Two cyclists ride between a pair of towns with and against the wind

[brotherbobby]

Homework Statement

A 10 km long straight road connects two towns A and B. Two cyclists simultaneously start one from town A and the other from town B. On reaching the opposite town, a cyclist immediately returns to his starting town whereas the other cyclist takes some rest and then returns to his starting town. Both of them can ride at speed 20 km/hr in absence of wind but during their whole journey uniform wind from town 𝐴 to 𝐵 increases the speed of the cyclist going into the wind by the same amount as it decreases the speed of the cyclist going against the wind. Both the cyclists meet twice, first at 2 km and then 6 km away from one of the towns. For what period does a cyclist rest?

Relevant Equations

1. In uniform motion with speed $v$, the displacement after a time is $\Delta t$ is $\Delta x=v\Delta t$.

2. If frame $S'$ moves with velocity $+v$ along the $+xx'$ axis in normal configuration, then the velocity of an object in frame $S$ is $u=u'+v$, where $u'$ is its velocity in frame $S'$.

Attempt : [I could do the problem but get different answers from that of the text and several other places on the internet where the problem's been solved. I haven't seen their solution(s) yet].

Meeting 1 : I start by drawing an image of the problem for the first meeting the position of which is shown by the blue cross mark ($\color{blue}{\boldsymbol{\times}}$). This meeting should happen 2 km from B because (cyclist) 1 should travel a greater distance (8 km) in the same time as 2 (2 km) due to wind. If this meeting happens at a time $t_1$, then
$\small{t_1=\dfrac{8}{20+v_w}=\dfrac{2}{20-v_w}\Rightarrow v_w = 12\,\text{km/h}}\quad{\Large{\color{green}\checkmark}}$. This answer matches with the text and I have put it in the diagram.

Meeting 2 : Here 2 has come to A and 1 has gone to B. When they start again, they meet at a point 6 km from B or 4 km from A shown by the blue cross mark ($\color{blue}{\boldsymbol{\times}}$). If neither cyclist had waited, they'd meet again at the same place as before, 2 km from B. But because they meet earlier than that, the cyclist from A (cyclist 2) must have lost ground and therefore it must have been him who rested at A ($\color{red}{\large{\boldsymbol\times}}$) for a time $\Delta t$ before starting his journey back to B.

This answer is wrong, as per the text.

During this time, 1 must have covered a distance $20\Delta t$ and reached a point shown as C in the diagram. Hence, when 2 is supposed to start from A, their distance of separation $\Delta x_2=10-20\Delta t$. But 2 had to travel 4 km, hence the time of the meet $\small{\Delta t_2=\dfrac{4}{20+12}=\dfrac{1}{8}\,\text{hr}}$. But this is also the time that 1 got to travel from B's side : $\small{\dfrac{1}{8}=\dfrac{10-20\Delta t-4}{8}\Rightarrow 6-20\Delta t=1\Rightarrow 20\Delta t=5\Rightarrow \boxed{\Delta t =\dfrac{1}{4}\,\text{hr}=15\,\text{mins}}}\quad\color{red}{\large{\boldsymbol\times}}$.

Correct answers : $\text{The cyclist 1 stops at B for a time of 18.75 mins}$.

Request : Where do you think am I going wrong? Thanks for the time and the trouble.

 

[haruspex]

We are not told which cyclist rests nor whether the 6km is from the same town as the 2km or the other town, right?

 

[brotherbobby]

We are not told which cyclist rests. The distances of meet, 2 km and 6 km, are from the same town, but we are not told which one. Yes.

 

[haruspex]

The distances of meet, 2 km and 6 km, are from the same town,

It does not say so in post #1. It is at least ambiguous.

If neither cyclist had waited, they'd meet again at the same place as before, 2 km from B

Why? Wouldn’t that be the case if they started out from the two towns at the same time? If neither rested, the first cyclist would start from B when the second is only 2.5km from B.

 

[PeroK]

Homework Statement: A 10 km long straight road connects two towns A and B. Two cyclists simultaneously start one from town A and the other from town B. On reaching the opposite town, a cyclist immediately returns to his starting town whereas the other cyclist takes some rest and then returns to his starting town. Both of them can ride at speed 20 km/hr in absence of wind but during their whole journey uniform wind from town 𝐴 to 𝐵 increases the speed of the cyclist going into the wind by the same amount as it decreases the speed of the cyclist going against the wind. Both the cyclists meet twice, first at 2 km and then 6 km away from one of the towns. For what period does a cyclist rest?

The wind is blowing from A to B, so the cyclist travelling from A to B (C1) goes faster than the other (C2). They must meet 2km from town B. That means that C1 travels 8km while C2 travels 2km. And the wind speed is 12km/h.

(Note that the assumption that wind speed can be added and subtracted like that is absurd and shows a fundamental lack of understanding of mechanics by the question setter. But, what's new? It would make more sense to have the the journey uphill one way. )

It's now clear that C1 must be the cyclist who rests. If C1 doesn't rest, then he is only 2.5km behind C2 and they will meet close to town A.

There are then two possibilies:

C1 rests and they meet 6km from town A
C1 rests and they meet 6km from town B

I can't see that either can be ruled out.

 

[haruspex]

C1 rests and they meet 6km from town A
C1 rests and they meet 6km from town B

I can't see that either can be ruled out.

Then go with the assumption the two distances are from the same town. (Gives the right answer.)

 

[Steve4Physics]

There are a couple of mistakes I can spot.

... If neither cyclist had waited, they'd meet again at the same place as before, 2 km from B.

