## Multiple Traveler Questions

### Transcript

Multiple travelers and multiple trips. Some motion based problems involve more than one traveler, or trips of more than one segment. We already saw a little of this in the previous video on average speed. The basic strategy is that each traveler, and each trip gets its own D=RT equation. And sometimes we have to set up multiple equations, and then use the techniques that we've learned for solving two equations with two unknowns.

Here's a practice problem. Pause the video and then we'll talk about this. All right, Martha and Paul started traveling from A to B at the same time. Martha traveled at a constant speed of 60 miles an hour, and Paul at a constant speed of 40. When Martha arrived at B, Paul was still 50 miles away.

What is the distance? So notice first of all that the time. We'll just say that time is, the time from the starting point to when Martha arrived. And so, Martha of course, traveled that whole distance D, but what about Paul? Paul didn't get all the way to D. He was still 50 miles away, so that is D minus 50.

And so the distance that we're going to use for Martha is D. The distance we're going to use for Paul is D minus 50. And so now we have something to plug in for distance, and rate, and time. So for Martha, distance D, rate 60 times T. For Paul, D minus 50, because at that same time he was 50 miles short of the city of B.

So he went D minus 50, and his speed was 40, and the same time T. Well notice now we've got two equations with two unknowns. The first one is solve for D, so just plug it into the second one. We get 60T minus 50 equals 40T. Add 50 to both sides, subtract 40T from both sides. We get 20T equals 50, divide by 20, we get T equals 5 over 2, or 2.5 hours.

Now we have to solve for D, now that we have the value of T. Plug this in, it makes sense to plug it into the first equation. D equals 60 times 2.5. Well 2 times 60 is 120. 0.5, or one half times 60 is 30. 120 plus 30 is 150, and that's the distance.

Here's a practice problem. Pause the video, and then we'll talk about this. Okay, Frank and Georgia started traveling from A to B at the same time. Georgia's constant speed was 1.5 times Frank's constant speed. When Georgia arrived at B, she turned around immediately and returned by the same route.

She crossed paths with Frank who was coming toward B when they were 60 miles away from B. How far away are A and B? So I'm gonna use the variable D for the distance between A and B. So that's the variable that we're looking for. For Frank and Georgia, Frank has a rate of R.

Georgia has a rate of 1.5 times that, so 1.5R. Georgia goes all the way from A to B, and then she comes back 60 miles. So the total distance that she covers is D plus 60. Frank starts out at A, but he falls short, he doesn't get to B. He's short 60 miles.

So the distance that he covers is D minus 60. And of course the time that we'll use, T, is gonna be from the starting point to the point that they cross paths again. So in that time, Georgia covers D plus 60, Frank covers D minus 60. So now we can set up our two D equals RT equations. D minus 60 equals RT, that's for Frank.

D plus 60 equals 1.5R times T, that's for Georgia. I'm going to rewrite that second equation as the right side is 1.5 times R times T. And now, since the first equation is solve for R times T, I can just plug in. Notice even though there are three unknowns here, I can eliminate two of them with one substitution. And I'm just left with an equation with D.

So that's very convenient, So I plug that in. I replace RT by D minus 60. That's the substitution from the first equation. So I get D plus 60 equals 1.5 times the quantity D minus 60. We get 1.5D minus, and then 1.5 times 60, 1 times 60, 0.5 is 30.

30 plus 60 is 90, so 1.5 times 60 is 90. Then what we'll do is subtract D from both sides. We'll add 90 to both sides. 0.5 is one half, we cancel that by multiplying both sides by 2, and we get D equals 300.

So the distance is 300. That's the distance between A and B. This is a slightly harder problem. Pause the video, and then we'll talk about this. Kevin drove from A to B at a constant speed of 60 miles an hour, turned around and returned at a constant speed of 80 miles an hour.

Exactly 4 hours before the end of his trip, he was still approaching B, and only 15 miles away from it. What is the distance between A and B? Let's think about this for a minute. So in the first segment, going from A to B, he was travelling at 60 miles an hour. Going back, he was traveling at 80 miles an hour.

And then starting from B, so from P all the way to B, and then all the way back to A, that whole segment took 4 hours. So P to B, back to A, is 4 hours. Well notice, I'm very interested just in this little interval right here, from P to B. We know the distance and we know the speed, so we could figure out the time.

The time would equal the distance, 15 miles divided by the speed, 60 miles an hour, 15 over 60 is one quarter. So that's one quarter of an hour. We could write that as 15 minutes, but let's actually leave it as a fraction. Well if P to B to A was 4 hours, and P to B is a quarter of an hour. Then the route from B back to A has to be, the time has to be the difference.

So that is 4 minus one quarter. So I'm gonna change 4 into an improper fraction. 16 over 4, minus one fourth, is 15 over 4. And I'm just going to leave that as an improper fraction right now. That is the time, 15 over 4 hours is the time of the trip from be back to A. Well now we know the speed as well as the time, so we can figure out the distance.

And that distance, of course, is the distance we're looking for, the distance from A to B. So the speed is 80, the time is 15 over 4. Remember the great trick, cancel before you multiply. 80 divided by 4 is 20. 20 times 15, well 2 times 15 is 30, 20 times 15 is 300.

And so that's actually the distance between A and B. In summary, when a word problem involves multiple travelers, multiple trips, or a trip with multiple legs, remember that each traveler, each trip, and/or each leg deserves its own D equals RT equation. Sometimes you will be able to solve for all quantities in one equation, and use those numbers to help solve for other equations.

