In reaching her destination, a backpacker walks with an average velocity of 1.22 m/s, due west. this average velocity results because she hikes for 6.44 km with an average velocity of 2.53 m/s, due west, turns around, and hikes with an average velocity of 0.445 m/s, due east. how far east did she walk?
can someone me ? i really really need on this

Wow! 2.95m/s is a mighty fast pace for a backpacker. Must have one of those Star Wars anti-gravity packs. Also, I would be curious as to why she passed her destination and then walked back. 
Anyway, it goes like this: 
Say the time walking east is 't', and the total time is 'T'. 
Then 5340 m + .511 t = 1.43 T 
(This assumes that velocity is positive in both directions) 
Two unknowns in one equation. But you also know that the time spent walking west is 
5340m/ 2.95m/s = 1810 sec. 
and the total time T = 1810 +t 
Substitute this into the first equation, and you can solve for t = 3092 sec. 
Then T = 4902 sec. and distance walked east is .511t = 1580m.

Iwannasay c is the most accurate