Two buses leave towns 304 miles apart at the same time and travel toward each other. One bus travels 14 mih slower than the other. If they meet in 2 hours, what is the rate of each bus?

The faster bus moves at 83mi/h and the slower one moves at 69mi/h.

Step-by-step explanation:

Let's define:

R₁ = rate of bus 1, this is the faster one.

R₂ = rate of bus 2, this is the slower one.

We know that one bus travels 14mi/h slower, then:

R₂ = R₁ - 14mi/h.

Now we know that:

Distance = Speed*Time.

If we add the distances that both busses travel in 2 hours, it should be equal to the initial distance between the buses, then:

R₁*2h + R₂*2h = 304 mi

Then we have the two equations:

R₂ = R₁ - 14mi/h

R₁*2h + R₂*2h = 304 mi

The first step is to replace the first equation in the second one, to get:

R₁*2h + (R₁ - 14mi/h)*2h = 304 mi

And now we can solve this for R₁.

R₁*2h + R₁*2h - 14mi/h*2h = 304 mi

R₁*4h - 28mi = 304mi

R₁*4h = 304mi + 28mi = 332mi

R₁ = 332mi/4h = 83mi/h

The faster bus moves at 83mi/h

And we know that the slower one moves at 14mi/h slower than this, then:

R₂ = R₁ - 14mi/h = 83mi/h - 14mi/h = 69 mi/h

the answer is 5/14.

hope this .

the correct answer is c