Distance Traveled by A: y = Vb.∆t2

[Zelderes]

Homework Statement

Three tourists have a bicycle, they must reach the tourist center in the shortest possible time (the time is counted until the last tourist arrives in the center). The bicycle can only carry two people and so the third tourist should go on foot. The cyclist leaves another tourist on the way and returns to pick up what was on foot. Walking speed is v_1 and biking is v_2 . What is the average speed of tourists for v_1 and v_2 ?

Relevant Equations

I thought like this: y is the distance traveled by A with velocity Vb over a time interval ∆t1, ie y = Vb.∆t2;

I thought like this: y is the distance traveled by A with velocity Vb over a time interval ∆t1, ie y = Vb.∆t2;

 

[WWGD]

You're not given actual numbers? Is the velocity assumed to be constant?

 

[Zelderes]

You're not given actual numbers? Is the velocity assumed to be constant?

I had used numbers, it was bad. Any idea?

 

[WWGD]

Not sure I understand about the use of numbers. It seems all you can go by is the total distance and total time. Do you have a formula for that?

 

[Zelderes]

I do not have

 

[hmmm27]

Have you considered thinking about it, writing it down in longhand first if you can't do it off the top of your head (I can't), then creating a formula ?

 

[WWGD]

Does this come from a book? Do you use , e.g., $x=v \cdot t$ or other formulas?

 

[PeterDonis]

Homework Statement: Three tourists have a bicycle, they must reach the tourist center in the shortest possible time (the time is counted until the last tourist arrives in the center). The bicycle can only carry two people and so the third tourist should go on foot. The cyclist leaves another tourist on the way and returns to pick up what was on foot. Walking speed is v_1 and biking is v_2 . What is the average speed of tourists for v_1 and v_2 ?

I'm not sure I understand what the question is asking for. If the distance to the tourist center is $D$ and the total time it takes for all three tourists to get there is $T$, then "average speed" could just mean $D / T$. Is that what the question is asking for?

Assuming it is, and taking $D$ as known, obviously you need to figure out what $T$ is as a function of $v_1$ and $v_2$. Have you tried to do that?

 

[GlenLaRocca]

The bicycle will get there in time tb = y/v2. Then the bicycle has to go back and get the walker at a point we will call y1. It will take until time t1 = tb + (y-y1)/v2 for the bicycle to get to the walker. We also know y1 = t1*v1 is how far the walker went until the bicycle made it back. Then it will take another (y-y1)/v2 for the bicycle to make it back to the center. I'll let you take a crack at the algebra.

 

[PeterDonis]

The bicycle will get there in time tb = y/v2. Then the bicycle has to go back and get the walker at a point we will call y1.

This assumes that the bicycle carries one person all the way to the center, then goes back to pick up the third person. That might not be the optimal solution given that the total time is to be minimized. It might be quicker for the bicycle to drop off one person at some point short of the center, letting them walk the rest of the way, and then go back and pick up the third person and bring them to the center.

 

[GlenLaRocca]

That will result in the same average speed. Just a lot more complicated math.

This assumes that the bicycle carries one person all the way to the center, then goes back to pick up the third person. That might not be the optimal solution given that the total time is to be minimized. It might be quicker for the bicycle to drop off one person at some point short of the center, letting them walk the rest of the way, and then go back and pick up the third person and bring them to the center.

 

[PeterDonis]

That will result in the same average speed.

Not if "average speed" means $D / T$ as I think it does. $D$ is constant, but $T$ depends on how far the bicycle takes the first two people before one is dropped off and the bicycle goes back to pick up the third person.

 

[GlenLaRocca]

Not if "average speed" means $D / T$ as I think it does. $D$ is constant, but $T$ depends on how far the bicycle takes the first two people before one is dropped off and the bicycle goes back to pick up the third person.

Average speed will always be D/T. Here D is fixed and T is total time to cover the course by both bike and walk. If you want, work it out the way I said, then work it out for going half way at a time. You will get the same answer. If that is the same it implies all strategies are the same--because half of each half must also be the same and so-on..

 

[PeterDonis]

Average speed will always be D/T.

Yes, but $T$ is not fixed; it depends on what I said before.

If that is the same it implies all strategies are the same

All strategies are not the same: that's obviously false. This is a homework thread, so please do not post misinformation. If you do so again you will be warned and banned from the thread.

 

[Antarres]

Since it is only the time when the last passenger arrives that counts, the optimal strategy would make sense to be, one where they all arrive at the same time. So the guy with the bike should ride until some point, then drop off one passenger to pick up the other, just in time to chase him at the goal the same moment he arrives there.

So to hint at the solution, we have three portions of travel. One where the biker transports first passenger. Second, when he is returning, and third when he is transporting the third passenger.

So we should set up distance/time relations for each of these portions, and use the relation that the sum of distances is fixed.

Now there is a possibility that my reasoning is not precise(although it makes the most sense to me), in a sense that maybe there is a more optimal strategy than this one, but in that case we should just apply the same treatment, except that the time wouldn't be fixed by this requirement that they arrive at the same time, so we would minimize it using derivatives. Since this is precalculus type of an exercise though, I assume that they expect you to arrive at the optimal strategy phenomenologically, instead of mathematically, though.

@GlenLaRocca All strategies can't be the same, because in the second portion of the travel, you have two people walking by foot and one on bike, it is not symmetrical.

 

[GlenLaRocca]

Yes, but $T$ is not fixed; it depends on what I said before.
All strategies are not the same: that's obviously false. This is a homework thread, so please do not post misinformation. If you do so again you will be warned and banned from the thread.

I am sorry that I did leave off one part of the solution. You should drop off the second person on the bicycle at a y2 prior to arrival at the center and have them walk such that they arrive at the same time as when the bike gets back.

 

[GlenLaRocca]

I apologize that I did not have the whole solution when I first posted. I made some misstatements because I left off part of the solution.

 

[PeterDonis]

I apologize that I did not have the whole solution when I first posted.

This is a homework thread, so you're not supposed to post the entire solution. But you haven't, since you haven't actually given a formula for the average speed in terms of known quantities. Please don't, since the point is not to tell the OP the answer but only to give hints.

 

[PeterDonis]

Since it is only the time when the last passenger arrives that counts, the optimal strategy would make sense to be, one where they all arrive at the same time.

You should drop off the second person on the bicycle at a y2 prior to arrival at the center and have them walk such that they arrive at the same time as when the bike gets back.

These are both good statements of a key hint. Please don't expand on them further since, as I've said, we don't want to tell the OP the answer. At this point I think we need to let the OP consider what we've said and post further if he needs more help.