A surveillance report concerning a suspected car details the...

Let the initial position of the car be the origin $(0,0)$.

The car travels $2.00 \text{ km}$ towards east, so the first displacement vector is $\vec{d}_1 = 2.00 \hat{i} \text{ km}$.

A perpendicular left turn from east is towards north. The car travels $500 \text{ m} = 0.50 \text{ km}$, so the second displacement vector is $\vec{d}_2 = 0.50 \hat{j} \text{ km}$.

A perpendicular right turn from north is towards east. The car travels $4.00 \text{ km}$, so the third displacement vector is $\vec{d}_3 = 4.00 \hat{i} \text{ km}$.

The total displacement vector is $\vec{D} = \vec{d}_1 + \vec{d}_2 + \vec{d}_3 = 2.00 \hat{i} + 0.50 \hat{j} + 4.00 \hat{i} = 6.00 \hat{i} + 0.50 \hat{j} \text{ km}$.

The magnitude of the displacement is $|\vec{D}| = \sqrt{(6.00)^2 + (0.50)^2} = \sqrt{36.00 + 0.25} = \sqrt{36.25} \approx 6.02 \text{ km}$.

The direction of the displacement is given by the angle $\theta$ north of east, where $\tan \theta = \dfrac{y}{x} = \dfrac{0.50}{6.00} = \dfrac{1}{12}$.

Thus, \theta = \tan^{-1} \left(\dfrac{1}{12}\right).