## Fundamentals of Physics Extended (10th Edition)

$3.189\,km$
We have three vectors representing the three parts of the path: $\vec{a} = 3.1\,km\, \hat{i}$ $\vec{b} = -2.4\,km\, \hat{j}$ $\vec{c} = -5.2\,km\, \hat{i}$ We add these up to get the total displacement: $\vec{t} = -2.1\,km\,\hat{i} - 2.4\,km\,\hat{j}$ The total distance of the displacement is: $|\vec{t}| = \sqrt{(-2.1\,km)^2+(-2.4\,km)^2} = 3.189\,km$.