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**1. Homework Statement**

Problem taken from Cutnell & Johnson 8th Edition Textbook (though it's also in the 7th edition as well, can't recall question # though). It is from chapter 2, question # 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?

**Known Data:**

v

_{avg}= 1.34 m/s, due west. For whole trip.

__1st Stretch__

x

_{1}= 6.44 km = 6440 m

v

_{avg}= 2.68 m/s, due west

t

_{1}= 6440 m / 2.68 m/s = 2402 s

__2nd Stretch__

x

_{2}= ?

v

_{avg}= 0.447 m/s, due east

t

_{2}= ?

I took the left direction as positive and the right negative.

**2. Homework Equations**

v

_{avg}= [tex]\Delta[/tex]x / [tex]\Delta[/tex]t

**3. The Attempt at a Solution**

This is where I get lost. I looked up some online solutions but was unable to follow them. One of them started:

v

_{avg}= 1.34 m/s = x

_{1}- x

_{2}/ t

_{1}- t

_{2}

1.34 m/s = 6440 - x

_{2}/ 2403 s + (x

_{2}/ 0.447 m/s)

The next step was to solve for x

_{2}but I'm stuck here as x

_{2}is in the equation twice. How do you solve for something that's in two parts of the equation? Factoring didn't seem to work, granted I may have done it incorrectly.

I hope this makes sense, it's a lot of work to type up everything in these forums. But the help is usually worth it. Any assistance would be greatly appreciated.

Thanks in advance!