Two tourists and a bicycle - uniform motion

[brotherbobby]

Homework Statement

Two tourists who are at a distance of 40 km from their camp must reach it together in the shortest possible time. They have one bicycle which they decide to use in turn. One of them started walking at a speed of $v_1 = 5\,\text{km/h}$ and the other rode off on the bicycle at a speed of $v_2 = 15\,\text{km/h}$. The tourists agreed to leave the bicycle at intermediate points between which one walks and the other rides.
(1) What will the mean speed of the tourists be?
(2) How long will the bicycle remain unused?

Relevant Equations

For uniform motion with velocity $v_0$ along a straight line where $x(t=0)=x_0$ and the motion starts at $t=0$, the position of the object at any time $t$ is $x=x_0+v_0t$.

Problem : I copy and paste the problem as it appeared in the text.





Attempt : $\texttt{I could make no significant attempt at solving the problem.}$

The only thing I realise is that if the two tourists be A and B, A travels on the bicycle for a while, leaves the bicycle and walks. B catches up with the bicycle and overtakes A who is on foot and stops at a distance further, leaving the bicycle and continues to walk. A catches up with the bicycle and cycles on. This continues till they both reach the camp at the same time.

However, every trial I made (manually), A and B don't arrive at the camp at the same time.

Or going backwards, taking arrival at the camp to be simultaneous, when I trace back, giving them times on and off the bicycle, they don't start off together.

$\texttt{How to decide how much time they spend on foot and how much time on the bicycle?}$

Request : Any help or hint would be welcome.

 

[pbuk]

$\texttt{How to decide how much time they spend on foot and how much time on the bicycle?}$

Please don't use math formatting to post text.

You may not know (yet) how much time each of them spends on the bicycle, but if they both arrive at the same time then they must each spend the same amount of time on the bicycle.

What does this tell you?

 

[brotherbobby]

You may not know (yet) how much time each of them spends on the bicycle, but if they both arrive at the same time then they must each spend the same amount of time on the bicycle.

What does this tell you?

It tells me that they must have walked for the same times (say some $t_W$) and cycled for the same times (say $t_C$).
However, I don't think $t_W = t_C$.

 

[haruspex]

What is the simplest pattern in which each uses the bike for part of the way?

 

[brotherbobby]

What is the simplest pattern in which each uses the bike for part of the way?

The simplest pattern would be that they travel either half the distance, or half the time, walking and cycling.
Since $v_{\text{cycle}}>v_{\text{walk}}$ and they want to reach in the shortest time, they should travel a greater distance on cycle than on walk.
So it seems that the times of travel $t_C=t_W$?

 

[haruspex]

The simplest pattern would be that they travel either half the distance, or half the time, walking and cycling.
Since $v_{\text{cycle}}>v_{\text{walk}}$ and they want to reach in the shortest time, they should travel a greater distance on cycle than on walk.
So it seems that the times of travel $t_C=t_W$?

No, that clearly doesn’t work logistically. The bicycle is will have to spend some time idle.
You wrote:

The only thing I realise is that if the two tourists be A and B, A travels on the bicycle for a while, leaves the bicycle and walks. B catches up with the bicycle and overtakes A who is on foot and stops at a distance further, leaving the bicycle and continues to walk. A catches up with the bicycle and cycles on. This continues till they both reach the camp at the same time.

What is the simplest such pattern?

You might find it helpful to draw a distance-time graph showing the trajectories of the two tourists and the bicycle.

 

[brotherbobby]

No, that clearly doesn’t work logistically. The bicycle is will have to spend some time idle.

Yes, I agree. Since the bicycle is left idle, walking times increase over cycle times making $t_W>t_C$.

What is the simplest such pattern?

I cannot think of a pattern. All I know is that they both cycled for the same times and walked for the same times and that they both walked longer than they cycled.

If I had to think of a simple pattern, I may suggest equal distances. Is $\Delta x_C = \Delta x_W$?

 

[haruspex]

Yes, I agree. Since the bicycle is left idle, walking times increase over cycle times making $t_W>t_C$.

I cannot think of a pattern. All I know is that they both cycled for the same times and walked for the same times and that they both walked longer than they cycled.

If I had to think of a simple pattern, I may suggest equal distances. Is $\Delta x_C = \Delta x_W$?

