Let me ask you this......will the intersection distance for each boat remain constant.....no matter what shore we start on?

Can you agree with that?

Edit: intersection distance is from starting shore to intersection with other boat. Each boat has it's own distance.

 

Let me ask you this......will the intersection distance for each boat remain constant.....no matter what shore we start on?

Can you agree with that?

Edit: intersection distance is from starting shore to intersection with other boat. Each boat has it's own distance.

Each boat has TWO "intersection distances" since the boats cross paths TWICE.

The first time they cross, each boat has traveled from its shore to the point of crossing. Thus each boat's first "intersection distance" is LESS than the width of the river.

The second time they cross, each boat has completed one complete crossing of the river to the opposite shore and is returning to the shore it originally left from but reaches the second crossing point before completing the return trip. Hence each boat's second "intersection distance" is GREATER than the width of the river but LESS than twice that width.

 

No...I don't think so. That's why the 10 minute break. There are two actions separated by 10 min.

On the first crossing, the intersection will be on one side of center river line. On the inverse or back crossing.....that intersection will be on opposite side of centerline. And equal distant from centerline.

No boat ever gets the opportunity to over take......they stop when they get to shore. They both have to start at same time. !0 min....takes all time out. This is strictly a length problem. And the ratio is the same both ways.

 

No...I don't think so. That's why the 10 minute break. There are two actions separated by 10 min.

Don't think what? We are not mind readers?

The 10 minutes is irrelevant for reasons already explained.

On the first crossing, the intersection will be on one side of center river line. On the inverse or back crossing.....that intersection will be on opposite side of centerline. And equal distant from centerline.

What is the basis for claiming that the first intersection point and the second intersection point are equidistant from the center line?

Again, going back to you and I crossing the country, if the first time we cross is 1000 miles east of the center of the country, why would the second time we cross have to be 1000 miles went of the center?

You are using "problem solving by happening" -- you keep throwing out different random things with no basis for key parts, hoping that at some point you will randomly generate a solution that happens to yield the correct answer.

How about this -- draw a graph showing the distance of each boat from the first bank as a function of time. Pick a width for the river and speeds for the two boats of whatever you want. Then start playing with those parameters until you satisfy the two crossing constraints. You will see that two things will have to happen -- the ratio of the two boats will have to be 13/9 and the width of the river has to be 1,760 meters.


Let's see if we can walk you through this step by step to see if you are able to at least recognize that the claimed solution really is a solution.

We have a river that is 1,760 meters across. Let's say that we have the left bank (Bank A) and the right bank (Bank B).

Both boats leave their respective banks at 12:00 noon.

Boat A travels at 104 meters/minute and Boat B travels at 72 meters/minute.

At 12:10 pm, how far at the boats from Bank A? How far are they from Bank B?

Which bank are they nearer to?

How far are they to the nearer bank?

Get that far and we'll go from there.

 

The river is 6" wide. Both Boats start(Boat A heading east), and meets Boat B at 4" from W shore. Boat A traveled 4" til meet and Boat B traveled 2" from E shore til meet. Now they both go on to other shore.

They have a quick beer. Both Boats start(Boat A heading west this trip), and meets Boat B at 2" from W shore. Boat A traveled 4' til meet and Boat B traveled 2" til meet. Now they go on to the other shore.

The reason being is the constant speed of the boats. Make sense?

You can use any boat velocities you want(as long as they are constant)......the symmetry will remain the same.

 

Both Boats take the same time to meet.....but different times to cross. Instead of 10 minute dock...make it 10 days.

 

Watch the first crossing from the north and the second crossing from the south.....and you will see what I mean.

 

The river is 6" wide. Both Boats start(Boat A heading east), and meets Boat B at 4" from W shore. Boat A traveled 4" til meet and Boat B traveled 2" from E shore til meet. Now they both go on to other shore.

They have a quick beer. Both Boats start(Boat A heading west this trip), and meets Boat B at 2" from W shore. Boat A traveled 4' til meet and Boat B traveled 2" til meet. Now they go on to the other shore.

The reason being is the constant speed of the boats. Make sense?

You can use any boat velocities you want(as long as they are constant)......the symmetry will remain the same.

You have artificially imposed a symmetry that just doesn't exist.

The boats are not going at the same speed. That is a GIVEN in the problem. Thus one boat will take LONGER to cross the river than the other. Each boat will stay at the opposite shore for ten minutes. The faster boat will arrive at the opposite shore sooner than the slower boat. Their ten minute holdover will start sooner than the slower boat's holdover. It will therefore end before the slower boat's holdover and the faster boat will start its return trip before slower boat starts its return trip. In fact, the faster boat may well start its return trip before the slower boat even arrives at its first stop.

So why are you assuming that both boats will start their return trip at the same time?

