## 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 a constant speed a 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- 50. And so the distance that we're gonna use for Martha is D, the distance we're gonna use for Paul is D- 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- 50 because at that same time he was 50 miles short of the city of B.

So he went D- 50 and his speed was 40. And the same time T. Well, notice now we have two equations with two unknowns. The first one is solve for D. So just plug it into the second one. We get 60T- 50 = 40T.

Add 50 to both sides, subtract 40T from both sides. We get 20T = 50. Divide by 20, we get T = 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=60*(2.5) well, 2x60 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 an hour away from B.

How far away are A and B? So I'm going to 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 + 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- 60. And of course, the time that we'll use, T is gonna be from the starting point to the point that they cross pass again.

So in that time, Georgia covers D + 60, Frank covers D- 60. So now we can set up our two D = RT equations, D- 60 = RT, that's for Frank, D + 60 = (1.5R)T, that's for Georgia. I'm gonna rewrite that second equation as the right side is 1.5(RT). And now since the first equation is solved for R times T, I can just plug in.

And notice I can eliminate, even though there three unknowns here. I can eliminate two of them with one substitution. And I'm just left with an equation with T so, that's very convenient. So, I plug that in I replace RT by D- 60, that's the substitution from the first equation. So I get D + 60 = 1.5(D- 60).

We distribute, we get 1.5D minus, and then 1.5 times 60. 1 times 60 is 60, 0.5 is 30. 30 plus 60 is 90. So 1.5 times 60 is 90. Then what we will 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 to 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 traveling at 60 miles an hour. Going back, he was traveling at 80 miles an hour.

And then starting from B, so from here, from P, all the way to B, and then all the way back to A, that whole segment took four 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 can figure out the time.

The time would equal the distance 15 miles divided by the speed 60 miles an hour 15 over 60 is 1/4. 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 four hours and P to B is a quarter of an hour, Then the route from B back to A, the time has to be the difference.

So that is 4- one-quarter. So I'm gonna change 4 into an improper fraction, 16 over 4, minus one-fourth, is 15 over 4. Now I'm just gonna leave that as an improper fraction right now. That is the time, 15 over 4 hours, is the time of the trip from B back to A. 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 15/4. Remember the great trick, cancel before you multiply. 80 divided by 4 is 20. 20 times 15, well 2 times 15 is 30, so 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 = 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 a constant speed a 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- 50. And so the distance that we're gonna use for Martha is D, the distance we're gonna use for Paul is D- 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- 50 because at that same time he was 50 miles short of the city of B.

So he went D- 50 and his speed was 40. And the same time T. Well, notice now we have two equations with two unknowns. The first one is solve for D. So just plug it into the second one. We get 60T- 50 = 40T.

Add 50 to both sides, subtract 40T from both sides. We get 20T = 50. Divide by 20, we get T = 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=60*(2.5) well, 2x60 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 an hour away from B.

How far away are A and B? So I'm going to 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 + 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- 60. And of course, the time that we'll use, T is gonna be from the starting point to the point that they cross pass again.

So in that time, Georgia covers D + 60, Frank covers D- 60. So now we can set up our two D = RT equations, D- 60 = RT, that's for Frank, D + 60 = (1.5R)T, that's for Georgia. I'm gonna rewrite that second equation as the right side is 1.5(RT). And now since the first equation is solved for R times T, I can just plug in.

And notice I can eliminate, even though there three unknowns here. I can eliminate two of them with one substitution. And I'm just left with an equation with T so, that's very convenient. So, I plug that in I replace RT by D- 60, that's the substitution from the first equation. So I get D + 60 = 1.5(D- 60).

We distribute, we get 1.5D minus, and then 1.5 times 60. 1 times 60 is 60, 0.5 is 30. 30 plus 60 is 90. So 1.5 times 60 is 90. Then what we will 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 to 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 traveling at 60 miles an hour. Going back, he was traveling at 80 miles an hour.

And then starting from B, so from here, from P, all the way to B, and then all the way back to A, that whole segment took four 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 can figure out the time.

The time would equal the distance 15 miles divided by the speed 60 miles an hour 15 over 60 is 1/4. 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 four hours and P to B is a quarter of an hour, Then the route from B back to A, the time has to be the difference.

So that is 4- one-quarter. So I'm gonna change 4 into an improper fraction, 16 over 4, minus one-fourth, is 15 over 4. Now I'm just gonna leave that as an improper fraction right now. That is the time, 15 over 4 hours, is the time of the trip from B back to A. 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 15/4. Remember the great trick, cancel before you multiply. 80 divided by 4 is 20. 20 times 15, well 2 times 15 is 30, so 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 = 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.