SOLUTION: 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

Let u be the speed of the boat in still water (in km/h).

and let v be the speed of the river ( == the speed of the current ).


(1)  First, person travels 1 km upstream.



(2)  At that time, he sees the log floating by.  Let call this position "the meeting point".



(3)  The person continue to travel upsteam for 1 more hour.


     During this hour, the person travels (u-v) km upstream.
     
     During this hour, the log travels v kilometers downstream with the current.



(4)  Then the person turns back, travels downstream and reaches the starting point at the same time as the log comes to this point.


         +----------------------------------------------+
         |   Let's consider kinematic, starting from    |
         |   the "meeting point" to the ending moment.  |
         +----------------------------------------------+


    During this time, the log travel 1 km with the current speed.



    The motorboat travels 1 hour u-v kilometers upstream, and then travels  ((u-v)+1)) kilometers DOWNSTREAM with the speed of (u+v) km/h.



    So, the log's travel time from the meeting point to the end is    hours;

        the motorboat's travel time from the meeting point to the end is    hours.



    The two travel times are the same, giving this "time equation"

         =  + 1   hours.



    OK. The setup is done.  Now our task is to solve the equation.

        For it, multiply both sides by v*(u+v).  You will get


            u + v = v*(u-v+1) + v*(u+v)

            u + v = vu - v^2 + v + vu + v^2

            u     = vu           + vu

            u     =  2vu

     Now divide both sides by u (which is not a zero, so we can divide safely)  and get

            1     = 2v

            v = .


The problem is just solved ( by a miraculous way (!) ).


ANSWER.  The speed of the river (== the speed of the current) is    km/h.

    As you understand from my post, I solved this problem INDEPENDENTLY from the solution to this problem

    given in the archive  MANY  YEARS  AGO  (question 870229; the link  

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.870229.html    )


    Now, couple hours after submitting my post, I looked at the archive' solution to this problem by @mananth.


    My  DIAGNOSIS  after looking at that post is that @mananth INCORRECTLY interprets the problem 

    and THEREFORE incorrectly solves it.



    This so called "tutor" is very well known at this forum as a person who REGULARLY produces wrong solutions.


    This problem is DEFINITELY of the much higher level than his (or her).


    So, simply ignore his / (her)  post, for your safety.

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 
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 1st spotted = 

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: 

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 1st seen, DOWNSTREAM, to the starting point = 

Time it took the log to travel from the 1st time itw’s seen, to the starting point: 
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: 
Therefore, we get the following TIME equation: 

                                                ------ Reducing fractions by dividing by numerator-GCF, 20
                                                ------- Multiplying by LCD, v(S + v)
                                               
                    River’s velocity/speed, or 

Note that both problems are the same; however, one uses velocity/speed in , while the other uses . 
My answer:  is the same, when converted, as .