Launch service between two stations along a flowing river

[brotherbobby]

Homework Statement

Two landing stakes $\text{M}$ and $\text{N}$ are served by launches that all travel at the same speed relative to water. The distance between the landing stages is $20\;\text{km}$. It is covered by each launch from $\text{M}$ to $\text{N}$ in one hour and from $\text{N}$ to $\text{M}$ in two hours. The launches leave the two landing stages at the same time at an interval of 20 minutes and stop at each of them also for 20 minutes.

Determine :

(1) The number of launches in service.
(2) The number of launches met by a launch travelling from $\text{M}$ to $\text{N}$.
(3) The number of launches met by a launch travelling from $\text{N}$ to $\text{M}$.

Relevant Equations

Relative velocity of A with respect to B : $v_{AB}= v_{AC}-v_{BC}$

Attempt : I start by compying and pasting the problem as it appeared in the text. Please note that for reasons of clarity, I replaced landing station $\text{K}$ with station $\text{N}$.

Let me draw a picture of the problem situation. The landing stations M and N are shown in red and the launches forward and backwards in green. The forward launches have a velocity $v_F$ relative to land and the backward launches a velocity $v_B$ relative to land. The times taken by the launches are $t_F$ and $t_B$ respectively. Each launch stops for a period $\Delta t$ at the stations and leave at intervals of 20 minutes also. At stations M and N, launches always leave simultaneously. Let the velocity of the river itself (w.r.t. land) be $v_R$, shown in blue.

[I could not answer the questions asked, but let me show you what I could do. Given how little it is, any help is welcome]

The time taken in the forward direction $t_F=\dfrac{d}{v_F}=\dfrac{d}{v_L+v_R}$, where $v_L$ is the velocity of the launch. Hence $1= \dfrac{20}{v_L+v_R}\Rightarrow v_L+v_R=20\qquad (1)$

Likewise, $t_B=\dfrac{d}{v_B}=\dfrac{d}{v_L-v_R}\Rightarrow 2= \dfrac{20}{v_L-v_R}\Rightarrow v_L-v_R=10\qquad (2)$.

Solving (1) and (2), $v_L=15\;\text{km/h}$ and $v_R=5\;\text{km/h}$.


I don't know how to proceed from here to answer questions (1) - (3).

Any help or suggestion is welcome.

 

[haruspex]

How long does it take a given launch to complete a cycle?

 

[brotherbobby]

How long does it take a given launch to complete a cycle?

3 hours and 20 mins = $3\dfrac{1}{3}\;\text{hr}$

 

[haruspex]

3 hours and 20 mins = $3\dfrac{1}{3}\;\text{hr}$

"stop at each of them also for 20 minutes."

 

[brotherbobby]

"stop at each of them also for 20 minutes."

Ok so you mean the cycle $M\leftrightarrow N$ and ready to depart from M? That would be 3hrs + 40 mins = $3\dfrac{2}{3}\;\text{hr}$.

 

[haruspex]

Ok so you mean the cycle $M\leftrightarrow N$ and ready to depart from M? That would be 3hrs + 40 mins = $3\dfrac{2}{3}\;\text{hr}$.

Right. And how many others depart M in that time?

 

[brotherbobby]

If each launch departs after $\dfrac{1}{3}\;\text{hr}$, then 11 $\left( =\dfrac{11}{3}\div \dfrac{1}{3} \right)$ launches (including this one) depart M in that time.

 

[haruspex]

If each launch departs after $\dfrac{1}{3}\;\text{hr}$, then 11 $\left( =\dfrac{11}{3}\div \dfrac{1}{3} \right)$ launches (including this one) depart M in that time.

Right. So what is the answer to (1)?

 

[brotherbobby]

Right. So what is the answer to (1)?

(1) is 11 launches.

 

[brotherbobby]

I am trying to think the answers for (2) and (3).

Any one launch will meet 3 of the other launches in its motion from M to N (1 hr).
On its way back, it is going to travel for 2 hours. It cannot meet the next launch that follows it - which is 20 minutes behind. So leaving itself and the next, it should encounter 11-(2) = 9 launches.

Is my reasoning correct?

How about the launch immediately preceding our launch? I think it won't cross its way either. In that case it will meet 11-3 = 8 other launches.

I wonder which one's right, if any.

Thank you @haruspex

 

[haruspex]

I am trying to think the answers for (2) and (3).

Any one launch will meet 3 of the other launches in its motion from M to N (1 hr).
On its way back, it is going to travel for 2 hours. It cannot meet the next launch that follows it - which is 20 minutes behind. So leaving itself and the next, it should encounter 11-(2) = 9 launches.

Is my reasoning correct?

How about the launch immediately preceding our launch? I think it won't cross its way either. In that case it will meet 11-3 = 8 other launches.

I wonder which one's right, if any.

Thank you @haruspex

It's tricky because the numbers are in exact ratios. We know that a launch will arrive at M exactly as it leaves, and another will depart N exactly as it arrives there. Should we count neither, one or both? I would count one, but I can’t tell what the problem setter had in mind.

 

[brotherbobby]

I think we won't be doing something wrong if I took a look at the solutions which the authors have posted. But I'd wait for your permission @haruspex

 

[brotherbobby]

It's tricky because the numbers are in exact ratios. We know that a launch will arrive at M exactly as it leaves, and another will depart N exactly as it arrives there. Should we count neither, one or both? I would count one, but I can’t tell what the problem setter had in mind.

