[Guest]

Jimbo leaves his home on the river and paddles his canoe at a steady constant pace upstream. He reaches a spot that he knows is one mile from his home. At this moment, a gust of wind blows his hat off his head into the water. He doesn't stop to pick it up, and continues to paddle upstream for an additional ten minutes. He then decides to turn and paddle home, paddling with the same steady methodical pace as before. When he reaches his home, he notices his hat being carried under his pier by the gentle current of the river.

How fast is the current of the river traveling?

[Guest]

Say that the man's paddling speed was V, river flow speed is X.

Man is 1 mile away from home, hat blows off and starts going to the home direction at speed of X.

Man continues on for 10 minutes at speed of V-X (let's marks speeds in miles per minutes)

So man is at 1+10(V-X)

And hat is at 1-10X

The time that the hat took to reach home at speed of X is same that took man to do at speed of X+V so

(1+10V-10X)/(X+V) = (1-10X)/X

X+10VX-10X²=X-10X²+V-10XV

20XV=V

X=1/20 miles per minute

X=3 miles per hour!

[Guest]

That is similar to a puzzle that was posted not long ago. Following the same logic, if he droped the hat and came back after 10 minutes, he's gonna take 10 minutes to get to the hat. And if the hat traveled 1 mile in that time, the speed of the stream is 6 mph

[Guest]

Oops I forgot to consider the first 10 minutes. Yeah I think you're right, Anza Power.

[Guest]

I got a different answer, but I also made it much more complicated by assuming

V_upriver does not equal V_downriver

Instead

V_upriver = V_paddle - V_river and V_downriver = V_paddle + V_river.

[Guest]

hello anja power

if we take the upstream speed X-V,

then at the end the equation becomes V=20*square(X).........

wats ur view???

[Guest]

the speeds are of course dependant on one another

the equation I get was

V_river=V_paddle(1-10*V_paddle)/(9+21V_paddle)

If the man is paddling at 3mph the river is running at 1.21mph

Edited June 8, 2010 by bgover4

[Guest]

If you take the river as a reference, you could say that the hat is sitting still while the man is moving at a certain speed. After 10 minutes, he turns back, and since the hat is "in the same place", he's going to take 10 minutes to reach it. At this point, they will be on the same water as when he dropped the hat, so to speak, but this water will be 1 mile away from where it was before. So 1 mile in 20 minutes means 3mph

[Guest]

hello anja power

if we take the upstream speed X-V,

then at the end the equation becomes V=20*square(X).........

wats ur view???

When he's moving upstream it's V-X and downstream it's V+X. (X is river speed)

[Guest]

If you take the river as a reference, you could say that the hat is sitting still while the man is moving at a certain speed. After 10 minutes, he turns back, and since the hat is "in the same place", he's going to take 10 minutes to reach it. At this point, they will be on the same water as when he dropped the hat, so to speak, but this water will be 1 mile away from where it was before. So 1 mile in 20 minutes means 3mph

I really like this solution!

[Guest]

Anza,

I am not sure your solution is accurate....He paddled 10 minutes up river, but that doesn't mean it took him 10 more minutes to get back to the one mile mark. Up river he would travel a shorter distance in ten minutes than he would downriver. Therefore it will take him less than 20 minutes to to make the extra distance while the hat is floating.

Edited June 9, 2010 by bgover4

[Guest]

The river is traveling at 3mph. As long as V, the rate at which the boat is being paddled in still water, is greater than 3mph any V will work, as the V drops out in the math.

Assume that D1 is the distance that the boat goes in the 10 minutes upriver. D1=1/6(v-x) : one sixth of an hour at v-x miles per hour.

Assume that D2 is the distance that the boat goes downriver. This is just D1+1 or D2=1+1/6(v-x)

Now the time it takes the boat to travel D1+D2 is just the time it takes to travel D2 + 1/6 of an hour, or, t-boat = 1/6+(((1+1/6(x-v))/(x+v)) : distance D2 divided by speed for D2 of x+v and add 1/6 of an hour for D1

Now the time it takes for the hat to go one mile at x miles per hour is just: t-hat=1/x

We know that t-boat plus t-hat are equal, so 1/x=1/6+(((1+1/6(x-v))/(x+v)).

Solve for x and the v drops out and x=3.

[Guest]

Anza,

I am not sure your solution is accurate....He paddled 10 minutes up river, but that doesn't mean it took him 10 more minutes to get back to the one mile mark. Up river he would travel a shorter distance in ten minutes than he would downriver. Therefore it will take him less than 20 minutes to to make the extra distance while the hat is floating.

I never said that, you have me confused with the other posts, my answer is the 2nd post on the first page...

[Guest]

Now the time it takes the boat to travel D1+D2 is just the time it takes to travel D2 + 1/6 of an hour, or, t-boat = 1/6+(((1+1/6(x-v))/(x+v)) : distance D2 divided by speed for D2 of x+v and add 1/6 of an hour for D1

mathboy,

Now the time it takes the boat to travel D1+D2 is just the time it takes to travel D2 + 1/6 of an hour, or, t-boat = 1/6+(((1+1/6(v-x))/(x+v)) : distance D2 divided by speed for D2 of x+v and add 1/6 of an hour for D1

1/x=1/6+(((1+1/6(v-x))/(x+v)

[Guest]

mathboy,

Now the time it takes the boat to travel D1+D2 is just the time it takes to travel D2 + 1/6 of an hour, or, t-boat = 1/6+(((1+1/6(v-x))/(x+v)) : distance D2 divided by speed for D2 of x+v and add 1/6 of an hour for D1

1/x=1/6+(((1+1/6(v-x))/(x+v)

Bgover4

I'm not sure if you are disagreeing with me or what, but if you check my solution it works. And I can't see where you have corrected me anywhere.

[Guest]

Drops hat, moves on 10 minutes, turns around and comes back.

It will take him 10 minutes to reach his hat again (because he and his hat are in the same environmental frame of reference i.e. the river).

This also happens to be how long it took for his hat to get home (and cross 1 mile).

The average speed of the river is 1 mile / 10 minutes = 1 mile/10 minutes * 60 minutes/hour = 6 mph

[Guest]

Drops hat, moves on 10 minutes, turns around and comes back.

It will take him 10 minutes to reach his hat again (because he and his hat are in the same environmental frame of reference i.e. the river).

This also happens to be how long it took for his hat to get home (and cross 1 mile).

The average speed of the river is 1 mile / 10 minutes = 1 mile/10 minutes * 60 minutes/hour = 6 mph

Lol i repeated the same mistake as someone else.

Drops hat, moves on 10 minutes.

After that, takes 10 minutes to get back to hat.

In whole time of 20 min, hat has traveled 1 mile ---> 3mph