Question

hapter 02, Problem 10 In reaching her destination, a backpacker walks with an average velocity of 1.15 m/s, due west. This average velocity results, because she hikes for 6.07 km with an average velocity of 2.78 m/s due west, turns around, and hikes with an average velocity of 0.675 m/s due east. How far east did she walk (in kilometers)?

Answer 1

The total average velocity $v=1.15 m/s$ (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
$v= \frac{S}{t}=\frac{S_1+S_2}{t_1+t_2}$

where:
-The total displacement is the algebraic sum of the displacement in the first part of the motion ($S_1 = +6.07 km=6070 m$, due west) and of the displacement in the second part of the motion ($S_2$, due east).
-The total time taken is the time taken for the first part of the motion, $t_1$, and the time taken for the second part of the motion, $t_2$. $t_1$ can be found by using the average velocity and the displacement of the first part:
$t_1 = \frac{S_1}{v_1}= \frac{6070 m}{2.78 m/s}=2183 s$
$t_2$, instead, can be written as $t_2= \frac{S_2}{v_2}$, where $v_2=-0.675 m/s$ is the average velocity of the second part of the motion (with a negative sign, since it is due east). 
Therefore, we can rewrite the initial equation as:
$v= 1.15 =\frac{6070+S_2}{2183- \frac{S_2}{0.675} }$
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
$S_2 = -1318 m=-1.32 km$