Question

 A student’s travel from his home to school is described as follows: 22 m westward, 35 m southeast, and 17 m, 15° north of east. Compute for his displacement.

Answer 1

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

$d_{res,x}=22\ m\times cos180^{\circ}+35\ m\times cos(360^{\circ}-45^{\circ})+17\ m\times cos15^{\circ}=19.2\ m,$$d_{res,y}=22\ m\times sin180^{\circ}+35\ m\times sin(360^{\circ}-45^{\circ})+17\ m\times sin15^{\circ}=-20.35\ m.$

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

$$d_{res}=\sqrt{d_{res,x}^2+d_{res,y}^2}=\sqrt{(19.2\ m)^2+(-20.35\ m)^2}=28\ m.$$

We can find the direction of student's resultant displacement from the geometry:

$$\theta=tan^{-1}(\dfrac{d_{res,y}}{d_{res,x}}),$$ $$\theta=tan^{-1}({\dfrac{-20.35\ m}{19.2\ m}})=46.7^{\circ}\ S\ of\ E.$$