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Explanation to this velocity equation

  1. Sep 3, 2009 #1
    I have already done this problem enough times forwards and backwards to get the answer, so this is more of a theory question rather than help with homework.

    The equation for Vavg.=displacement/time or deltaX/deltaT.
    This equate to V=(X2-X1)/(T2-T1), correct?

    In working through a problem, the correct version of this equation ended up being:
    V=(X1-X2)/(T1+T2).

    Here's the question: In reaching her destination, a backpacker walks with an average velocity of 1.34m/s, due west. This average velocity results because she hikes for 6.44km with an average velocity of 2.68m/s, due west, turns around, and hikes with an average velocity of 0.447m/s, due east. How far east did she walk?

    In fact, here's a link to the solution to the problem that I found:
    http://www.cramster.com/physics-answers-5-315022-cpi0.aspx

    My question is this: How was that form of the equation derived? I solved this many times using the standard equation and was doing the entire procedure correctly...but was getting the wrong answer as I was using my initial equation rather than the version I have mentioned here. Can someone explain to me what the reasoning is behind this?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Sep 3, 2009 #2
    [tex]v_{avg}=\frac{x_2-x_1}{t_2-t_1}[/tex]
    [tex]x_2[/tex] is your final position and [tex]x_1[/tex] is your initial position.
    [tex]x_1=0[/tex]
    [tex]x_2=6.44-d[/tex]
    For time [tex]t_1[/tex] is your initial time and [tex]t_2[/tex] is your final time.
    [tex]t_1=o[/tex]
    [tex]t_2=d_1v_1+d_2v_2[/tex]
     
  4. Sep 3, 2009 #3
    Are you interchanging x and d for displacement?
    I'm still not following how Vavg=(deltaX)(deltaT) or V=(X2-X1)/(T2-T1) turned into:
    V=(X1-X2)/(T1+T2)
     
  5. Sep 3, 2009 #4
    |----------x-----|
    You walk 10m right then 3m back.
    Now we find displacement, (10-3)-0 = 7.

    Another thing to point out. The guy who solved the formula has X1 = distance traveled in the first walk and X2 equals distance travled in the second walk. So if you walk X1 units right, and X2 units left, the displacement would be X1-X2. In the formula you're so interested in you have an X2 and X1 but they do not equal the X1 and X2 he has defined. The same with t, t1 = time of first walk and t2 = time of second walk.
     
  6. Sep 3, 2009 #5
    Okay, your diagram somewhat clears the water for me. So the displacement is 7 because your initial location was 0 and your final location was 7...which would work for Xf - Xo also. However, when writing the motion your final position is the 10-3 because you come back over the path you've already traveled, nulling its quantity? That clears up the displacement aspect of it for me.

    Now, for the average time. Why is it added together in this example rather than being Tf-To like normal? Is it because your final travel time is actually the result of the time it takes you to make the westward hike + the eastward walk? Similarly to the displacement equation, above, Tf=(T1+T2)-To?
     
  7. Sep 3, 2009 #6
    You first have to decide on a sign convention; which direction is positive and which is negative? Since the question looks only for a magnitude, it does not matter which convention you choose, just so long as you stick to it.
     
  8. Sep 3, 2009 #7
    Yeah.
    [tex]\Delta t = t_f - t_o [/tex]
    Your total time, tf, is the time of walk 1 + walk 2. And the time that you start this at is 0.
    [tex]\Delta t = T_1+t_2[/tex]
     
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