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Applied Linear Equations: Distance Problem - Problem 1 3,386 views

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company


So, here is an example of another distance, rate, time problem that you may come across, the good old train problems that everybody loves to hate. A train leaves San Francisco and travels North, at 40 miles per hour. Another train leaves at the same time, and travels South at 80 miles per hour, how long until they are 540 miles apart.

So, this is a classic example of two rates, sort of going opposite each other and trying to figure out when they're distance apart. There are two ways of doing this. I will show you the way I sort of think about it, which is a very logical approach, and then, we can also think about it, from a more mathematical mind as well.

A train travels North at 40 miles per hour, after an hour, how far is it it gone? 40 miles. A train travels South at 80, after an hour, it has gone 80 miles, so in hour one, how far apart are they? One goes 40, one goes 80 the opposite directions they go 120 total.

So, I want to sort of see a visual. One goes North, one goes South, here is 40 and here is 80, together they've gone 120 miles. After two hours, this one has gone another 40, this one has gone another 80, so basically, they're going to double this distance, 240, so every hour, they actually move 120 miles apart.

Going to the equation that relates everything, distance is equal to rate times time, the rate that they're moving apart is the sum of these two things which is just 120 and we're asking when they are 540 miles apart, equal to distance.

From here just a very straight forward relationship, divide by 120, let's go to our calculator 540 divided by 120, t is 4 point 5 hours. So that's from a logic perspective. If you want to do a little bit more of a mathematical approach. How you can look at this is distance of train one, the distance of train two is equal to the total distance, we somehow solve this.

We know that the total distance needs to be 540, we know that distance is equal to rate times time, so this is rate times time train one, rate times time train two. The rate of the first train is 40, the rate of the second train is 80 and we know that they leave at the same time so this is the same t, t and t equal to 540. Combine like terms 40t plus 80t will end up giving us 120t and we end up solving it out the same exact way.

So one way a little bit more logical, one way a little bit more mathematical, either way we'll get the same answer.