Two launches ferrying between two landing stations along a river

  • #1
brotherbobby
699
163
Homework Statement
Mail is carried between two landing stages M and K by two launches. At the appointed time the launches leave the landing stages, meet each other, exchange the mail and return. If the launches leave their respective launching stages at the same time, the launch departing from M spends 3 hours both ways and that from K 1.5 ##-## hours. The speeds of both launches are the same relative to water. Determine ##\underline{\text{graphically}}## how much later should the launch depart from M after the other one leaves K for the two to travel the ##\underline{\text{same time}}##. (my emphasis)
Relevant Equations
If the speed of river is some ##v## and the speed of launch relative to river ##u,## then motion downstream is ##u+v## and upstream is ##u-v.\quad (u>v)##
1716617822272.png
Attempt :
I copy and paste the problem from the text.

As it happened, I could solve the problem analytically but not graphically. I describe both methods below.


##(\rm{I})-\text{Analytical approach}##

The problem is divided into two parts. In the first part, both launches leave their stations at the same time.

1716617904261.png
##(\rm{a}) -\text{Simultaneous leaving}##

I draw the situation of the problem alongside. Let the stations be a distance ##d## apart and the river to flow with a speed ##v##. Let each launch move with speed ##u## relative to river. Let the launching station M also be te origin, O. Both launches meet at a distance of ##x## (from O).
Launch 1 leaves M and 2 leaves K - since the launch from M takes longer to go and return, 1 it must be travelling downstream first for a greater distance also to meet 2. Let us use single primes ##'##s for forward motion and double primes ##''##s for return.
Time taken by both launches are the same for them to meet : ##\small{t'_1=\dfrac{x}{u+v}=\dfrac{d-x}{u-v}=t'_2\Rightarrow x=\dfrac{u+v}{u}\dfrac{d}{2}=x_1\quad \left(>\dfrac{d}{2}\right)}##. The distance travelled by launch 2 ##\small{x_2=d\left( 1- \dfrac{u+v}{2u} \right)=\dfrac{u-v}{u}\dfrac{d}{2}}##. The common time for them both ##t'_1=t'_2=\dfrac{d}{2u}##. To return, the first launch will take a time of ##\small{t''_1=\dfrac{u+v}{u-v}\dfrac{d}{2u}}## and the second launch ##\small{t''_2=\dfrac{u-v}{u+v}\dfrac{d}{2u}}##. Adding the forward and return times for both launches and equating them to the times given, we find the total time for launch 1 ##t_1=\dfrac{d}{u-v}=3\quad\color{blue}{(1)}\quad## and for launch 2 ##\quad t_2= \dfrac{d}{u+v}=\dfrac{3}{2} (=1.5)\quad\color{blue}{(2)}##

1716617953802.png
##(\rm{b}) -\text{1 leaving M after a time }\Delta t##

Launch 2 here travels for a time ##\Delta t## and is at a position ##(x_2)_{\Delta t}=d-(u-v)\Delta t## and arrives at N. If they both meet at P at a distance ##x## (from O or M), then the time taken by 1 ##\small{t'_1=\dfrac{x}{u+v}}##. This the same as the time taken by 2 to travel the remaining distance to P, ##\small{t'_2=\dfrac{d-(u-v)\Delta t-x}{u-v}}## which, after some algebra, simplifies to ##x=\dfrac{u+v}{2u}\{d-(u-v)\Delta t\}=x_1##. The distance covered by launch 2 ##x_2=d-x=d-\dfrac{u+v}{2u}\{d-(u-v)\Delta t\}=\dfrac{u-v}{2u}\{d+(u+v)\Delta t\}##, after some algebra.
The time for 1 forwards to meet : ##\small{t'_1=\dfrac{x_1}{u+v}=\dfrac{d-(u-v)\Delta t}{2u}}\quad (3)##. Likewise, the time for launch 2 forwards to meet 1 ##\small{t'_2=\dfrac{x_2}{u-v}=\dfrac{d+(u+v)\Delta t}{2u}}\quad (4)##.
Note that these times are different despite a "meeting" taking place., because launch 2 travelled for a longer time ##\Delta t##.
The time for 1 to return to his station M : ##\small{t''_1=\dfrac{x_1}{u-v}=\dfrac{u+v}{2u(u-v)}\{d-(u-v)\Delta t\}}\quad (5)##. The total travel time for launch 1 can be found by adding ##(3)+(5)##, which leads to ##\small{t_1=t'_1+t''_1=\dfrac{d-(u-v)\Delta t}{u-v}}\quad (6)##.
The time for launch 2 to return to his station K : ##\small{t''_2=\dfrac{x_2}{u+v}=\dfrac{u-v}{2u(u+v}\{d+(u+v)\Delta t\}}\quad (7)##. Hence the total travel time for launch 2 can be found by adding ##(4)+(7)\Rightarrow\small{t_2=t'_2+t''_2=\dfrac{d+(u+v)\Delta t}{u+v}}\quad (8)##.
But the two times are given to be equal, ##t_1=t_2\Rightarrow\small{\dfrac{d-(u-v)\Delta t}{u-v}=\dfrac{d+(u+v)\Delta t}{u+v}\Rightarrow \dfrac{d}{u-v}-\dfrac{d}{u+v}=2\Delta t\Rightarrow \Delta t=\dfrac{vd}{u^2-v^2}}\quad (9)##.

