How Far East Did the Backpacker Walk?

[crono_]

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

x₁ = 6.44 km = 6440 m

v_avg = 2.68 m/s, due west

t₁ = 6440 m / 2.68 m/s = 2402 s

2nd Stretch

x₂ = ?

v_avg = 0.447 m/s, due east

t₂ = ?

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

Homework Equations

v_avg = \Deltax / \Deltat

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 = x₁ - x₂ / t₁ - t₂

1.34 m/s = 6440 - x₂ / 2403 s + (x₂ / 0.447 m/s)

The next step was to solve for x₂ but I'm stuck here as x₂ 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!

 

[rl.bhat]

1.34 m/s[2403 s + 2.237*x2s] = 6440 - x2.
1/0.447 = 2.237
Now remove the bracket in LHS and collect the terms of x2 from both sides and solve for x2.

 

[crono_]

Ok...

Why are we 1/0.447? Where does the 1 come from?

 

[rl.bhat]

x2/O.447 = x2*(1/0.447)