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?
Answer
Final Answer: The speed of the faster bus is \(\boxed{68 \frac{\mathrm{mi}}{\mathrm{h}}}\) and the speed of the slower bus is \(\boxed{50 \frac{\mathrm{mi}}{\mathrm{h}}}\).
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 \(\boxed{68 \frac{\mathrm{mi}}{\mathrm{h}}}\) and the speed of the slower bus is \(\boxed{50 \frac{\mathrm{mi}}{\mathrm{h}}}\).
