This is a solution to problem 23 on the hard interview questions node. If you have not read the question, the following will make no sense to you:

It is easiest to think of everything relative to the stream.

Say the woman meets the log at 12:00, then paddles away from it until 1:00, at which point she paddles toward it at the same speed. Clearly she will reach the log again at 2:00.

So two hours have elapsed, during which time the log (which travels with the current), has travelled one mile to the dock. Thus the current is travelling at 1/2 mph.

dtaylor: what I meant by "constant" was that, for example, the woman doesn't start to paddle faster at any point... of course, when she turns around, she has to change her speed, but we assume this is an instantaneous change and she continues paddling at the same speed in the other direction. This is all in the spirit of the principle of well-constructed examples: if it starts to get ugly, you've strayed from the intended path.

daleks: I guess the problem statement is a little ambiguous on this point, but the woman does not necessarily paddle upstream for a total of 2 hours, or 2 miles. She paddles upstream for one mile, then one hour.

(In reply to the above writeup in the Canoe solutionnode)

That would mean that:

Travelling upstream the woman is doing v mph relative to the stream.
Travelling downstream the woman is doing v mph relative to the stream.

The log is always doing 0 mph relative to the stream.

Therefore, relative to the land:

(Assume the current is flowing at c mph)

Upstream, the woman is doing v - c mph.
Downstream, the woman is doing v + c mph.
The log is always doing c mph relative to the land.

Yet the question stated all the speeds were constant. However, relative to the land, the speeds were most definately not constant.

If you assumed the speeds were constant relative to the land you would get a different answer:

Relative to the land:

Upstream, the woman is doing v mph.
Downstream, the woman is doing v mph.
The log is always doing c mph.

Relative to the river:

Upstream, the woman is doing v + c mph.
Downstream, the woman is doing v - c mph.
The log is always doing 0 mph.

So, if she meets the log at 00:00, and paddles for an hour, she would travel v+c miles away from the log by 01:00. She would then meet the log again once the stream/log reached the dock - i.e. after 1/c hours, travelling at v-c mph.
Therefore, she would have travelled a total of (v + c + v/c - 1 miles, in 1 + 1/c hours, and the log would have travelled 1 mile at c mph.
But, she had an overall displacement of 0m, so:

1/(v + c) * (v + c) + 1 * (v + c) - 1/c(v - c) = 0
1 + v + c - v/c - 1 = 0
v + c - v/c = 0
c/v - 1/c = 0
c/v = 1/c

c2 = v
c = sqrt(v)

I'm not entirely sure, (due to my brain becoming mush), but I don't think its possible to actually find a value for c, but I'm happy to be corrected.

But basically, thanks to everything being relative to something else, the question gives enough information to work out one possible answer, but leaves enough details out to make another answer possible but indeterminate.

If this writeup makes sense, it's a miracle. If this writeup is actually correct... well, there's no point worrying about it, because I'm sure I made a mistake somewhere.

Begin with d=rt, where distance is in miles, rate is in miles per hour, and time is in hours. It took the woman 2 hours to paddel 2 miles up stream. Since the effect of her paddling on her overall net speed is constant and the river current rate is constant, we have the equation

2=(r_w-r_c)2 (1)

Solving for r_w yields

r_w=1+r_c (2)

Again, return to d=rt to model her downstream path. She paddled downstream for 2 miles this time making the stream current additive, but did so for an unspecified amount of time. The equation to represent this is

2=(r_w+r_c)t (3)

Substituting (2) for r_w yields

2=(1+r_c+r_c)t
2/t=1+2r_c
2r_c=2/t-1
r_c=1/t-1/2 (4)

Now examine the log. Return to d=rt and model what happens. The log travels 1 mile, but for 1+t hours. 1+t comes from the fact that the log is floating downstream while the woman is still paddling upstream for 1 hour and then paddling downstream for the unspecified amount of time, t. The equation for this is

1=r_c(1+t) (5)
1/r_c=1+t
1/r_c-1=t (6)

Equations (4) and (6) can now be used to solve for r_c. The work is below

1/(1/r_c-1)=r_c+1/2
1=(r_c+1/2)(1/r_c-1)
1=1-r_c+1/(2r_c)-1/2
0=-r_c-1/2+1/(2r_c)
0=-r_c^2-r_c/2+1/2
r_c^2+r_c/2-1/2=0
r_c=-1/4+sqrt(1/4+2)/2
r_c=-1/4+3/4
r_c=1/2

Thus the rate of the current is 1/2 miles per hour, or at least I think so.

This post is with the speed being constant relative to the bank.

A previous post (the one by daleks) stated that the stream would be going at half a mile per hour. But this must be incorrect, for the woman went 1 mile, met the log, and then went for another hour upstream. By now the log would have been half the way back to the start. The woman would then have to go back downstream for an hour to get to the 1 mile mark (for she would take 1 hour to go that same distance). When she reaches the 1 mile mark for the second time, she would have taken two hours to get back there and if the log was travelling at half a mile an hour, the log would have reached the start point but she would still be at the 1 mile mark, so this answer must be incorrect.

She travels 1 mile and meets a log.
From now the log travels 1 mile to the start.
She now travels at speed v for one hour upstream, travelling v miles.
She turns around and travels at speed v for one hour downstream, travelling v miles downstream.
She is now back at the 1 mile mark.
She now takes 1/v hours to get back to the start.
Therefore, both of them (the log and her) travel for 2 + 1/v hours before they get to the starting point.

Let the speed of the water be w.
speed = distance / time
w=1/(2+1/v)

It can be seen from this that if the speed is taken as constant relative to the bank, the speed of the river is linked to the speed of the woman.

If she was travelling at 1 mile per hour, the trip would take 3 hours. Hence, the stream must be travelling at 1/3 miles per hour.
If she was travelling at 2 miles per hour, the trip would take 2 and 1/2 hours. Hence, the stream must be travelling at 2/5 miles per hour.
If she was travelling at 10 miles per hour, the trip would take 2 1/10 hours. Hence, the stream must be travelling at 10/21 miles per hour.

The boundaries of the speed of the stream are:
upper 1/2 a mile per hour (as v tends to infinity)
lower 0 miles per hour (as v tends to 0)
Of course, neither of these speeds can actually be reached, as they are the limits. Also, the speed of the stream must always be slower than that of the woman.

" Clearly she will reach the log again at 2:00." - this is absolutely incorrect because the canoe speed is different in case of upstream and downstream travelling.

Here is the solution to the canoe problem:
==========================================

A---------B----------C----------------------D

A - start/end point
B - point where the log is at after the canoe wraps around the point D
C - point where the canoe meets the log
D - wrap around point of canoe

AC = 1 mile
CD = 1 hour
V - ground speed of the canoe
V1 - river stream speed
(V+V1) - canoe speed upstream
(V-V1) - canoe speed downstream

From the above drawing, it's obvious that the time it takes from log to get travel from B to A is the same as the time it takes canoe to travel from point D to A. So, all we need to get the answer is solve this equation:
AB/V1 = DA / (V+V1)
DA = CD + AC
AC = 1 miles
CD = 1hour * (V-V1)= V-V1
AB = 1 - V1
So, DA=1+V-V1.

After substituting the above values and solving the equation, we get the log speed as 0.5 miles/hour.
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