How Far East Did the Backpacker Walk?

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The discussion revolves around calculating the distance a backpacker walked east after initially traveling west. The backpacker walks 6.44 km west at an average velocity of 2.68 m/s, then turns around and walks east at 0.447 m/s. The average velocity for the entire journey is 1.34 m/s due west. Participants clarify the equations needed to find the eastward distance, emphasizing the correct formulation of the relationship between the distances and velocities. The final expression for the eastward distance is provided, correcting a previous misunderstanding in the calculations.
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Homework Statement



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 0.447m/s, due east. How far east did she walk?

Homework Equations


The Attempt at a Solution



Okay so I had some help from cramster, but I don't get the last part.

<------------ avg velocity = 1.34m/s
<-------------------- avg velocity = 2.68 m/s & distance = 6440m
-------------> v=.447m/s

Let east be positive direction

1) [avg v = xw+xe/tw+te

where tw is the time spent along the west

2) tw=xw/vw
te=xe/ve

The step I don't get is when they combine equation (1) and (2) to get

xe = ((1-v/vw) xw) / (1-v/ve)Please help!
 
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Where did you get the last equation? It is wrong. This is the right expression.

x_{e}=\frac{x_{w}(\frac{v}{v_{w}}-1)}{(1-\frac{v}{v_{e}})}

You get it from combining 1) and 2)
 
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