More often, you will have to use the techniques for solving two equations with two unknowns. And we talked about substitution and elimination. If those are new to you, you can find out more about them in the algebra module.

Read full transcriptHere's a practice problem. Pause the video and then we'll talk about this. All right, Martha and Paul started traveling from A to B at the same time. Martha traveled at a constant speed of 60 miles an hour, and Paul at a constant speed of 40. When Martha arrived at B, Paul was still 50 miles away.

What is the distance? So notice first of all that the time. We'll just say that time is, the time from the starting point to when Martha arrived. And so, Martha of course, traveled that whole distance D, but what about Paul? Paul didn't get all the way to D. He was still 50 miles away, so that is D minus 50.

And so the distance that we're going to use for Martha is D. The distance we're going to use for Paul is D minus 50. And so now we have something to plug in for distance, and rate, and time. So for Martha, distance D, rate 60 times T. For Paul, D minus 50, because at that same time he was 50 miles short of the city of B.

So he went D minus 50, and his speed was 40, and the same time T. Well notice now we've got two equations with two unknowns. The first one is solve for D, so just plug it into the second one. We get 60T minus 50 equals 40T. Add 50 to both sides, subtract 40T from both sides. We get 20T equals 50, divide by 20, we get T equals 5 over 2, or 2.5 hours.

Now we have to solve for D, now that we have the value of T. Plug this in, it makes sense to plug it into the first equation. D equals 60 times 2.5. Well 2 times 60 is 120. 0.5, or one half times 60 is 30. 120 plus 30 is 150, and that's the distance.

Here's a practice problem. Pause the video, and then we'll talk about this. Okay, Frank and Georgia started traveling from A to B at the same time. Georgia's constant speed was 1.5 times Frank's constant speed. When Georgia arrived at B, she turned around immediately and returned by the same route.

She crossed paths with Frank who was coming toward B when they were 60 miles away from B. How far away are A and B? So I'm gonna use the variable D for the distance between A and B. So that's the variable that we're looking for. For Frank and Georgia, Frank has a rate of R.

Georgia has a rate of 1.5 times that, so 1.5R. Georgia goes all the way from A to B, and then she comes back 60 miles. So the total distance that she covers is D plus 60. Frank starts out at A, but he falls short, he doesn't get to B. He's short 60 miles.

So the distance that he covers is D minus 60. And of course the time that we'll use, T, is gonna be from the starting point to the point that they cross paths again. So in that time, Georgia covers D plus 60, Frank covers D minus 60. So now we can set up our two D equals RT equations. D minus 60 equals RT, that's for Frank.

D plus 60 equals 1.5R times T, that's for Georgia. I'm going to rewrite that second equation as the right side is 1.5 times R times T. And now, since the first equation is solve for R times T, I can just plug in. Notice even though there are three unknowns here, I can eliminate two of them with one substitution. And I'm just left with an equation with D.

So that's very convenient, So I plug that in. I replace RT by D minus 60. That's the substitution from the first equation. So I get D plus 60 equals 1.5 times the quantity D minus 60. We get 1.5D minus, and then 1.5 times 60, 1 times 60, 0.5 is 30.

30 plus 60 is 90, so 1.5 times 60 is 90. Then what we'll do is subtract D from both sides. We'll add 90 to both sides. 0.5 is one half, we cancel that by multiplying both sides by 2, and we get D equals 300.

So the distance is 300. That's the distance between A and B. This is a slightly harder problem. Pause the video, and then we'll talk about this. Kevin drove from A to B at a constant speed of 60 miles an hour, turned around and returned at a constant speed of 80 miles an hour.

Exactly 4 hours before the end of his trip, he was still approaching B, and only 15 miles away from it. What is the distance between A and B? Let's think about this for a minute. So in the first segment, going from A to B, he was travelling at 60 miles an hour. Going back, he was traveling at 80 miles an hour.

And then starting from B, so from P all the way to B, and then all the way back to A, that whole segment took 4 hours. So P to B, back to A, is 4 hours. Well notice, I'm very interested just in this little interval right here, from P to B. We know the distance and we know the speed, so we could figure out the time.

The time would equal the distance, 15 miles divided by the speed, 60 miles an hour, 15 over 60 is one quarter. So that's one quarter of an hour. We could write that as 15 minutes, but let's actually leave it as a fraction. Well if P to B to A was 4 hours, and P to B is a quarter of an hour. Then the route from B back to A has to be, the time has to be the difference.

So that is 4 minus one quarter. So I'm gonna change 4 into an improper fraction. 16 over 4, minus one fourth, is 15 over 4. And I'm just going to leave that as an improper fraction right now. That is the time, 15 over 4 hours is the time of the trip from be back to A. Well now we know the speed as well as the time, so we can figure out the distance.

And that distance, of course, is the distance we're looking for, the distance from A to B. So the speed is 80, the time is 15 over 4. Remember the great trick, cancel before you multiply. 80 divided by 4 is 20. 20 times 15, well 2 times 15 is 30, 20 times 15 is 300.

And so that's actually the distance between A and B. In summary, when a word problem involves multiple travelers, multiple trips, or a trip with multiple legs, remember that each traveler, each trip, and/or each leg deserves its own D equals RT equation. Sometimes you will be able to solve for all quantities in one equation, and use those numbers to help solve for other equations.

More often, you will have to use the techniques for solving two equations with two unknowns. And we talked about substitution and elimination. If those are new to you, you can find out more about them in the algebra module.