Answer
$15^{\circ} south\ of\ west$, $3.5km$
Work Step by Step
We can find the required direction and magnitude as follows:
$r_x=v_xt$
$\implies r_x=(-27)(125)=-3375m$
Now the displacement towards south is
$r_y=v_yt=(-14)(66)=-924m$
and the direction is given as
$\theta=tan^{-1}(\frac{r_y}{r_x})=tan^{-1}(\frac{-924}{-3375})=15^{\circ}+180^{\circ}=195^{\circ}$ or $15^{\circ}$ (south of west)
The magnitude is given as:
$r=\sqrt{(-3375)^2+(-924)^2}=3500m=3.5Km$
