# Kinematics Problem, seemingly missing information

1. Aug 31, 2008

### mcode

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

In reaching her destination, a backpacker walks with an average velocity of 1.36 m/s, due west. This average velocity results, because she hikes for 5.18 km with an average velocity of 2.05 m/s due west, turns around, and hikes with an average velocity of 0.699 m/s due east. How far east did she walk (in kilometers)?

2. Relevant equations

avg. v = distance/time

3. The attempt at a solution

the closest i was able to get to a solution was

1.36 = (5180 meters + distance of second part of walk)/(2526.829268 + time of second part of walk)

as far as i can tell, there doesn't seem to be enough information for me to figure out the question.

Last edited: Aug 31, 2008
2. Aug 31, 2008

### rl.bhat

Rewrigh the above step as
1.36 = (5180 meters + distance of second part of walk)/(2526.829268 +
distance of second part of walk/0.699) and solve the equation for distance of the second part of walk.

3. Sep 1, 2008

### LowlyPion

You need to remember that velocity is a vector, and that Avg velocity in this case is total displacement from the start - not total distance - divided by the total time.

The eastward walk after the westward should be treated as minus not plus, even though its contribution to total time will be plus.

$$1.36 = \frac{5180 - Dist}{\frac{5180}{2.05} + \frac{Dist}{.699}}$$

Last edited: Sep 1, 2008
4. Sep 1, 2008

### mcode

thanks for the help guys