Hello, symmetry!
Two cars are approaching an intersection.
One is 2 miles south of the intersection and is moving at 30 mph.
The other car is 3 miles east of the intersection and is moving at 40 mph.
Express the distance $\displaystyle d$ between the cars as a function of time $\displaystyle t$. Code:
C 3  40t B 40t Q
*         *     *
 *
 *
2  30t  * d
 *
 *
A*

30t 

P*
The first car starts at $\displaystyle P$ and drive north to the intersection $\displaystyle C$.
In $\displaystyle t$ hours, it has gone $\displaystyle 30t$ miles to point $\displaystyle A$.
. . Hence, $\displaystyle AC \:=\: 2  30t$
The other car starts at $\displaystyle Q$ and drives west to $\displaystyle C$.
In $\displaystyle t$ hours, it has gone $\displaystyle 40t$ miles to point $\displaystyle B$.
. . Hence, $\displaystyle BC \:=\:3  40t$
Using Pythagorus, the distance between them is: .$\displaystyle d^2\;=\;(230t)^2 + (340t)^2$
Simplifying, we get: .$\displaystyle d^2\;=\;2500t^2  360t + 13$
Therefore: .$\displaystyle \boxed{d \;=\;\sqrt{2500t^2  360t + 13}} $