No, you don’t get what I am asking.
Take what you wrote in post #1, as I quoted in post #6, and trim it down to the simplest sequence of that type.

 

[brotherbobby]

Take what you wrote in post #1, as I quoted in post #6, and trim it down to the simplest sequence of that type.

By simplest do you mean the number of turns they take on the bicycle?
In that case I can think of a single turn walking and bicycle for both.

 

[BvU]

In that case I can think of a single turn walking and bicycle for both.

And the distance vs time plot is ...?

 

[haruspex]

By simplest do you mean the number of turns they take on the bicycle?
In that case I can think of a single turn walking and bicycle for both.

Right, so plug in a variable and write an equation.

 

[pbuk]

It tells me that they must have walked for the same times (say some $t_W$) and cycled for the same times (say $t_C$).

Think about distance instead of time.

 

[kuruman]

My personal recommendation would be to draw a velocity vs. time graph for each tourist noting that the biker moves three times as fast as the walker and that distance traveled is the area under each curve.

Also note that, in order to reach camp simultaneously, neither walks (or bikes) for a longer time interval than the other. The time to reach camp is minimized if they keep moving without stopping to rest or to take selfies.

 

[brotherbobby]

Thank you all for your comments and apologies for the almost 24 hour delay in my reply.

For now @haruspex 's method seems the most do-able to me, not least because of my familiarity of equations over space-time diagrams.

Right, so plug in a variable and write an equation.

The total distance $x = v_C t_C+v_W t_W\qquad (1)\quad$ where $C,W$ stand for biCycle and Walking respectively.

We established that both tourists cycle and walk for the same times. Meaning if tourist A walks for 1 hour in total, so does tourist B. However, for a given tourist, say A, walking time should be greater than cycle time, meaning $t_W>t_C$. Instead, for minimum time of travel, the distance travelled for a given tourist for walking and cycling should be the same : $x_C = x_W\Rightarrow v_Ct_C = v_Wt_W\;^{\color{red}{\Huge{*}}}\qquad(2)$

From (1) and (2), $x = 2v_Ct_C\Rightarrow t_C=\dfrac{x}{2v_C}=\dfrac{40}{2\times 15}=\dfrac{4}{3}\,\text{hr}\qquad(3)$

Time for walking $t_W = \dfrac{v_C}{v_W}t_C=3\times\dfrac{4}{3}=4\,\text{hr}\qquad (4)$

Total time of travel $t_T = 4+1\dfrac{1}{3}= \dfrac{16}{3}=5\dfrac{1}{3}\,\text{hr}\qquad (5)$

[1] What will the mean speed of the tourists be?
Average velocity : $\boxed{\bar{v}} = \dfrac{x}{t_T}= \dfrac{40}{16/3}= \dfrac{40\times 3}{16}= \boxed{7.5\,\text{km/hr}}\quad{\color{green}{\large{\checkmark}}}$, which answers the first of the two questions asked.

[2] How long will the bicycle remain unused?
Let's say tourist A cycles first, covering a distance $(x_C)_A=\dfrac{4}{3}\times 15=20\,\text{km}\;\text{using (3)}$. In that time, B has walked a distance of $(x'_W)_B = \dfrac{4}{3}\times 5=\dfrac{20}{3}\,\text{km}$. The bicycle is left idle at a distance of 20 km, which is at a distance of $(\Delta x_W)_B = 20-\dfrac{20}{3}=\dfrac{40}{3}\,\text{km}$ which B is yet to cover. This distance will be covered by B in a time $(\Delta t_W)_B = \dfrac{40}{3\times 5}=\dfrac{8}{3}\,\text{hr}=2\dfrac{2}{3}\,\text{hr}$.

The bicycle will be idle for this length of time : $\boxed{(\Delta t_C)_{\text{idle}}=\text{2 hr and 40 min}}\quad{\color{green}{\large{\checkmark}}}$.

The book solutions show that these answers are right.




Doubts :

(1) $\color{red}{\huge{*}}$Before I solve the problem using space-time diagrams, I confess that I haven't understood the argument for equal distances covered by a given tourist on cycle and walk for minimal time of travel [$x_C = x_W$], which I have marked with a red asterisk. The authors make use of the same point, which I underlined in red ($\color{red}{\rule[0.4pt]{20pt}{1pt}}$).

