Unit
Linear Equations and Inequalities
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Now we're going to look at another distance, rate, time problem, but, this one, we're actually dealing with things, going towards each other. So, a train leaves New York city, headed towards LA, travelling 120 miles per hour. At the same time, another train leaves LA, heads towards New York at 180. If the two cities are 2400 miles apart, how long until they meet.
So, let's look at this. We're first going to do this from a logical standpoint. We have basically two cities, pretty far away. We have, a train that leaves LA at 180 miles per hour, a train that leaves New York city, at 120 miles per hour. After one hour, this train has gone 180, after one hour, this train has gone 120 miles, so after one hour, they're actually 300 miles closer, after two hours, double that, 600, after three hours, triple, 900. Every hour, they are getting 300 miles closer together.
So, what we actually end up to solve, is just, distance is equals to rate times time. The distance we're concerned about is, the distance they're apart 2400. The rate that they're going together, is just the sum of their two individual rates, which is 300, and then the time. Solve this out, divide by 300. 2400 divide by 300 is 8, unit of time, hours.
So this is more of a logical approach, if you also do more of the mathematical approach which we know that the distance of one train plus the distance of the other is equal to the distance in total. The distance of one train is just rate times time, distance of the second train is just rate times time and the distance in total we know to be 2400.
We know that these times are equivalent because they leave at the same time and they meet at the same time. So our rate of our first train is 120 and our time is just t, then the rate of our second train is 180 and our total distance is still 2400. If we then combine like terms 120 plus 180 is actually 300t which actually takes us back to where we were over here divide by 300, once again we would end up with t equals 8. Two approaches, both are completely valid, either one works.
