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Velocity and distance of walking backpacker

  1. Sep 12, 2007 #1
    1. The problem statement, all variables and given/known data
    In reaching her destination, a backpacker walks with an average velocity of 1.32m/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 .447m/s due east. How far east did she walk?

    2. Relevant equations
    Unsure. However,
    V^2 = V(initial)^2 + 2ax seems like it would yield an answer if you treat walking in the opposite direction like slowing down (as far as average velocity is concerned).

    3. The attempt at a solution

    I tried using the equation mentioned above, thinking that turning around and walking east at .447m/s was like having an acceleration of -.447m/s because her positive average velocity due west is being decreased as she walks east. Unfortunately doing so yields answers of 24.34m which seems unreasonably low.

    The other method I tried which yielded results was guess and check, using that she walked 6440m west at 2.68m/s which means for a total of 2403 seconds. I tried out a few distances east she could walk and checked to see what kind of impact that would have on her average velocity, this was the best match i could find:

    6440m - 805m / 2403s + (805m/.447m/s) = 1.34m/s

    i picked 805meters east, so i subtracted that from the 6440m west since it's like she's "unwalking" that distance, for time I add the 2403seconds it took her to get to 6440m west and the time it would take her to walk those 805m east which I got by dividing the 805m by her easterly velocitiy to come up with the time it would take her to achieve that distance.

    I am wondering if the second method of guess and check and the answer it yielded seems reasonable. I am also curious if there is an easier way to do it but I was unable to get anything significant using the equations provided thus far in the course (which havent been many).
  2. jcsd
  3. Sep 13, 2007 #2


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    You shouldn't use that equation... that's only for constant acceleration...

    Hint: average velocity = (total displacement)/(total time)

    find the time she walked east first... then from that you can get how far she walked east.
  4. Sep 13, 2007 #3
    I cant find any equations that could tell me time given only an initial and final velocity, they all seem to include either the final distance (which I dont know) or the acceleration. I am sorry if I am completely missing the boat here, but I am not at all sure how I would figure out time walking east without knowing distance walking east (or vice versa).
  5. Sep 13, 2007 #4


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    Let t = the time walking east...

    In terms of t, what is the total displacement?
  6. Sep 13, 2007 #5
    It would be her velocity going east multiplied by the length of time she walks east so,
  7. Sep 13, 2007 #6


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    yes, but what is her total displacement for the entire trip...
  8. Sep 13, 2007 #7
    6440m - .447m/s(t) / 2403s + t
  9. Sep 13, 2007 #8


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    cool... that's exactly what you need... it's not the displacement... but that's the average velocity of the entire trip taking west as positive... now just solve:

    [tex]\frac{6440m - .447m/s(t)}{ 2403s + t} = 1.32[/tex]
  10. Sep 13, 2007 #9
    Thank you! That's the same equation I was using to guess and check except I never let the variable be time, always used distance instead. It's kind of funny how your mind can paint itself into a corner. Thanks again, for the help.
  11. Sep 13, 2007 #10


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    no prob.
  12. Oct 8, 2007 #11
    Can anyone explain the solving for t part?
  13. Oct 8, 2007 #12
    What exactly is initial and final velocity?
  14. Oct 8, 2007 #13

    Doc Al

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    Staff: Mentor

    First things first: Do you understand what velocity is?

    Initial and final just refer to the start and end of some interval. You start out with some initial velocity and you end up with some final velocity.
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