(2) My derivation of the idle time of the bicycle was long and arduous. The authors do it quickly, claiming that the bicycle remains idle for half the time of total travel, but how? I have underlined in red ($\color{red}{\rule[0.4pt]{20pt}{1pt}}$) the author's statement in the solutions above.

Request : An explanation for the above doubts.

 

[brotherbobby]

I have drawn the distance time graph of the problem below for tourists A and B, and the cycle C. But it hasn't helped me in my solution of the problem. In fact, having solved the problem analytically, I used the values to draw the graph.

Crucially, I don't see why the distances travelled by each tourist on cycle and foot should be the same.

 

[haruspex]

I confess that I haven't understood the argument for equal distances covered by a given tourist on cycle and walk for minimal time of travel

The bicycle covers the whole distance, never reversing, so at any point on the road one must walk there and the other cycles there. So the distance one walks is the distance the other cycles.
If the time for both to have arrived is minimum, can you see that means they arrive at the same time?

(2) My derivation of the idle time of the bicycle was long and arduous. The authors do it quickly, claiming that the bicycle remains idle for half the time of total travel,

It doesn’t look to me that they deduce that by some unstated means and then use it to solve the question. Rather, they deduce it as a result of solving the question, and it is merely how it happens to be with the given numbers. (It is only true for a velocity ratio of 3:1.)
If you make the bicycle extremely fast then it is unused for almost all the time.

 

[pbuk]

Good grief, you are overcomplicating this to the extent that you have lost your way.

It tells me that they must have walked for the same times (say some $t_W$) and cycled for the same times (say $t_C$).

So if they have cycled for the same times, they must have cycled for the same distances.

And if they want to get there as quickly as possible, they want to cycle as much as possible.

So if they have 40 km to travel and they each need to use the bicycle for as much of this as possible, how much distance do they each need to use the bicycle for?

 

[Steve4Physics]

Crucially, I don't see why the distances travelled by each tourist on cycle and foot should be the same.

I think it’s already established that the tourists spend equal times on the bicycle.

To minimise the overall journey time, each tourist must cover as much of the 40km as possible on the bicycle.

The bicycle is a shared resource. The best case is that each tourist covers half the distance on the bicycle.

The simplest way to achieve this is for one tourist to cycle to the midpoint (20km), leave the bicycle there rady for the other tourist, and walk the remaining 20km.

Edit. Aha. I see @pbuk beat me to it.

 

[kuruman]

A velocity vs. time graph provides an at-a-glance solution and summarizes the points made above.

The initial biker gets drops the bike off at time $\Delta t$ (to be determined) and continues on foot. His velocity is represented by the solid red line below. The initial walker needs time $3\Delta t$ to get to the bike and reaches the end after an additional $\Delta t$. His velocity is represented by the dashed blue line.

Consider a unit of displacement represented by the area of a rectangle $\Delta x=5~(\text{km/hr})\times \Delta t.$ Both tourists have covered the same distance $6\Delta x$ in the same amount of time $4\Delta t$.
From the graph we see that

1. The distance traveled by each tourist is $~6\Delta x=40~\text{km}\implies \Delta x = 20/3~\text{km}.$
2. The bike is dropped off at distance $~3\Delta x=3\times (20/3~\text{km})=20~\text{km}~$ from the starting point.
3. The bike is dropped off at time $t_d=\Delta t=\dfrac{\Delta x}{15~\text{km/hr}}=\dfrac{4}{3}~\text{hr}.$
4. The bike remains idle for an interval $(\Delta t)_{\text{idle}}=2\Delta t=\dfrac{8}{3}~\text{hr}.$
5. The total time for the trip is $T=4\Delta t=\dfrac{16}{3}~\text{hr}.$
6. The average speed is $v_{\text{avg.}}=\dfrac{40~\text{km}}{T}=\dfrac{40~\text{km}}{16/3~\text{hr}}=7.5~\text{km/hr}.$

 

[brotherbobby]

A velocity vs. time graph provides an at-a-glance solution and summarizes the points made above.