For the part 1 journeys, the cyclists leave A and B simultaneously, meeting 2km from B. But do the cyclists arrive at A and B simultaneously? So (if neither cyclist waits) would they start their part 2 (return) journeys simultaneously? Does this affect the position of their 2nd meeting?

During this time, 1 must have covered a distance $20\Delta t$

You have forgotten that cyclist 1 is not travelling at 20km/hr.

 

[brotherbobby]

It does not say so in post #1. It is at least ambiguous.

It says so in the problem statement itself.

If neither rested, the first cyclist would start from B when the second is only 2.5km from B.

If neither cyclicts rested, they Motion 2 would be identical to Motion 1 and they'd meet again the way they did in Motion 1, 2 km from Town B.

 

[brotherbobby]

But do the cyclists arrive at A and B simultaneously?

Thank you. Crucial error on my part.

You have forgotten that cyclist 1 is not travelling at 20km/hr.

Thank you. Yes I spotted that soon after my post.

 

[PeroK]

It says so in the problem statement itself.

If neither cyclicts rested, they Motion 2 would be identical to Motion 1 and they'd meet again the way they did in Motion 1, 2 km from Town B.

You mean from town A.

 

[brotherbobby]

You mean from town A.

My statement was a serious mistake. They don't reach the towns simultaneously, so the second leg of their journey won't begin simultaneously, unless 1 decides to rest at B for exactly the time it takes 2 to reach A.

 

[kuruman]

I opted for a graphical solution that illustrates the points made earlier. I first plotted the position vs. time round trips for the two cyclists assuming that neither rested. These are represented by the blue and red lines in the plot below. The first meeting is at 2 km from B at t = 5/16 hr = 18.75 min. If neither rests, the second meeting will occur at the symmetric point of 2 km from A or 8 km from B. For the second meeting to occur at 6 km from B, the cyclist starting at A (red line) must be delayed by resting. To find the resting interval, note that the red line from B to A is parallel to the blue line from B to A and that a delay of 5/16 hrs results in a position difference of 2 km. Thus, if the red line cyclist is delayed for an additional 5/16 hrs by resting, the gray line representing his position will drop by an additional 2 km when it intersects the blue line. This can also be shown, perhaps more formally, by considering similar triangles.

To summarize, the blue line represents the round trip of the cyclist who starts at B and does not rest. The red line from t = 0 to t = 5/16 hrs represents the "go" trip for the cyclist starting at A; the gray line between square markers represents his resting interval; the gray line from t = 5/8 hrs to t = 1 9/16 hrs represents his "return" trip.

 

[haruspex]

It says so in the problem statement itself.

It's ambiguous.
"first at 2 km and then 6 km away from one of the towns."
= "first at 2 km from one of the towns and then 6 km away from one of the towns."

If neither cyclicts rested, they Motion 2 would be identical to Motion 1

I see from post #11 you have withdrawn that, yes?

 

[brotherbobby]

It's ambiguous.
"first at 2 km and then 6 km away from one of the towns."
= "first at 2 km from one of the towns and then 6 km away from one of the towns."

Yes. Both distances measured from the same town.

I see from post #11 you have withdrawn that, yes?

Yes, I was horribly mistaken.

 

[brotherbobby]

Thank you all, I have solved the problem. The method was manual and painstaking. I took time and distances covered, a step at a time.
If people like @haruspex, @PeroK and others would like, I'd post it here.
If you have a smarter way to do the problem, please share with me.

 

[pbuk]

It's ambiguous.

No it isn't.

"One of the towns" uniquely identifies one of the towns, without loss of generality let this be B. The phrase is then "first at 2 km and then 6 km away from B".

If the question setter had intended to express the meaning that you want the phrase would have been written "first at 2 km and then 6 km away from one or other of the towns".

 

[pbuk]

Thank you all, I have solved the problem. The method was manual and painstaking. I took time and distances covered, a step at a time.

I don't know how you solved it but I wouldn't describe it as "manual and painstaking":

1. Your diagram and workings in the OP for the first meeting are correct.
2. You can then see that it takes C1 $\frac{2}{32} = \frac{1}{16}$ hour to get to A and C2 $\frac{8}{8} = 1$ hour to get to B.
3. Your diagram for the second meeting is also correct, and from this you can see that it takes C1 $\frac{6}{8} = \frac{3}{4}$ hour and C2 $\frac{4}{32} = \frac{1}{8}$ hour to get to the second meeting.
4. The total travel time between meetings for C1 is therefore $\frac{1}{16} + \frac{3}{4} = \frac{13}{16}$ hour and for C2 $1 + \frac{1}{8} = \frac{9}{8}$ hour.
5. C1 therefore rests for $\frac{9}{8} - \frac{13}{16} = \frac{5}{16}$ hour or 18.75 minutes.

 

[kuruman]

"One of the towns" uniquely identifies one of the towns, without loss of generality let this be B.

I don't think that there is generality to be lost. The author of the problem, although a bit cagey, provides enough information to identify the town as B. This became apparent to me when I constructed the plot in post #12. Here is what we are told.

1. The wind blows from A to B. Therefore the "go" trip for cyclist A takes less time than the "go" trip for cyclist B. It follows that the first encounter at 2 km "from one of the towns" identifies the "from" town as B. OP expressed as much in post #1.
2. The cyclists depart simultaneously which sets clock time $t = 0$ when this event occurs. If one is to produce a position vs. time graph for both cyclists, the first encounter point must be closest to the origin of position/ time coordinates which sets town B as that origin.

If we were to set town A as the origin of coordinates, then we would have to change the direction of the wind.