# Kinematics in 1D Question

1. Sep 21, 2009

### crono_

1. The problem statement, all variables and given/known data

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:

vavg = 1.34 m/s, due west. For whole trip.

1st Stretch

x1 = 6.44 km = 6440 m

vavg = 2.68 m/s, due west

t1 = 6440 m / 2.68 m/s = 2402 s

2nd Stretch

x2 = ?

vavg = 0.447 m/s, due east

t2 = ?

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

2. Relevant equations

vavg = $$\Delta$$x / $$\Delta$$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:

vavg = 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!

2. Sep 21, 2009

### 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.

3. Sep 21, 2009

### crono_

Ok....

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

4. Sep 21, 2009

### rl.bhat

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

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