I counted both in both of my (manual) attempts in post#10 above. Yes I think we should because it is also a meeting of two boats, never mind that it didn't happen on water but at one camp or another.

 

[kuruman]

How about the launch immediately preceding our launch? I think it won't cross its way either. In that case it will meet 11-3 = 8 other launches.

If the total number of launches is 11, isn't it true that over a complete cycle the total number of launches encountered by any given launch would be 10 because itself doesn't count?

Also, the launch under consideration will cross the preceding launch at the landing. If there is a 20 min. difference, the launch under consideration will be leaving while the preceding launch will be arriving.

 

[kuruman]

Homework Statement: Two landing stakes $\text{M}$ and $\text{N}$ are served by launches that all travel at the same speed relative to water. The distance between the landing stages is $20\;\text{km}$. It is covered by each launch from $\text{M}$ to $\text{N}$ in one hour and from $\text{N}$ to $\text{M}$ in two hours. The launches leave the two landing stages at the same time at an interval of 20 minutes and stop at each of them also for 20 minutes.

Solving (1) and (2), $v_L=15\;\text{km/h}$ and $v_R=5\;\text{km/h}$.

These numbers are incorrect. The distance is 20 km and the trip times are 1 hr and 2 hr. So the speeds should be respectively 20 km/hr and 10 km/hr. However, the cycle time of 3:40 hr is correct.

 

[Steve4Physics]

It may be worth noting that there is no need to find the speed of the water or the speed of the boats relative to the water. E.g. the problem could equally well be asked about lorries moving on dry land:

The distance between 2 lorry-depots, M and N, is 20km.
Lorries travelling from M to N take one hour.
Lorries travelling from N to M take two hours.
(E.g. because it’s downhill from M to N.)
Lorries leave the two depots at the same time, at intervals of 20 minutes, and stop at each of the depots also for 20 minutes.

This is equivalent to the original problem-statement (at least in terms of the information needed to answer questions 1-3).

 

[brotherbobby]

If the total number of launches is 11, isn't it true that over a complete cycle the total number of launches encountered by any given launch would be 10 because itself doesn't count?

Also, the launch under consideration will cross the preceding launch at the landing. If there is a 20 min. difference, the launch under consideration will be leaving while the preceding launch will be arriving.

Yes, that is correct and I stand corrected. The correct answer, so far as I think, is that, in both directions to and fro, each launch encounters the remaining 10 other launches plying between the landing stations M and N.

Of course, we are assuming that two launches, one leaving M and one arriving at M, constitute what the authors of the text call a "meet".

 

[brotherbobby]

These numbers are incorrect. The distance is 20 km and the trip times are 1 hr and 2 hr. So the speeds should be respectively 20 km/hr and 10 km/hr. However, the cycle time of 3:40 hr is correct.

Yes but what speeds? The speed of a launch relative to earth, you mean.
Relative to water, the speed of a launch is 15 km/h and relative to earth, the river flows at 5 km/h.
Hence, a launch has a speed 20 km/h towards, and 10 km/h backwards, in the earth's frame.

 

[brotherbobby]

I am the OP for the problem and I think the time has come for me to look at the solution as given by the authors. Thank you all, especially @haruspex and @kuruman, for your inputs.

I want to start with the problem statement.

Problem statement :











Solution (ours) :

$\quad\text{(1) There are }\mathbf{11}\;\text{launches operating between towns M and N}$
$\quad\text{(2) and (3) From both M to N and back, each launch encounters } \mathbf{10}\;\text{launches}$

Solution (text) :

Revision : I can see that the authors have done it via a space-time diagram. I have drawn the motion of the first launch from M to N and back in red ink ($\color{red}{\rule[0.4pt]{20pt}{1pt}}$) till it is ready to travel again. In that time, we can count four launches that start from N (or K) and six launches that start from M which verify that there are a total of 11 (=1+4+6) launches in operation.

The authors say that each launches encounters 8 launches during its motion towards or return. Hence, it means that only a "meeting" on water is counted as a meet and not one which happens at either station or M or N.

I think our method of doing it was easier. More so, why draw so many space time lines when they could have stopped where the red line that I drew ended?

Thank you all.

 

[kuruman]

I just figured out the positions of each launch on a spreadsheet over a 3 hr interval which is the time for a round trip (not a cycle which is 40 min. longer) and made a plot (see below). The calculations were done in 20 min increments. Then I counted dots. Note that there are two dots on top of one another at $t=1~$hr. One is just leaving for the return trip and the second one is just arriving.

 

[Steve4Physics]

FWIW here’s another approach.

Suppose a launch (call it X) leaves stage M at 12:00.

At the same moment, a launch from N is arriving; this X’s first ‘meet’. Note that the arriving-launch left N at 10:00.

X reaches N at 13:00 and meets a launch leaving; this is X’s final meet.

So X meets every launch that left N between 10:00 and 13:00 inc. Since launches leave every 20 mins, there are ten such meets. (Because the launches which X meets left N at the following ten times: 10:00, 10:20, 10:40, 11:00, 11:20, 11:40, 12:00, 12:20, 12:40, 13:00.)

 

[jbriggs444]

A simple counting argument suffices. If a boat moves from one end of the course to the other, it will necessarily encounter every other boat at some point. Exactly once each.