From ##(1), (2)## above, ##\small{\dfrac{d}{u-v}-\dfrac{d}{u+v}=3-\dfrac{3}{2}\Rightarrow \dfrac{2vd}{u^2-v^2}=\dfrac{3}{2}\Rightarrow \dfrac{vd}{u^2-v^2}=\dfrac{3}{4}}##. But from ##(9)## this is the value of ##\Delta t##. Hence ##\Delta t = \dfrac{3}{4}\,\text{hr}\Rightarrow\boxed{\Delta t= 45\,\text{min}}\quad{\color{green}{\Large\checkmark}}##

This answer matches with the text.


##(\rm{II})-\text{Graphical approach}##

As before, there are two cases - when the launches leave simultaneously and when launch 1 leaves M after a delay of time ##\Delta t.##

1716618041305.png
##(\rm{a}) -\text{Simultaneous leaving}##

Launches 1 and 2 leave stations M and K shown in red. They meet at even P and return to their stations. We are given launch 1 takes a total of 3 hours, so MR = 3, while launch 2 takes 1.5 hours, so KQ = 1.5. The triangles KPQ and RPM are similar. Since base MR is twice KQ, the distance to P (and back) must be twice by M as for K. Hence we have ##x_1=2x_2##. Thus 1's speed must be twice 2's speed : ##u+v=2(u-v)\Rightarrow u=3v##, where ##u,v## are the speeds of the launches and river respectively.

1716618094967.png


##(\rm{b}) -\text{1 leaving M after a time }\Delta t##

Here launch 1 leaves M after a delay, they meet at P and return to their landing stations. In doing so, it's given that they take the same times, so KQ = NR. Hence triangles KPQ and RPM are congruent, implying that distances ##x_1=x_2##, so they both meet midway between the landings.


I could not glean further information from the graphs in order to answer the question : what is ##\Delta t=?##

Request : A hint as to how to go about solving the problem graphically.
 
Last edited:
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  • #2
brotherbobby said:
I could not glean further information from the graphs in order to answer the question : what is Δt=?
Does it help to consider what it looks like if you rotate triangle NPR about the vertical axis through P?
 
  • #3
haruspex said:
Does it help to consider what it looks like if you rotate triangle NPR about the vertical axis through P?
1716655805612.png





By how much? A ##180^{\circ}## rotation would align NPR with QPK, both triangles being congruent.
 
  • #4
brotherbobby said:
View attachment 345956




By how much? A ##180^{\circ}## rotation would align NPR with QPK, both triangles being congruent.
Sorry, I was unclear. By vertical axis I meant the line through P parallel to KM.
Rotate 180°.
 
  • #5
haruspex said:
Sorry, I was unclear. By vertical axis I meant the line through P parallel to KM.
Rotate 180°.
1716708448529.png
Same response from me, I'm afraid, as before.

If you rotate ##\triangle## NPR about a vertical axis parallel to KM by ##180^{\circ}## it would coincide with ##\triangle## QPK.

