Question

Susan hikes 6 miles east, 15 miles north, 1 mile west and 3 miles south. What is her displacement?

Answer 1

Let's first find $x$- and $y$-components of the resultant displacement:

$d_x=6\ mi\cdot cos0^{\circ}+15\ mi\cdot cos90^{\circ}+1\ mi\cdot cos180^{\circ}+3\ mi\cdot cos270^{\circ}=5\ mi,$

$d_y=6\ mi\cdot sin0^{\circ}+15\ mi\cdot sin90^{\circ}+1\ mi\cdot sin180^{\circ}+3\ mi\cdot sin270^{\circ}=12\ mi.$

We can find the magnitude of the resultant displacement from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(5\ mi)^2+(12\ mi)^2}=13\ mi.$$

We can find the direction of the resultant displacement from the geometry:

$$\theta=tan^{-1}({\dfrac{d_y}{d_x}}),$$ $$\theta=tan^{-1}({\dfrac{12\ mi}{5\ mi}})=67.4^{\circ}\ N\ of\ E.$$