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The problem is as follows:

An air force pilot navigates across the Amazon river aboard an inflatable rescue lifeboat. He spends $\textrm{4 hours}$ by going in the same direction of the current to cover $\textrm{72 kilometers}$ between his landing site to the nearest town. Later on, he returns to his landing site spending $\textrm{6 hours}$ by going against the river stream. What is the speed of the river if we know the speed of the lifeboat and the river are constant?

In my attempt to solve this problem I sketched the situation in Figure A.

Sketch of the problem revised

I defined the speed of the lifeboat as:

$$\textrm{lifeboat=}v_{b}$$

and the speed of the river stream as:

$$\textrm{river stream=}v_{r}$$

For constant speed the equation is the following:

$$x=x_{0}+vt$$

Since I'm assuming it departs from the center of the axis $\textrm{(0,0)}$ the equation becomes into:

$$x=vt$$

Initially I thought that the speed of the lifeboat was this:

$$v_{b}=\frac{72\,\textrm{km}}{4\,\textrm{h}}=18\,\frac{\textrm{km}}{\textrm{h}}$$

However after some deeper thought I said that this cannot be as the speed of the boat is influenced by the stream as it mentions going downstream and not in water at rest so I changed the equation to take into account the speed of the river.

Since both are in the same direction I thought to summing them up and by doing this the new equation became into:

$$\left ( v_{b}+v_{r} \right )\times 4 = 72$$

and when the lifeboat goes upstream would be the difference in other words the speed of the stream subtracted from the lifeboat, therefore becoming into:

$$\left ( v_{b}-v_{r} \right )\times 6 = 72$$

By solving the system:

$$\left ( v_{b}+v_{r} \right )\times 4 = 72$$ $$\left ( v_{b}-v_{r} \right )\times 6 = 72$$

$$v_{b}+v_{r} = 18$$ $$v_{b}-v_{r} = 12$$

$$2v_{b} = 30$$

$$v_{b}=15$$

$$v_{r}=3$$

Therefore the speed of the river stream is $3\,\frac{km}{h}$. However I'm not sure if what I did was correct. Also, is it okay to assume velocities going in the same direction should be summed and when they're in opposite should be subtracted? The answer I got is somewhat reasonable as I do not expect that the river has a really big speed. But what if the stream speed is larger than the speed of the boat? can this situation happen? would this affect the way how I did approached this problem?. Can someone help me to clear these ideas?.

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    $\begingroup$ Your calculation is correct. if the speed of the river is greater than the speed of the boat in water, he cannot make any progress upstream but will be forced downstream. Yes that could happen. $\endgroup$ – David Quinn Mar 31 '18 at 17:27
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Your calculation is indeed correct.

Velocity is a vector, i.e. it comes with a magnitude and direction. When they are in the same direction, we add them, when they are in the opposite direction, we subtract them.

To be able to travel in the opposite direction, your boat needs a faster speed than the river. Otherwise, imagine someone who push you backward $2$ steps whenever you try to move forward a step, if the destination is in a straight line and not a loop, you can't reach the destination. Hence you won't have the second equation.

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  • $\begingroup$ I seem to have missed that velocity is a vector, I thought that the equations $x=x_{0}+vt$ held true only for scalars so I was not so sure to use it in there. $\endgroup$ – Chris Steinbeck Bell Apr 1 '18 at 10:59
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"However after some deeper thought I said that this cannot be as the speed of the boat is influenced by the stream "

"The speed of the boat" is actually ambigious. "The speed of the boat" in a swimming pool is different then the "speed of the boat" when it on a river, is different then "the speed of the boat" on airplane, is different from the speed of the boat in a rocket ship being crashed into the sun.

There are three different speeds alluded to in the problem

Speed 1: $v_1$ "the speed of the boat with the current" when it went with the current and traveled $72$ miles in $4$ hours. This speed is indeed $v_1 = \frac {72}{4} = 18 kmh$ but this will not necessarily be what most people consider "the speed of the boat" as it does not indicated the boat traveling solely on its own power.

Base Speed: $v_b$ how fast the boat would travel on its own solely under its own power.

Clearly $v_1 = v_b + v_r$. So $18 = v_b + v_r$.

Speed 2: $v_2$ "the speed of the boat against the current" when it went $72$ kilometers in $6$ hours. That speed is indeed $v_2 = \frac {72}6 = 12 kmh$. (Why does your picture show "16"?)

And Clearly $v_2 = v_b -vr$. So $12 = v_b - v_r$.

So you have $18 = v_b + v_r$ and $12 = v_b - v_r$ and must solve for $v_r$.

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  • $\begingroup$ That's an interesting observation. When I tried to solve this problem I did not took in consideration the intricacies of the language but at you pointed out, when I used those words it was not very specific. What I meant was that the speed of the boat in the river going downstream cannot be treated as if the boat was in water at rest, to figure this out it took me a while, therefore I couldn't just make the division as mentioned above. Now that we have that clear, I seem to have missed the concept of "own power" that you mentioned. $\endgroup$ – Chris Steinbeck Bell Apr 1 '18 at 10:55
  • $\begingroup$ Next time I'll be carefull with that because by using this concept it all makes sense by summing and subtracting one from the other. Regarding the picture I changed it accordingly, sorry it was a typo due rushed drawing what I meant was $12 \frac{\textrm{km}}{\textrm{h}} $. $\endgroup$ – Chris Steinbeck Bell Apr 1 '18 at 10:56

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