Question 1889
Sol: 
      Start       A(point meeting the log)
      <---------------------------------->
 
     Let Vb be the speed of the rower,and Vc be the speed of the current.
     So,the upstream rowing speed is Vb-Vc, and the downstream rowing speed is 
     Vb + Vc.
     Let A be the point where the rower met the log(first time).
     One hour later, the rower reached Vb-Vc miles from point A while the log 
  reached Vc  miles from A. Hence, their distance is Vb miles at that time.  
     Next,we see that, it took Vb/(Vb+Vc-Vc) = 1 hr for the rower back 
    (downstream) to  reach the log at the starting point. Totally,the log 
     floating 1 mile for 1+1= 2 hrs.
     Therefore,the speed of current flowing is 1/2 miles/hr.

     The key point ofsolving this problem is to use
     when two running in the opposite direction with speed V1(=Vb-Vc) & v2(=Vc)
      Distance/(V1+V2)  = time (that is why we got Distance =(Vb-Vc+Vc) * 1 hr 
     = Vb miles)
     when two running in the same direction with speed v1 & v2(if V1=Vb+Vc > 
     V2=Vc)
     Distance/(V1-V2) = time (that is why we got Time = Vb/(Vb+Vc-Vc) = 1 hr)
  
     By the way,if you are able to use the idea about time, then you may 
     not solve this problem.


 Kenny