Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,
<em>r</em> = (-200 km) <em>j</em>
then 80 km southwest,
<em>s</em> = (-80/√2 km) <em>i</em> + (-80/√2 km) <em>j</em>
so that its total displacement is
<em>r</em> + <em>s</em> = (-80/√2 km) <em>i</em> + ((-200 - 80/√2) km) <em>j</em>
<em>r</em> + <em>s</em> ≈ (-56.6 km) <em>i</em> + (-256.6 km) <em>j</em>
This vector has magnitude
√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km
and direction <em>θ</em> such that
tan(<em>θ</em>) = (-256.6 km) / (-56.6 km) ==> <em>θ</em> ≈ -102.4º
relative to east, or about 12.4º west of south.
