Solving Speed & Velocity: Backpacker Walks 5.58km West & East

[pookisantoki]

Homework Statement

In reaching her destination, a backpacker walks with an average velocity of 1.09m/s, due west. This average velocity results, because she hikes for 5.58km with an average velocity of 2.30m/s due west, turns around, and hikes with an average velocity of .430 m/s due east. How far east did she walk ( in Km)

I have no idea on how to even go about this problem. I do know that average velocity = displacement/ elasped time... Like would i be solving it out three different times or plug it all in? Please let me knwo how to approach this problem Thank you!

 

[rl.bhat]

Find the time taken by her to hikes for 5.58km with an average velocity of 2.30m/s due west. Let d be the distance she hikes due east. Average velocity is given. Find the time to hike this distance in terms of d.
Now what is her net displacement?
What is the total time?
Net average velocity is given. Find d, and then the displacement.

 

[pookisantoki]

So this is what i have following what you wrote, I'm not sure if I understood it all.

2.30m/s due west= 5580meter/t
t=.000412186

.430m/s=d/.00412186
d=.00017724

Did I do this correctly?? If so, I am not sure on what to do after this step.

 

[rl.bhat]

v = d/t.
Then t = d/v
Check your maths.

 

[pookisantoki]

I'm still getting it wrong...
.0023km/sec= 5.58/t
t=5.58/.0023
t=2426.0869sec

.00043=d/2426.0869sec
d=1.04321