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Basic Kinematics Problem -- Backpacker's average velocity and distance...

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem reads:
    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 2.68 m/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
    Vavg = delta X/ delta t


    3. The attempt at a solution
    I let
    vw= average velocity from hiking west
    xw = west displacement
    tw = the time it took to travel in the west direction

    ve = average velocity from hiking east
    xe = east displacement
    te = the time it took to travel in the east direction


    vw = 2.68 m/s
    xw = 6.44 km = 6440 m
    tw = 6440 m / 2.68 m/s = 2403 s

    ve = 0.447 m/s
    xe = ?
    te = ?

    using the avg. velocity equation as ve = xe / te I solve for xe where ve = 0.447 m/s
    xe = te * 0.447 m/s


    therefore according to the equation Vavg = xe-xw / te-tw
    However, I'm looking at other solutions and it shows the equation to be Vavg = xw + xe / te + te
    I thought average velocity is a change in displacement over a change in time? So why are they adding the displacement & time as opposed to subtracting? So confused...
     
    Last edited by a moderator: May 18, 2015
  2. jcsd
  3. May 18, 2015 #2

    Orodruin

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    Because the total change in time is the sum of the two times and not the difference as the times are defined:
     
  4. May 18, 2015 #3
    But every Average Velocity equation is x2-x1 / t2-t1
    at least that's what my book defines it...the change in time is the final time minus the initial time.

    Same applies for displacement but yet they are adding it here as well.
     
  5. May 18, 2015 #4

    Orodruin

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    You need to look beyond symbols and realise what they are representing rather than simply try to apply a formula.

    In this equation:
    t1 and t2 are the initial and final times, respectively.

    In the equation in your OP, the times given are not the start and end times, it is the travel times.
     
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