Question 1182081
I have tried to understand the answer given to question 870229 but I am struggling with the initial translation of words into algebraic relationships. It doesn’t look correct to me, for example, it looks like the given solution falsely assumes that the motorboat travels 1km in one hour but there is no statement to indicate the time actually taken to travel that 1km. 
The question does say the motorboat does travel “one more hour upstream at the same speed” but this doesn’t mean the first km took one hour to traverse. Eg. If the first km took 10hr, it is completely possible to “travel one more hour at the same speed “.  Clearly an assumption of 1km/hr is wrong in such a case. 

What am I missing here?  How can the given solution claim that the first km was traveled in one hour?

I am very grateful if someone can help me understand this as I find the translation of problems in words to algebraic relationships to be my biggest challenge. 

The original question is repeated below:
A river flows with uniform velocity v. A person in a motorboat travels 1km upstream, at which time a log is seen floating by. The person continues to travel upstream for one more hour at the same speed and then returns downstream to the starting point, where the same log is seen again. Find the velocity of the river. (Hint: The time of travel of the boat after it meets the log equals the time of travel of the log.)
<pre>The response below is for Problem # 870229. Note that you got a response from one of the WORST persons in this forum. That's why you were confused because that person has ABSOLUTELY no idea how to solve simple problems, much less a somewhat complex TRAVEL problem such as this. It's as they say, "You get what you paid for!" Luckily, you didn't pay for the LOUSY/RIDICULOUS answer you received.

Now, getting to the problem at hand, it should be noted that velocity/speed is in {{{meters/minute}}}
Let speed of boat be S
With the river’s speed being v, the time taken to travel 1,000 m upstream, when log is 1<sup>st</sup> spotted = {{{"1,000"/(S - v)}}}

Time taken to travel further upstream: 60 m
Distance traveled upstream, after log was spotted: 60(S - v) = 60S - 60v
Total distance traveled, one-way upstream (also distance downstream, back to starting point) = 1,000 + 60S - 60v

Returning to the starting point, after turning around means that another 1,000 + 60S  -  60v was traveled. However, the return distance, being DOWNSTREAM, was at an average speed of S + v
We then get the time taken to travel DOWNSTREAM as: {{{("1,000" + 60S  -  60v)/(S + v)}}}

The log was again spotted at the starting point, so, in the time it took the boat to go further upstream - after spotting log the 1st time - turn around and then back to the starting point, the log had traveled 1,000 m, DOWNSTREAM

The SPEED of the log is the speed of the river’s current, or v. Therefore, time taken for the log to travel from the spot itw’s 1<sup>st</sup> seen, DOWNSTREAM, to the starting point = {{{"1,000"/v}}}

Time it took the log to travel from the 1<sup>st</sup> time itw’s seen, to the starting point: {{{"1,000"/v}}}
Time taken for the boat to travel upstream, after seeing log: 60 minutes
Time taken by boat to travel downstream, back to starting point, after turning around: {{{("1,000" + 60S  -  60v)/(S + v)}}}
Therefore, we get the following TIME equation: {{{matrix(1,3, "1,000"/v, "=", 60 + ("1,000" + 60S  -  60v)/(S + v))}}}

                                               {{{matrix(1,3, 50/v, "=", 3 + (50 + 3S  -  3v)/(S + v))}}} ------ Reducing fractions by dividing by numerator-GCF, 20
                                               {{{matrix(1,3, 50(S + v), "=", 3v(S + v) + v(50 + 3S  -  3v))}}} ------- Multiplying by LCD, v(S + v)
                                               {{{matrix(3,3, 50S + 50v, "=", 3Sv + 3v^2 + 50v + 3Sv  -  3v^2, 50S, "=", 3v^2  -  3v^2 + 3Sv + 3Sv + 50v - 50v, 50S, "=", 6Sv)}}}
                    River’s velocity/speed, or {{{highlight_green(matrix(1,8, v, "=", 50S/(6S), "=", 25/3, "=", 8&1/3, meters/minute))}}}

Note that both problems are the same; however, one uses velocity/speed in {{{meters/minute}}}, while the other uses {{{kilometers/hour}}}. 
My answer: {{{matrix(1,2, 8&1/3, meters/minute)}}} is the same, when converted, as {{{matrix(1,4, 1/2, or, .5, km/hour)}}}.</pre>