Barge Problem from 26 Years of Posing Problems

Question

I'm struggling to find the solution this problem:

A boy at the stern of a canal barge leaps off onto the tow path and while the barge keeps moving, runs along the path until he gets to the bow, where he instantly picks up a thermos flask of coffee and runs back until he gets to the stern. An observer noticed that while the boy was running, the barge moved forward a distance equal to its length. How far does the boy run compared to the length of the barge?

My initial thought was that the boy traveled two barge lengths (there & back) end of story. But that seems incorrect because it implies he would be traveling at twice the speed of the barge. If he were traveling at only twice the speed, he would only reach the bow of the boat by the time the boat travels a distance equal to its length.

Any help?

Answer 1

Very nice question! A diagram is usually helpful with this sort of thing, so...

The red line represents the stern of the barge; the pink line the bow; and the green line is the boy.

Let's say the barge, whose length is $L$, moves at $v_{\text{barge}}$ and the boy at $v_{\text{boy}}$. Also say he reaches the bow (point $P$) after a time $\tau$ and returns to the stern (point $Q$) at time $T$.

When the boy reaches the bow, the barge has travelled a distance $v_{\text{barge}} \tau$; so $$v_{\text{boy}}\tau=v_{\text{barge}} \tau + L \tag{1}$$

By time $T$, the barge has travelled its length; so $$L=v_{\text{barge}}T \tag{2}$$

After getting to the bow, the boy runs in the opposite direction; so we also have $$L=v_{\text{boy}}\tau-v_{\text{boy}}(T-\tau) \tag{3}$$

We want to know the total distance the boy ran, $d = v_\text{boy} T$, in terms of $L$.

After a bit of manipulation, we find $$L= \frac{2dL} {d-L} - d$$

which we can solve to find the distance ran

$$d = \left(1+\sqrt2 \right) L$$

Answer 2

The time it takes for the boy to reach the other end of the barge is the length of the barge divided by their relative speed. If the boy runs at speed $v$ and the barge has length $L$ and moves at speed $V$, then the time it takes the boy to reach the bow is $t_b=L/(v -V)$ and the time it takes to return to the stern is $t_s=L/(v+V)$. Meanwhile, the barge took $t_{tot}=L/V$ to move one barge length. Setting $t_b+t_s =t_{tot}$ gives $$ {L\over v-V}+ {L \over v+V}= {2v L\over v^2-V^2} = {L\over V} \Longrightarrow v^2 -2vV -V^2= 0. $$ Solving gives $v= (1+\sqrt{2})V$. So the distance the boy ran to reach the bow was $d_b=v t_b = vL/(v-V) = (1+1/\sqrt{2})L$. His endpoint was $L$ away from the start, so he must have run back $L/\sqrt{2}$ to get there. Thus the total distance is $(1+\sqrt{2})L$.

Answer 3

Assuming the words mean these :
stern : rear end of ship.
barge : boat.
towpath : path along a canal or river used by animals towing boats.
bow : front end part of ship.

|S########B~~~~~~~~~~~~~~~~
@@@@@@@@@@@@@@@@@@@@@@@@@@@

Here , I am taking "|" to be the Initial Position , from where the Boy runs.
"@@@" indicates tow path.
"~~~" indicates water.
"S########B" indicates the boat , from Stern to bow.

Terminating Position is :

|~~~~~~~~~~S########B~~~~~~~~~~~~~~~~
@@@@@@@@@@@@@@@@@@@@@@@@@@@

Intermediate Position (when Boy has reached the Boat front) is :

|~~~~~S########B~~~~~~~~~~~~~~~~
@@@@@@@@@@@@@@@@@@@@@@@@@@@

Boy : Distance moved forward is $x+B$ , taking time $t$.
Boat : At that moment , Boat has moved Distance $x$ forward.

Boy : Distance moved backward will $x+B$ , taking time $t$ , because Boy is running at Constant velocity.
Boat : Distance Boat has moved is $B$ , taking time $2t$.

Hence , $B$ must be $2x$.
Boy : Total Distance is $2(x+B)=2(B/2+B)=3B$.

Boy must have run 3 times the Barge length.

Answer 4

There is a simple but serious mathematical point in this problem. That is: when solving an equation, understand what all the different solutions might mean.

In this case, it leads us to ask the engineering question: can a barge move backwards? Obviously if powered by horse, the answer would be yes, but even modern barges with water jets can use a “reverse bucket” to redirect the jet slightly forward. This system is similar to how thrust reversers on jet aircraft work.

If $u$, $v$, $t_{0}$, $t_{1}$ & $L$ represent respectively the boat's speed, boy's speed, duration of first sprint, duration of second sprint & length of the boat, then:

$v*t_{0} = L + u*t0$

$v*t_{1} = L - u*t0$

$u*(t_{0}+t_{1}) = L$

Eliminating $t_{0}$, $t_{1}$ & $L$ we get:

$v^2 - 2vu - u^2 = 0$

& hence:

$v/u = \sqrt{2} + 1$ when the boat is going forwards, or

$v/u = \sqrt{2} - 1$ when the boat is going backwards