The total average velocity [tex]v=+1.41 m/s[/tex] (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
[tex]v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2} [/tex]
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion ([tex]S_1=6.30 km=6300 m[/tex], due west) and of the displacement in the second part of the motion ([tex]S_2[/tex], due east).
-The total time taken t is the time taken for the first part of the motion, [tex]t_1[/tex], and the time taken for the second part of the motion, [tex]t_2[/tex]. [tex]t_1[/tex] can be found by using the average velocity and the displacement of the first part:
[tex]t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s [/tex]
[tex]t_2[/tex], instead, can be written as [tex] \frac{S_2}{v_2} [/tex], where [tex]v_2=-0.630 m/s[/tex] 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:
[tex]v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} } [/tex]
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
[tex]S_2=-844 m=-0.844 km[/tex]
