Speed of current?

Question

Two miles upstream from his starting point, a canoeist passed a log floating in a river's current. After paddling upstream for one more hour, he paddled back and reached his starting point just as the log arrived. Find the speed of the current.

I don't get it, how do I calculate it from this info?

Please be clear, thanks!

Answer 1

let c = speed of the canoe
let r = speed of the current

let's call point B is where the canoeist passes the log

the canoeist passes a log 2 miles away from his starting point or at point B. When traveling upstream, the speed of the canoeist is
c - r. He keeps paddling for 1hr. So the conoeist is now
1*(c-r) = c - r miles away from point B

Meanwhile, the log floats downstream at the same speed of the current. So the speed of the log is r. During one hour, the log travels 1 * r = r miles away from point B.


We know that the log is 2 - r miles from the starting point. The log will reaches the staring point in (2 - r)/r hrs.

Given that the canoeist overtake the log at the starting point. So (2-r)/r is also the time it takes the canoeist to travel to the starting point.

The distance the canoeist has to travel is 2 + c - r. When traveling downstream, the c + r is the speed of the canoeist

distance = speed * time
2 + c - r = (c + r) (2 - r) / r

now solve
2r + cr - r^2 = (c + r) (2 - r)

2r + cr - r^2 = 2c - cr + 2r - r^2

cr = 2c - cr

divide c for both sides
r = 2 - r
2r = 2
r = 1

so the speed of the current is 1 mil/hr

hope this helps!

Answer 2

Let's use some notations the first answer used.

let c = speed of the canoe
let r = speed of the current

let's call point A the starting point, point B where the canoeist passes the log and point C where the canoeist turns back.

The log travels from B to A, the distance is 2 miles, the time is 2/r.

During this time 2/r, the canoeist travels from B to C and from C to A.
from B to C the distance is 1 hour times speed (c-r) so it is c-r (miles). Then he turns back, from C to B the distance is c-r, the speed is c+r, so the time is (c-r)/(c+r). From B to A, the distance is 2 miles, the speed is c+r, so the time is 2/(c+r). Therefore in total it takes 1 + (c-r)/(c+r) + 2/(c+r) for the canoeist to travel from B to C to A. It should equal to the time for the log, 2/r,
so
1 + (c-r)/(c+r) + 2/(c+r) = 2/r
[(c+r)+(c-r)+2]/(c+r) = 2/r
(2c+2)/(c+r) = 2/r
(2c+2)r = 2(c+r)
2cr + 2r =2c + 2r
2cr = 2c
so r = 1 (mile/hour)

An easier way to solve this question is like this.
Imagine you were standing on the log, then you would not see the water flowing. you would see the canoeist went away for 1 hour then turned back for another 1 hour. During this 2 hours, the starting point would go towards you for 2 miles. So the speed of the starting point was 2miles/2hour = 1 mile/hour. Of course the starting point can't move. The above speed actually is the speed of the log, or that of the current.