A pretty simple physics problem.

133794m3r · Post by **133794m3r** » Fri 19 Oct, 2007 3:19 am

Ok well here we go it's a relatively simple problem. It took me about six hours to do but after I did the answer seemed so simple. And yes there is an answer.

A man is on a river in a canoe. He is going against the current. There is a bridge up stream of him. When he goes under the bridge the bottle that's sitting on the very front of his boat is knocked off. He continues going for exactly thirty minutes until he realizes that he's lost the bottle. He turns around and catches up with the bottle exactly one mile downstream from where it fell of under the bridge. What is the exact velocity of the water.

(You may use the English system for the problem.)

P.S.
This problem has stumped many mathematicians, if you can't solve don't worry take a few minutes off and come back to it later. I shall revile the answer only after everyone one has chimed in or a week has passed.

P.P.S

The six hours was six complete hours. I found this equation in the book A Biography of Physics .

Code: Select all

 010000000 01000001 010000000 0110000010

Homestar · Post by **Homestar** » Fri 19 Oct, 2007 4:26 am

I assume the man is moving the exact speed of the water meaning he moves nowhere underneath the bridge. Highlight post for the workout and Answer
M = speed of man, W = Speed of water, B = position/distance of the bottle

|M| = |W|, and M + W = 0 {as W is negative}

(0.5hr)W = B

B + 1mi = -M + W {W - M = 2W or -2M}

{exchange B for (0.5hr)W} (0.5hr)W + 1mi = 2W

{subtract (0.5) W from both sides} 1mi = (1.5hr)W

{divide both sides by (1.5hr)} 1mi/(1.5hr) = W

(Highlight for spoiler vvv)
_________________________________
|W = 2mi/3hr or 2/3 MPH|
_________________________________
Solved in 10 min and 20 seconds including the explanation

CoBB · Post by **CoBB** » Fri 19 Oct, 2007 7:38 am

Hmm, I have a different solution, but I’m notoriously bad at basic maths, so take it with a pinch of salt.

v_w - velocity of water
v_m - the relative velocity of man to the water (I’m assuming it is the same in both directions)
p_b - position of bottle (the origin is the bridge, and the water flows in the positive direction)
p_m - position of man

We have three notable moments:
0 s: p_b = p_m = 0
1800 s: p_b = 1800v_w, p_m = 1800(v_w - v_m)
t s (some moment in the future): p_b = 1mi = t*v_w, p_m = 1mi = 1800(v_w - v_m) + (t - 1800)(v_w + v_m)

Let’s look at the equations at time t:

1mi = t*v_w = 1800(v_w - v_m) + (t - 1800)(v_w + v_m)
t*v_w = 1800v_w - 1800v_m + t*v_w + t * v_m - 1800v_w - 1800v_m
0 = t * v_m - 3600 * v_m = (t - 3600)v_m

There are to solutions: either the man is not rowing at all, i.e. v_m is 0, or excluding that, t is 3600 seconds, i.e. the bottle travelled 1 mile in an hour.

Hopefully I didn’t misunderstand anything, because I don’t see anything in this problem that should stump a mathematician.

coelurus · Post by **coelurus** » Fri 19 Oct, 2007 8:50 am

Isn't this one of those "hard" exercises you do in junior high school?

Homestar · Post by **Homestar** » Fri 19 Oct, 2007 2:13 pm

M = speed of man, W = Speed of water, B = position/distance of the bottle in 30 minutes, and T = time of man to retrieve the bottle

|M| = |W|, and M + W = 0 {as W is negative} W - M = 2W

(0.5hr)W = B

B + TW = 1mi

(0.5hr)W + TW = 1mi

WT = 1mi - (0.5hr)W

1mi = (W - M)T

1mi = 2WT {replace WT with 1mi - (0.5hr)W}

1mi = 2mi - (1hr)W {add 2mi to both sides}

-1mi = -(1hr)W

1mi/1hr = W

Therefore W = 1MPH

I made that post just before I went to bed, and as soon as I hit the pillow I went "wait a minute, did I just add the distance of the bottle to the total distance?"

and then I was like "ohwell, I'll figure out in the morning"... and it turns out that I did,

133794m3r · Post by **133794m3r** » Fri 19 Oct, 2007 4:50 pm

Yeah you got it right. Good for you.

thecheat · Post by **thecheat** » Sat 27 Oct, 2007 1:14 pm

LOL, I got it wrong, but I ended up with 2mph... probably because I forgot to factor in something...

my idea is because this somehow is a physics problem, the only way for the bottle to get knocked off is if the bottle has a higher velocity then the boat. Bridges accelerate the stream a bit and I assumed that the man went 30 minutes without moving under the bridge. He slowed down, the bottle fell off (inertia) then he went after it.

133794m3r · Post by **133794m3r** » Thu 01 Nov, 2007 12:29 am

Ok I guess that I forgot that woops. Ok the bottle gets knocked off as he goes under it. He I guess hit a pole. That shouldn't matter. I mean there's Vs, As, and time. Thus its physics.