Two buses leave a station at the same time and travel in opposite directions. One bus travels $18\frac{\mathrm{mi}}{\mathrm{h}}$ slower than the other. If the two buses are 472 miles apart after 4 hours, what is the rate of each bus?

Step 1 ：Let's denote the speed of the faster bus as $x$ and the speed of the slower bus as $x-18$.

Step 2 ：Since the two buses are moving in opposite directions, their speeds are added together. Therefore, in 4 hours, they will cover a distance of $4x+4(x-18)$ miles.

Step 3 ：We know that this distance is 472 miles, so we can set up the equation $4x+4(x-18)=472$.

Step 4 ：Solving this equation gives us $x=68$. This is the speed of the faster bus.

Step 5 ：Subtracting 18 from this speed gives us the speed of the slower bus, which is $68-18=50$.

Step 6 ：Final Answer: The speed of the faster bus is $\boxed{{\displaystyle 68\frac{\mathrm{mi}}{\mathrm{h}}}}$ and the speed of the slower bus is $\boxed{{\displaystyle 50\frac{\mathrm{mi}}{\mathrm{h}}}}$.