The total average velocity [tex]v=1.15 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 is the algebraic sum of the displacement in the first part of the motion ([tex]S_1 = +6.07 km=6070 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 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{6070 m}{2.78 m/s}=2183 s [/tex]
[tex]t_2[/tex], instead, can be written as [tex]t_2= \frac{S_2}{v_2} [/tex], where [tex]v_2=-0.675 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= 1.15 =\frac{6070+S_2}{2183- \frac{S_2}{0.675} } [/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 = -1318 m=-1.32 km[/tex]
