2009 #3
    Distance/time = average velocity

    D/T = 1.31

    t1 = time walking west
    t2 = time walking east
    d1 = distance walked west = 6.44 km = 6440 meters
    d2 = distance walked east

    T = t1 + t2
    D = d1 - d2

    t1 = distance/speed = 6440/2.54 = 2535.4 seconds

    t2 = d2/v2 = d2/0.475
    so d2 = 0.475*t2

    Now use all this in the equation for D/T.
    D/T = (d1 - d2)/(t1 + t2) = 1.31
    (6440 - 0.475*t2)/(2535.4 + t2) = 1.31
    6440 - 0.475*t2 = 3321.42 + 1.31*t2
    1.786*t2 = 3118.58
    t2 = 1747.11 seconds

    d2 = t2*v2 = 829.87

    Check by calculating the displacement and time and then seeing if we get the average velocity we started with.
    displacement = d1 - d2 = 5610.13 meters
    time = t1 + t2 = 4282.54 seconds
    average speed = displacement/time = 5610.13/4282.54 = 1.31 m/s

    So she walked 5610.13 meters (or 5.61 km) east



    but my answer is still wrong... :(
     
  5. Sep 9, 2009 #4
    What you did is correct. The question is asking for how far east she traveled. That would be d2, 829, not 5610 which is the total displacement which 5610m west.
     
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