ok i feel really stupid for asking this question but i cant figure this out...?

Question

ive drawn like a billion diagrams to model this situation but it still doesnt help. and NO, this not hw. it IS in my math textbook but im doing it as practice and not to get good grades...and hopefully you all believe me.

anyways:
two miles upstream from his starting point, a canoeist passed a log floating in the rivers current. After paddling upstream for one more hour, he finally paddled back and reached his starting point just as the log arrived. Find the speed of the current.

the part i get screwed up at is when i have to take into consideration the guy going "upsteam" so against the current which will obviously affect his speed. and downstream which will make him go faster.

oh and please show me how to solve it STEP BY STEP so i acutally KNOW how to do it.

wtf...y!a wanted to put this in singles and dating. haha.

no. it can be done. i asked my teacher and he said there was and answer.

i even SAW the answer in the teachers textbook with the answer key. but i have no idea how they got it...and plus,i forgot the answer.

omg. thnx you guys!!!

i feel kinda dumb now cuz i was going along that path but must've screwed up so two variables didnt cancel out.

anyways THNX!!!!

or oops. lemme be mathematically correct and not say "cancel out"

i just meant that i kept getting an answer in terms of the other variable and not as a number.

Answer 1

Call the speed of the current r. Then the log travels the 2 miles downstream in time 2/r. Meanwhile the canoeist, paddling along at speed c relative to the river, has travelled upstream at speed c-r (relative to the bank) for 1 hour, or c-r miles, to a turning point which is therefore c-r+2 miles from his starting point, and then downstream for the same c-r+2 miles back to his starting point at a speed of c+r relative to the bank, which therefore takes him (c-r+2)/(c+r) hours. When he gets back, the total time since he passed the log is therefore 1+ (c-r+2)/(c+r) hours. It is a trivial matter, left to the reader in the traditional way, to solve for the two times to be equal, ie 2/r = 1+(c-r+2)/(c+r), to obtain r=1, ie the current is 1mph. A neat little puzzle.

PS We don't know anything about the canoeist's speed, in fact he could as easily have been in a sports car travelling at 70mph along the riverside road or a combat aircraft travelling at 1000mph above the river.

Answer 2

Let P be the speed of the paddler (canoeist)
Let L be the speed of the log (current)

Going upstream, the canoeist goes at a speed of P-L for 1 hour, or a total distance of P-L.

Going downstream, the canoeist travels this same distance plus an additional 2 miles (P-L + 2), in a time of T.
And the log goes downstream for 2 miles for a time of T + 1.

Canoeist's distance downstream: P-L+2
Canoeist's rate downstream: P+L
Canoeist's time: T

Log's distance downstream: 2
Log's rate downstream: L
Log's time: 1 + T

Using D = RT for the log, we get:
2 = L(1 + T)
L = 2/(1 + T)

And using T = D/R for the canoeist, we get:
T = (P-L+2)/(P+L)

Now substitute T back into L = 2(1 + T)
L = 2/(1 + (P-L+2)/(P+L))

Inside the parentheses, replace 1 with (P+L)/(P+L), so you have a common denominator:
L = 2/((P+L) + (P-L+2)) / (P+L)

Switch P+L to the numerator:
L = 2(P+L) / (2P + 2)
Factor out a 2 and then cancel:
L = 2(P+L) / 2(P+1)
L = (P+L)/(P+1)

Multiply both sides by P+1:
L(P+1) = (P+L)

Multiply L through on the left:
LP + L = P + L

Subtract L from both sides:
LP + L - L = P
LP = P

Divide both sides by P:
L = P/P
L = 1

So the log was traveling at 1 mph with the current.

Edit: I then spent a long time trying to solve for P but realized P could be anything. For example, if P=2, then his rate upstream would be P-1 = 1 mph and his rate downstream would be P+1 = 3 mph. Thus he would travel 1 mile upstream in the hour, and 1+2 miles downstream in another hour. So I thought I had an answer.

But the canoeist could just have easily paddled at 5 mph. Upstream he would go 4 miles, then he has to go downstream at 6 mph to make up the 4 miles + 2 miles. But this all makes sense. So maybe he was traveling at 5 mph, we don't know.

We can know the speed of the current (a.k.a. the speed of the log) as 1 m.p.h. but the speed of the canoeist could be most anything (as long as it is greater than 1). Fortunately your question only asked for the speed of the current.

Answer: The current is flowing at 1 m.p.h.

Answer 3

At 1st glance, this seems to probably be an over-simple problem. In other words, it probably really *can't* be worked out.

My initial reaction is that the log is traveling at 2 miles an hour. The log traveled 2 miles, and the paddler for 1 hour. However, the way the problem is worded, the paddler was traveling *upstream* for 1 hour, and we have no idea how quickly he paddled back. If the paddler *met* the log after 1 hour (during which he continued to paddle upstream, and then paddled home) then we can say with certainty that the log is traveling at 2 miles per hour.

