Solving a distance, rate, time problem using a linear equation

Question

Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels 18 miles per hour slower than the westbound train. If the two trains are 680 miles apart after 4 hours, what is the rate of the eastbound train? Do not do any rounding.

Answer 1

Hi Dwight. The first thing you want to do in this problem is make a variable representing the speed of one of the trains. It doesn't matter which one you pick, but you need to be sure that you add (or subtract) the 18 mph where appropriate.

Let's call the speed of the westbound train "x"

Here is the information we've been given:

1. distance between trains at 4 hours = 680 miles
2. elapsed time = 4 hours
3. eastbound train is 18 mph slower than westbound train

Looking at point 3, we can write an equation for the speed of each train:

1. Westbound speed = x
2. Eastbound speed = x-18

Now we have to write an equation that brings this all together so we can solve for x. Because the trains are travelling in opposite directions, we can add their speeds together to get a rate at which the distance between them changes.

d = (Eastbound speed + Westbound speed) * (elapsed time)

d = (x + x - 18) * t

d = (2x - 18)*t

Plugging in the values at t = 4 hours:

680 = (2x - 18)*4

Divide by 4...

170 = 2x - 18

Add 18...

188 = 2x

Divide by 2...

94 mph = x

Wait! We're not done yet! Remember that x is the speed of the westbound train, and we are looking for the speed of the eastbound train. Fortunately, we have already done the hard part and can use this equation from earlier:

Eastbound speed = x-18

Thus, the speed of the eastbound train is x - 18 = 94 - 18 = 76 mph