Please let me know if am wrong.
 
  • #6
brotherbobby said:
View attachment 345982Same response from me, I'm afraid, as before.

If you rotate ##\triangle## NPR about a vertical axis parallel to KM by ##180^{\circ}## it would coincide with ##\triangle## QPK.

Please let me know if am wrong.
Rotate about the thin black line in your diagram, next to where you have written ##x_1, x_2##. R moves to M.
 
  • #7
haruspex said:
R moves to M.
1716720190281.png




Why do you say so? I can't understand.
Of course RN = KQ.

(do you mean flip about the thin black line marked with distances ##x_1,x_2##?)
 
  • #8
brotherbobby said:
(do you mean flip about the thin black line marked with distances ##x_1,x_2##?)
Yes, is that not what I wrote?
 
  • #9
I suppose I'd like to clear this. Is rotation of a 180 degrees about the y-axis the same as a reflection about the y-axis? (It's a digression from the problem but an important point)
 
  • #10
brotherbobby said:
I suppose I'd like to clear this. Is rotation of a 180 degrees about the y-axis the same as a reflection about the y-axis? (It's a digression from the problem but an important point)
If it is a 2D plane being reflected or rotated 180°, they're the same.
For a 3D object, rotation 180° about the y axis is also the same as a reflection through the y axis ((x,y,z)->(-x,y,-z)) but is different from a reflection in a plane containing the y axis (e.g. (x,y,z)->(-x,y,z)).
 
  • #11
haruspex said:
If it is a 2D plane being reflected or rotated 180°, they're the same.
For a 3D object, rotation 180° about the y axis is also the same as a reflection through the y axis ((x,y,z)->(-x,y,-z)) but is different from a reflection in a plane containing the y axis (e.g. (x,y,z)->(-x,y,z)).
I'd like to share with you a demo I just did to verify what you said above. I took a toy model of a truck and subjected it to a reflection about the y-axis and then a 180 degree rotation counter-clockwise. These were carried out in a drawing software called ##\text{EdrawMax}^{\circledR}##.

Below are the results. The reflected truck and the rotated truck look different.

1716722219259.png



So I'd like you tell me what's going on here.
 
  • #12
Your rotation is about the wrong axis. It should be about the dashed line. Think of it as a page in a book. Turn the page over.
 
  • #13
Ok so I rotated about an axis going in through the page, perpendicular to it.
You're asking me to rotate about the dashed line, which is in the page. Got you. Yes I believe the two will coincide.
 
  • #14
1716722908058.png



Ok to our problem. If triangle NPR is flipped (or rotated) about the vertical line AB, N will move to the right of B as far as it is to the left. R will move to the left of B as far as it is to the right.

N will come vertically below Q and yes, R will coincide with M. I agree with you.

How does it help in finding ##\Delta t##?

EDIT : The above means B is the midpoint of MR. That is all I could conclude.
 
  • #15
brotherbobby said:
View attachment 345987


Ok to our problem. If triangle NPR is flipped (or rotated) about the vertical line AB, N will move to the right of B as far as it is to the left. R will move to the left of B as far as it is to the right.

N will come vertically below Q and yes, R will coincide with M. I agree with you.

How does it help in finding ##\Delta t##?

EDIT : The above means B is the midpoint of MR. That is all I could conclude.
You worked out the ratio of the two slopes, no?
 
  • #16
haruspex said:
You worked out the ratio of the two slopes, no?
1716739237850.png



Yes, the slope of NQ is twice that of KR.
 
  • #18
haruspex said:
So what is the ratio NB:BR?
1716972796847.png
Sorry for coming in late.

Let's have the problem statement again.




Let's have the graphs again for the launches leaving simultaneously and the launch from M leaving after some delay ##\Delta t##.

1716971551672.png
1716973128931.png












Let's refer to the graphs as G1 and G2.

From G1, we conclude that ##u = 3v## where ##u## is the speed of the launch in still water and ##v## is the speed of the river. The slope of the line MQ is twice the slope of KR.

From G2, after flipping the triangle NPR about the line AB, R would coincide with M, making MB half of MR. Using the slopes of NQ and KR, we find BR = 2NB (from definition of slope = rise divided by run). These give MN = NB and so, we have ##\Delta t = NB=\dfrac{1}{4}MR##.

