Logic problem?

Question

John rows his boat at the same speed. He left his front porch one morning rowing upstream. When he was one mile from his house his hat fell off and landed in the water. It was an old hat so he continued rowing. Ten minutes later he remembered that he had a $1 bill stuck inside that old hat so he turned (without losing any speed) down river. He caught up with his hat just as it bumped up against the front of his porch. What was the speed of the river current that morning in mph?? and why? I can't solve this problem and would like some help

Answer 1

Answer: 3 mph.

Actually, it is quite simple, if you happen to know about reference frames.

Assume you're drifting downstream in a raft in the same river. Just when John overtakes you, his hat falls onto the water. You don't know about the dollar bill, so you don't make for it. As the raft and the hat drift downstream with the same speed, you see the hat always at the same distance; it doesn't get closer, nor go away. The stream is your reference frame, and from this standpoint, both the raft and the hat are at rest.

As for John, you see him rowing at some unknown speed for 10 minutes, and then you see him turning around and approaching. How much will it take for John to catch with the raft? Clearly, another 10 minutes, since boat's speed, relative to the stream, is always the same.

But he catches up with his hat in front of his porch, 1 mile downstream of the spot where he lost his hat. Now, if in 20 minutes the stream traveled 1 mile, in a lapse of time 3 times longer, the stream would have traversed 3 times 1 mile...

I think you owe me 10 points...

Answer 2

Let
b = speed of boat in still water
c = speed of current
dup = distance traveled upstream beyond 1 mile
tup = time traveled upstream beyond 1 mile
tdown = time rowing down river

We have
b - c = speed up river
b + c = speed down river

tup = 10 minutes = 1/6 hour
dup = (b + c)*(1/6) = (b + c)/6

tdown = (time hat in water) - 1/6 = 1/c - 1/6

tdown = 1/c - 1/6 = (1 + (b - c)/6) / (b + c)
(6 - c)/(6c) = (6 + b - c)/{6(b + c)}
(6 - c)/c = (6 + b - c)/(b + c)
(6 - c)(b + c) = c(6 + b - c)
6b + 6c - bc - c² = 6c + bc - c²
6b - bc = bc
6b - 2bc = 0
2b(3 - c) = 0
c = 3 miles/hour

The speed of the river current was 3 miles/hour.

Answer 3

wow this is actually hard

well lets start off setting the velocity of the boat as v and the velocity of the current as vc. so going upstream his velocity is v-vc while downstream is v+vc. the distance the hat traveled at velocity vc is 1 mile in a time in hours we'll call t. t also equals the distance the boat traveled at v-vc for 10 min + that distance and 1 mile at v+vc. the distance the boat covered in the 10 min after the mile is (1/6)(v-vc). we are trying to solve for vc.

so

vc = 1/6 + (1+[1/6](v-vc))/(v+vc)
and 1/6(v-vc) = t(v+vc) - 1
and t = vc

theres ur two variables, two equations
plug and chug in ur grapher to get vc!

Answer 4

Try using the formula D=RxT (Distance = Rate x Time)

His rate is a constant.
The stream rate is a constant, although it works with him sometimes and against him sometimes.
The distance up stream is a mile, plus whatever he went in the ten minutes.
The distance downstream is the same
The only time you have is the ten minutes.

You need to figure out how to put all of these together into a single equation or a systems of equations.

Good luck

Answer 5

Well the answer could be either really simple or quite difficult. Firstly we need to establish the fact that the question is kinda misleading as it gives information on John and him rowing his boat, where the question asks for the SPEED OF THE RIVER not the SPEED OF THE BOAT.


basiclly it would be 1mile per 10mins as he dropped the hat at 1mile and it took the hat 10mins to return to the house
or 0.167mph

Answer 6

I can get a solution but it is not unique; for example, the lack of sufficient information allows for more then one solution.

Answer 7

the same speedrowing...
i cant get an answer exactly in numbers

Answer 8

without some assumptions the question can not be answered