Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors [tex]\vec{i}\textrm{ and }\vec{j}[/tex] as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors [tex]\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}[/tex] in the directions North, [tex]45^{o}[/tex] South of West and [tex]60^{o}[/tex] North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector = [tex]4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}[/tex]
Magnitude of Displacement = [tex]|-4.25\vec{i}+3.165\vec{j}|=5.3km[/tex]
∴ Total displacement = [tex]5.3km[/tex]