## In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west.

Question

**In reaching her destination, a backpacker walks with an average velocity of 1.27 m/s, due west. This average velocity results, because she hikes for 5.89 km with an average velocity of 2.18 m/s due west, turns around, and hikes with an average velocity of 0.555 m/s due east. How far east did she walk (in kilometers)?**

solved
0

General Physics
4 years
1 Answer
1078 views
0
## Answer ( 1 )

You need to find the distance east, you know the velocity east so you just need to find the time spent traveling east, using the average velocity formula

take west as positive direction, east as negative

V = 1.27m/s

v1 = 2.18m/s

t1 = 5890/2.18 = 2702s …..time traveling west

v2 = -0.555m/s

t2 = ?………………………….time traveling east

d1 = v1t1 = 5890m ……….displacement west

d2 = v2t2 = -0.555t2………displacement east

final displacement is

D = d1 + d2 = 5890 – 0.555t2

the total time of travel is

T = t1 + t2 = 2702 + t2

the average velocity is

V = 1.27 = D/T= (5890 – 0.555t2)/ (2702 + t2)

gives

1.27(2702 + t2) = (5890 – 0.555t2)

expand and rearrange for t2

t2 = (5890 – 1.27*2702) / (1.27 + 0.555) = 1347s

is the time spent traveling east

so the distance traveled east is

|d2| = |v2t2| = 0.555*1347 = 748m = 0.748km