Step 1: Set up the relative velocity picture.
The rain falls straight down with some speed $v_r$. The car moves horizontally at $40$ km/h. Sitting inside the car, the passenger sees the rain as the rain velocity minus the car velocity, which adds a backward horizontal piece to the downward rain.
Step 2: Build the right triangle.
The velocity of rain with respect to the car has a vertical side equal to the rain speed $v_r$ and a horizontal side equal to the car speed $40$ km/h. The trace on the window leans at $30^\circ$ from the vertical.
Step 3: Use the angle to find $v_r$.
The angle from the vertical satisfies $\tan 30^\circ=\dfrac{\text{horizontal}}{\text{vertical}}=\dfrac{40}{v_r}$. Since $\tan 30^\circ=\dfrac{1}{\sqrt3}$, we get $\dfrac{1}{\sqrt3}=\dfrac{40}{v_r}$, so $v_r=40\sqrt3$ km/h.
Step 4: Find the magnitude of the relative velocity.
The relative speed is the hypotenuse of the triangle: \[ V=\sqrt{(40)^2+(40\sqrt3)^2} \]
Step 5: Compute the value.
Inside the root, $40^2=1600$ and $(40\sqrt3)^2=1600\times 3=4800$. Adding, $1600+4800=6400$, and $\sqrt{6400}=80$.
Step 6: Conclude.
The rain moves relative to the car at a speed of $80$ km/h.
\[ \boxed{80\ \text{km h}^{-1}} \]