Question 1889: One mile upstream from his starting point, a rower passed a log floating with the current. After rowing upstream for one more hour, he rowed back and reached his starting point just as the log arrived. How fast was the current flowing??
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 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
|
|
|