Question is, what is MR? Is MR the same in both graphs G1 and G2?

Thank you @haruspex for your time.
 
  • #19
brotherbobby said:
Is MR the same in both graphs G1 and G2?
Clearly not since the difference is ##\Delta t##. But KR is the same in both.
 
  • #20
haruspex said:
Clearly not since the difference is ##\Delta t##. But KR is the same in both.
1716996322746.png
1716996353018.png












In both the graphs above, G1 and G2, drawn for equal times of departure and later departure respectively, the distances MR look suspiciously equal. MR from G1 is 3 hours. Though looks can deceive.
Howover, I proved earlier than ##\Delta t = MN=NB=\dfrac{1}{4}MR##. If MR in G2 is the same as MR in G1, then ##\Delta t = \dfrac{1}{4}\times 3\,\text{hr}=\dfrac{3}{4}\,\text{hr}=45\,\text{min}##, which agrees with the answer.

1716996652711.png


Please note that in the first part of the problem, both launches leave simultaneously but take different times. In the second part, one leaves earlier so that they both take same times. Hence NR = KQ in G2, but that says nothing about MR is G1 and G2. They may be equal. But as I argued above, for us to get the right answer, they have to be equal. I'd like you to help me find out why.

In the second part of the problem (G2), both launches travel equal distances. The distance between the landing stations M and K are fixed. Can that have a bearing on why MR is the same in the two graphs? Note in G2, MR = \Delta t + total time of travel for Launch 1. In G1, MR = Total time of travel for Launch 1.

Thanks for your time and patience.


EDIT - I'd like to rethink the problem in the river's frame to see whether the total times for launch 1 in the two cases, MR, are equal.
 
  • #21
brotherbobby said:
Is MR the same in both graphs G1 and G2?
You are right. Not sure what I was thinking of. Seems I looked at NR, maybe.
 
  • #22
haruspex said:
You are right. Not sure what I was thinking of. Seems I looked at NR, maybe.
I only made a guess. I can think of no proof why MR should be the same in both cases.
 
  • #23
brotherbobby said:
I only made a guess. I can think of no proof why MR should be the same in both cases.
See the second part post #19.
 
  • #24
haruspex said:
See the second part post #19.
Sorry for coming in late. I think I have got the answer, but let me revisit if it's ok.

Problem statement :
1717422937311.png


I already solved this problem analytically in my original post #1. The dofficulty I had was regarding the graphical solution to the problem.


Below are the two graphs showing the problem situation for the two launches leaving simultaneously and launch 1 leaving M a time ##\Delta t## later so that the travel times of both launches 1 and 2 are the same.

Let's call them graphs G1 and G2.
1717423206876.png
1717423244706.png













From G1, using similarity of triangles MRP and QPK, distance ##x_1=2x_2\Rightarrow (u+v)t'_1= 2(u-v)t'_2##, were ##u,v## are velocities of launch and river and the primed times are those the launches took to meet, which must be equal since they left simultaneously. Hence ##u+v = 2(u-v)\Rightarrow u=3v##. Also the ##\text{slope of NQ}=\dfrac{x_1}{t'_1}=2\dfrac{x_2}{t'_2}=2\times\text{slope of KR}##. In G2, these values can be seen to remain the same.

In G2, the triangles NPR and QPK become congruent. If R is flipped about AB, it coincides with M. So MB is half of MR. Using the slopes of NQ and KR and the common side PB of triangles PNB and PRB, we find that the ratio of ##\text{NB:BR}=1:2##. So MN = NB and therefore ##\Delta t=\text{MN}=\dfrac{1}{4}\times \text{MR}##. However, KR is the same in both graphs G1 and G2. So MR from G1 must be the same as MR from G2. The time MR from G1 is 3 hours = the time MR from G2. Hence ##\boxed{\Delta t} = \dfrac{3}{4}\times 3\,\text {hr} = \boxed{\text{45 min}}\quad{\color{green}{\Large\checkmark}}##

I haven't seen their solutions yet. I hope you wouldn't mind @haruspex if I return to you.
 
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