Kinematics Problem, seemingly missing information

[mcode]

Homework Statement

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)?

Homework Equations

avg. v = distance/time

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.

 

[rl.bhat]

1.
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.

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.

 

[LowlyPion]

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.

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.

[tex]1.36 = \frac{5180 - Dist}{\frac{5180}{2.05} + \frac{Dist}{.699}}[/tex]

 

[mcode]

thanks for the help guys