Otherwise, we have this (typing as I think):

paddler speed: P
river speed: R
log speed: L = river speed = R

speed of paddler upstream = P - R

(P - R) * 1 hour = additional distance paddler went upstream = dU
total distance of paddler = dP
= 2 + dU = 2 + (P-R) = 2 + P - R

speed of paddler downstream is P + R

the time spent paddling downstream = tD
this is also the time (additional to 1 hour) that it took the log to return to "home"

L = R = 2 miles / (1 hour + tD)
solve for tD, R * (1 + tD) = 2
R + tD * R = 2
tD * R = 2 - R
tD = (2 - R) / R (tD equation #1)

P + R = speed of paddler downstream = dP/tD
solve for tD
tD = dP / (P + R)
insert dP equation
tD = (2 + P - R) / (P + R) (tD equation #2)

equate tD equations
(2 - R) / R = (2 + P - R) / (P + R)

(2 - R) * (P + R) = (2 + P - R) * R
2P + 2R - RP - R ^ 2 = 2R + RP - R ^ 2
2P - RP = RP
2P = 2RP
R = 1 = 1 mile per hour

Wow, it worked! Who knew....


Check...
paddler paddled upstream 1 hour
dU = P - R = P - 1
P = paddler speed, so
(P - R) * 1 hour = dU = P - 1
R = L = 1 so it works!!!!!

Jim, http://www.jimpettis.com/wheel/

Answer 4

Initially, I suspect a piece of info is missing.

Let x mph be his paddling speed in still water.
Let y mph be the speed of the current.

For the log, it traveled 2 miles downstream at y mph to reach the canoeist's starting point. Thus the time taken is 2/y h.

For the canoeist, he travels upstream at x-y mph and downstream at x+y mph. Another hour upstream covered a distance of (x-y) m. Thus, he traveled 2+(x-y) miles downstream at x+y mph to reach his starting point. Thus, time taken is (2+x-y)/(x+y).

So, you have the equation
1 + (2+x-y)/(x+y) = 2/y

Normally, to find y, or x, there should be another equation.
However, in this case, the above equation can be simplified to
xy = x

Although we still can't know x, we know y = 1.

Answer 5

the crux of the problem is that in the time the log covered 2 miles the canoeist covered 2 times the distance upstream ( for which he took 1 hour going up) + 2 miles extra down( which he had already covered when he saw the log on his way up, so he covers that distance only once in the same time as the log covered the 2 miles).
now, let the canoeists speed be x, the streams speed be y
the speed of canoest upstraem is x-y and downstream is x+y
speed of log = speed of stream, which is y
time taken foor log to come down to the end = distance/speed = 2miles/y = 2/y
distance covered further upstream by canoeist = (x-y)mph * 1 hr = x-y
time taken to come downstream will be distance/speed going down = x-y / x+y sp total time taken by canoeist can be calculated in 3 parts
1) time taken to go up = 1 hr
2)time taken to come down the same distance = x-y/x+y
3)time taken to go down further 2 miles to the end = 2/x+y
total time for the canoeist = 1+x-y/x+y + 2/x+y
total time for the log = 2/y
these two times are equal
so, 1+x-y/x+y + 2/x+y = 2/y
x+y+x-y+2 /x+y =2/y
2x+2 /x+y = 2/y
cancelling 2 on both sides numerator
x+1/x+y = 1/y
cross multiply
xy+y=x+y
xy=x
xy-x=0
x(y-1) = 0

hey, i know my concept is write but i guess i m going wrong with my math...maybe if u can tell us the answer we can get this right

Answer 6

1 mile / hr ?

try searching for the problem in a search engine

u ----speed of canoe in still water
v----speed of stream

upstream --------u-v
downstream ----------u+v

let t hours be the time taken by the log to reach the starting point (2miles) after being seen first
therefore t = (2 / v) ----------(1)
let the canoe travels k more miles upstream after seeing the log in the next 1 hour
therefore 1 = [ k /(u- v) ]
or k = u -v ----------(2)

therefore the canoe travels k miles upstream in one hour and (k+2) miles downstream in the t hours
therefore t = 1 + [(k+2) / (u+v)]

t = 1 + [(u -v +2) / (u+v)] using equation (2)

t = (2u+2) / (u+v)

substitute in equation(1)

(2/v) = (2u+2) / (u+v)
2(u+v) = v(2u+2)
simplifying
v = 1

Answer 7

If they need a concrete answer, I'm not sure it can be done - you don't have the time he paddled to reach the two mile marker, you don't have the time he arrived at the starting point.

You can probably devise a relative equation as an answer but it will be solely relational, without a "real" number answer

Answer 8

Assuming that the canoe is moving at a fixed speed, then the speed of current is 1 mph