Question

In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, because she hikes for 6.30 km with an average velocity of 2.49 m/s due west, turns around, and hikes with an average velocity of 0.630 m/s due east. How far east did she walk (in kilometers)?

Answer 1

The total average velocity $v=+1.41 m/s$ (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 S is the algebraic sum of the displacement in the first part of the motion ($S_1=6.30 km=6300 m$, due west) and of the displacement in the second part of the motion ($S_2$, due east).
-The total time taken t 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:
$t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s$

$t_2$, instead, can be written as $\frac{S_2}{v_2}$, where $v_2=-0.630 m/s$ 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:
$v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }$
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
$S_2=-844 m=-0.844 km$