Even if you thought this was the case, then they would cross the second time 720 m from the other shore (i.e., the shore that the slower boat is departing from now). But they don't. They meet 400 m from that shore, which should have shown you that you are doing something wrong.

If you would bother to answer the questions I gave you, you will see how it all works out. But it seems like even getting you to do that much math is a fool's errand.

 

Watch the first crossing from the north and the second crossing from the south.....and you will see what I mean.

You will not see the error of your interpretation until you walk through it carefully and completely. But it is becoming quite obvious that you have no intention of doing that and will instead keep insisting on your fantasy version. Whether it's because you can't do the math or simply won't really doesn't matter. But unless and until you are willing to do so, there is not point trying to show you where you are wrong, so I'm not going to waste any more time trying.

When and if you choose to start working your way through it by answer the questions I asked in a recent post, we can continue.

 

Ok....but only one step at a post. If you state multiple steps...I can't stop and ask you why you assume this and that.

Can we both start with agreeable assumptions?

You start.

 

Ok....but only one step at a post. If you state multiple steps...I can't stop and ask you why you assume this and that.

Can we both start with agreeable assumptions?

You start.

Fine.

The first thing we need to agree on is whether we are going to try to solve the problem from the information given, or whether we are going to start from the proposed solution of 1,760 meters for the width of the river and proceed to verify that is IS a solution to the problem.

Which do you prefer?

 

Thanks...I have a proposed solution also. Can we take the statement problem, line by line and see what we do agree on....and on what we question?

 

1. I assume that on the first crossing, both boats start at same time, both boats are at constant but different speeds.

2. I assume the second crossing, like the first, start at same time, because of 10 min. break.

3. I assume the speeds of boats are same for both crossings. By same....I me boat A speed is same on both crossings and boat B speed is same on both crossings. Even thought A and B speeds are different.

Do you agree with these?

 

2. I assume the second crossing, like the first, start at same time, because of 10 min. break.

No! Just because there is a 10min break, boat A has started the journey back well before boat B... Boat B's 10 mins starts when it gets to shore!!

This is the part where you get hung up...

 

I wonder if whoever set the question based it on the answer being exactly one mile?

 

I wonder if whoever set the question based it on the answer being exactly one mile?

Indeed, in the original question the distances were given in yards – giving the width of the river as 1 mile.

I metricated the question, changing to metres [note the correct spelling of the SI unit of length, and not ‘meter’ – which is used to measure something].

 

Well then, poopie on it.

 

1. I assume that on the first crossing, both boats start at same time, both boats are at constant but different speeds.

2. I assume the second crossing, like the first, start at same time, because of 10 min. break.

3. I assume the speeds of boats are same for both crossings. By same....I me boat A speed is same on both crossings and boat B speed is same on both crossings. Even thought A and B speeds are different.

Do you agree with these?

No. Your assumption #1 is inconsistent with your assumption #2, as already pointed out.

This is what the problem states:

After arriving at their destination, each boat remains for 10 minutes before starting on its return trip.

This is what has previously been explained to you once it was clear that you were making this wrong assumption.

The boats are not going at the same speed. That is a GIVEN in the problem. Thus one boat will take LONGER to cross the river than the other. Each boat will stay at the opposite shore for ten minutes. The faster boat will arrive at the opposite shore sooner than the slower boat. Their ten minute holdover will start sooner than the slower boat's holdover. It will therefore end before the slower boat's holdover and the faster boat will start its return trip before slower boat starts its return trip. In fact, the faster boat may well start its return trip before the slower boat even arrives at its first stop.

So why are you assuming that both boats will start their return trip at the same time?

Do you agree that

1) Because both boats are traveling at different speeds, it take s each boat a different amount of time to cross the river?

If so, then let's determine when the two boats leave for the second crossing.

Define:

T0 = the time at which both boats start the first crossing.

Ta = the time that boat A takes to cross the river.

Tb = the time that boat B takes to cross the river.

T1a = the time at which boat A starts the second crossing.

T1b = the time at which boat B starts the second crossing.

Since boat A starts the second crossing 10 minutes after it completes the first crossing, it starts the second crossing at.

T1a = To + Ta + 10 minutes

Since boat B starts the second crossing 10 minutes after it completes the first crossing, it starts the second crossing at.

T1b = To + Tb + 10 minutes

You are assuming

T1a = T1b

But, this requires

Ta = Tb

which requires that both boats cross the river in the same amount of time which we know is not true since the problem states that the boats are traveling at different speeds.

Since the consequence of your assumption cannot be true, the assumption cannot be true.

 

WBahn.....I yield, I yield....I misinterpreted the stated problem. I'm still wondering what the nearest shore is.

No problem...haven't crossed a river in five years.

 

WBahn.....I yield, I yield....I misinterpreted the stated problem. I'm still wondering what the nearest shore is.

No problem...haven't crossed a river in five years.

What is so hard about the concept that the shore that is closest to the point of interest is the nearer shore to that point?