The initial biker gets drops the bike off at time $\Delta t$ (to be determined) and continues on foot. His velocity is represented by the solid red line below. The initial walker needs time $3\Delta t$ to get to the bike and reaches the end after an additional $\Delta t$. His velocity is represented by the dashed blue line.

Consider a unit of displacement represented by the area of a rectangle $\Delta x=5~(\text{km/hr})\times \Delta t.$ Both tourists have covered the same distance $6\Delta x$ in the same amount of time $4\Delta t$.
From the graph we see that

1. The distance traveled by each tourist is $~6\Delta x=40~\text{km}\implies \Delta x = 20/3~\text{km}.$
2. The bike is dropped off at distance $~3\Delta x=3\times (20/3~\text{km})=20~\text{km}~$ from the starting point.
3. The bike is dropped off at time $t_d=\Delta t=\dfrac{\Delta x}{15~\text{km/hr}}=\dfrac{4}{3}~\text{hr}.$
4. The bike remains idle for an interval $(\Delta t)_{\text{idle}}=2\Delta t=\dfrac{8}{3}~\text{hr}.$
5. The total time for the trip is $T=4\Delta t=\dfrac{16}{3}~\text{hr}.$
6. The average speed is $v_{\text{avg.}}=\dfrac{40~\text{km}}{T}=\dfrac{40~\text{km}}{16/3~\text{hr}}=7.5~\text{km/hr}.$

View attachment 343957

Brilliant. I was about to do it myself but thanks a lot.
It was a tricky problem and I have clarified my doubts.

 

[pbuk]

It was a tricky problem

Perhaps, but the solution (as hinted at by @haruspex and @Steve4Physics) is much more simple than set out above. We are not supposed to post complete solutions here, but as @kuruman has already let the cat out of the bag:

• From the information in the question they both arrive at the same time so they must both cycle the same distance, and beacause cycling is quicker than walking in order to arrive as early as possible this must be 20 km.
• A cycles for 20 km at 15 km/h, taking 1h20, parks the bicycle and then walks 20 km at 5 km/h, taking 4h00: a total of 5h20 for 40 km, 7.5 km/h.
• B walks for 20 km, taking 4h00 to reach the parked bicycle (the bicycle is therefore unused for 2h40) and then cycles the remainder, reaching the destination at the same time and with the same average speed as A.

 

[kuruman]

We are not supposed to post complete solutions here, but as @kuruman has already let the cat out of the bag $\dots$

I was under the impression that OP had already let the cat out with post #14 which looks like a complete solution to me. After OP has posts a complete solution, there is no restriction on postings of complete "streamlined" solutions using alternative approaches. By consolidating in a single post all that has been said and explained, they deepen OP's and future readers' understanding of the problem at hand.

 

[brotherbobby]

I was under the impression that OP had already let the cat out with post #14 which looks like a complete solution to me. After OP has posts a complete solution, there is no restriction on postings of complete "streamlined" solutions using alternative approaches. By consolidating in a single post all that has been said and explained, they deepen OP's and future readers' understanding of the problem at hand.

Yes I had already written out the solution in post#14, even though I had conceptual doubts.

 

[Mister T]

Crucially, I don't see why the distances travelled by each tourist on cycle and foot should be the same.

Because there is only one bicycle. If A starts by riding the bicycle a distance $d$ and then starts walking, he will have left the bicycle a distance $d$ from the starting position. Thus B, who starts at the same position as A, will have to walk the distance $d$ to get to the bicycle where he can then begin to ride it.

 

[brotherbobby]

Because there is only one bicycle. If A starts by riding the bicycle a distance $d$ and then starts walking, he will have left the bicycle a distance $d$ from the starting position. Thus B, who starts at the same position as A, will have to walk the distance $d$ to get to the bicycle where he can then begin to ride it.

Yes. Which means if A cycles a distance $d$ greater than he walks, B has to walk the distance $d$ greater than he gets to cycle and the two won't reach the destination simultaneously.

 

[Mister T]

Yes. Which means if A cycles a distance $d$ greater than he walks, B has to walk the distance $d$ greater than he gets to cycle and the two won't reach the destination simultaneously.

Right. And if A cycles a distance less than he walks, B will have to cycle a distance greater than he walks. The only possibility left is that A cycles a distance equal to what he walks, and so does B.