A car travels 50 km East, then 30 km North, then 25 km at 30 degrees East of North. Sketch the vector diagram and determine the total displacement.

Let East be the positive x-direction and North be the positive y-direction.

For the first displacement, $\mathbf{D_1}$: 50 km East

$$D_{1x} = 50 \text{ km}$$ $$D_{1y} = 0 \text{ km}$$

For the second displacement, $\mathbf{D_2}$: 30 km North

$$D_{2x} = 0 \text{ km}$$ $$D_{2y} = 30 \text{ km}$$

For the third displacement, $\mathbf{D_3}$: 25 km at $30^\circ$ East of North.
This angle is measured from the North axis. To find the x-component (East), we use $\sin(30^\circ)$, and for the y-component (North), we use $\cos(30^\circ)$.

$$D_{3x} = 25 \sin(30^\circ) = 25 \times 0.5 = 12.5 \text{ km}$$ $$D_{3y} = 25 \cos(30^\circ) = 25 \times \frac{\sqrt{3}}{2} \approx 25 \times 0.8660 = 21.65 \text{ km}$$

Total East displacement ($D_x$):

$$D_x = D_{1x} + D_{2x} + D_{3x} = 50 + 0 + 12.5 = 62.5 \text{ km}$$

Total North displacement ($D_y$):

$$D_y = D_{1y} + D_{2y} + D_{3y} = 0 + 30 + 21.65 = 51.65 \text{ km}$$

The magnitude of the total displacement, $D_{total}$, is found using the Pythagorean theorem:

$$D_{total} = \sqrt{D_x^2 + D_y^2}$$ $$D_{total} = \sqrt{(62.5)^2 + (51.65)^2}$$ $$D_{total} = \sqrt{3906.25 + 2667.7225}$$ $$D_{total} = \sqrt{6573.9725}$$ $$D_{total} \approx 81.08 \text{ km}$$

The direction $\theta$ is found using the arctangent function:

$$\theta = \arctan\left(\frac{D_y}{D_x}\right)$$ $$\theta = \arctan\left(\frac{51.65}{62.5}\right)$$ $$\theta = \arctan(0.8264)$$ $$\theta \approx 39.56^\circ$$

Since both $D_x$ and $D_y$ are positive, the direction is North of East.

The total displacement of the car is $\boxed{81.08 \text{ km at } 39.56^\circ \text{ North of East}}$.