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Kinematics in one dimension

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data
    10. 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?
    (Cutnell, John D.. Physics, 8th Edition. John Wiley & Sons, 012009. p. 53).
    <vbk:9780470458754#page(53)>

    2. Relevant equations

    3. The attempt at a solution
    2.68 m/s2 = 6.44 km/tw
    .447 m/s2 = De/te
    Avg V = (X - X0)/(t - t0)

    The answer is 0.81 km, but I can't figure out how to solve this.
     
    Last edited: Oct 7, 2011
  2. jcsd
  3. Oct 6, 2011 #2
    To solve this prob we have to use concept of avg velocity and its formula avg velocity =total distance by total time.1st convert all kms into metres and use the equation 1.34=6440-x/2400+x/0.447 where x is distance travelled in east.solve x and you will get x=.81 approx.Now you can head for my question
     
  4. Oct 7, 2011 #3
    i got 2400=6440/2.68
     
  5. Oct 7, 2011 #4
    I'm not quite sure what you're saying. Can you break it down a bit more and maybe format better? I can't tell if you are saying 1.34 = [(6440-x)/(2400+x)]/0.447 or 6440 - (x/2400) + (x/0.447) or some other variation
     
  6. Oct 7, 2011 #5
    I M SAYING 1.34=6440-x/[2400+x/0.447]
     
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