River Crossing

Geometry Level 4

John is located on the south bank of a river. He has a small raft that he must use to get across the river to the north bank. There is only one small location on the north bank that is suitable for landing with his raft.

Using the best strategy possible, how long will it take John to arrive at the safe landing point on the north bank of the river?

Give your answer in minutes: for example, 51.5 minutes.

Assumptions:

The river current is 1.4 km/hr, running to the west.
The river current is constant and even across the entire river.
The river has straight banks on the north and south and is running due west.
The river is 2 km wide.
John will maintain a constant forward speed through the water of 2 km/hr while paddling across the river.
John can maintain a constant forward speed of 3 km/hr along the beach while carrying his raft.
The south river bank is clear and open both to the west and east. John can carry his raft and launch his raft for the crossing from any location he wants along the south bank.
John’s starting location is directly across the river from the safe landing location, i.e. he is due south of the landing spot.

The answer is 78.45.

2 solutions

Darryl Dennis
Dec 13, 2018

Calculate paddling velocities components in the North and east directions.

In order to cross the river and arrive at the safe landing spot John most paddle in such a way that when he has made the 2km trip North and has also walked or paddled East a distance equal to the distance the current will carried him downstream while he was paddling.

${ D }_{PaddlingEast }=1.4{ T }_{ paddling }-3{ T }_{ walking }$

${ D }_{ PaddlingNorth }=2km$

$Velocity=\frac{ Distance }{ Time }$

${ { V }_{ PN } }=\frac { 2 }{ { T }_{ P } }$

${ { V }_{ PE } }=\frac { 1.4{ T }_{ P }-3{ T }_{ walking } }{ { T }_{ P } }$

${ { V }_{ PTot } }=2$

${ { V }_{ Total } }^{ 2 }={ { { V }_{ PN } }^{ 2 } }+{ { V }_{ PE } }^{ 2 }$

$4=\frac { { 2 }^{ 2 } }{ { { T }_{ P } }^{ 2 } } +\frac { { \left( 1.4{ T }_{ p }+3{ T }_{ W } \right) }^{ 2 } }{ { { T }_{ P } }^{ 2 } }$

$4{ { T }_{ P } }^{ 2 }=4+{ \left( 1.4{ T }_{ P }-3{ T }_{ w } \right) }^{ 2 }$

$4{ { T }_{ P } }^{ 2 }=4+1.96{ { T }_{ p } }^{ 2 }-8.4{ T }_{ P }{ T }_{ w }+9{ { T }_{ w } }^{ 2 }$

$0=-2.02{ { T }_{ P } }^{ 2 }-8.4{ T }_{ P }{ T }_{ w }+9{ T }_{ W }+4$

Quadratic equation ${ T }_{ P }=-\frac { 8.4{ T }_{ W }+\sqrt { 144{ { T }_{ W } }^{ 2 }+32.64 } }{ 4.08 }$

And ${ T }_{ P }=-\frac { 8.4{ T }_{ W }-\sqrt { 144{ { T }_{ W } }^{ 2 }+32.64 } }{ 4.08 }$

${ T }_{ T }={ T }_{ P }+{ T }_{ W }$

By determining the first derivative and then equating it to zero (i.e. no slope) it should be possible to determine the shortest possible time. To cross the river. It has been more than 40 years since I have done any calculus and I am not comfortable with my current skills there. I decided to determine the minimum value using a excel spreadsheet solver. I realize that this approach is not pure math and is not really computer science either.

I entered these expressions into excel sheet cells and used a excel solver to find the lowest possible value of total time

The time values of the speed i.e. 2km/hr, 3km/hr and 1.4km/hr are expressed in Hours. The answer is to be in given minutes ,therefore the calculated answer must be changed to minutes.

Minimum travel time 78.48 minutes total

12.53 minutes walking east for .637KM

Then 65.94 minutes paddling to arrive at the safe landing spot

Excel chart with total trip time on y Axis and walk time on X axis

Hi, could it not be walking East for 1.02km before paddling? I think we could do it under 79 minutes.

Saya Suka - 2 years ago

The answer I got by using lagrange multipliers was approx 78.38 minutes.

Tristan Goodman - 1 year ago

I have submitted a report on this problem. I now agree that the answer accepted is not correct. I don't know what can be done to rectify the solution and give credit for yourself and perhaps others. I am leaving it up to Brillian.org. For now, I have posted a solution that I believe is correct I cannot change the value that is accepted as correct. Your answer is close to my current solution the difference may be a rounding error.

It has been more than 40 years since I have done any calculus and don't feel comfortable in my abilities there. I decided to use a excel solver to find the minimum value once I had derived a function relating the total trip time to the walking time,

Thanks for the post. It was interesting.

Darryl Dennis - 1 year ago

If he walked east for 1.02 km before starting to paddle across the river

Walking time to cover the 1.02 km at 3.0 km/hr

1.02/3.0=0.34 hr or 20.4 minutes of walking time.

In order to reach the safe landing spot, he must paddle his raft so that he reaches the north shore directly north of his start location. . As he paddles the river current will be moving west at 1.4 km/hr. If he begins his river crossing from a point 1.02km east of the landing point his river crossing will be a trip that takes him 2km to the north and 1.02 km to the west to reach the landing spot on the north riverbank.

If he paddled directly north for the entire 2km distance to the north riverbank the river crossing would take him 1hr. (2.0km at 2.0 km/hr). John would arrive at the north riverbank 80.4 minutes after he started his walk to the east. His position on the north riverbank would be 1.4km west of the location that he began to paddle across the river due to the river current moving his raft west for the entire time while he was making the crossing. Since John had walked east 1.02km and then drifted west 1.4km during his crossing he would miss the point on the north bank that was safe to land his raft. He would reach the north riverbank 0.38km to the west of the safe landing spot.

He must paddle east as well as north in order to arrive at the safe land point.on the north bank.Having to paddle with a easterly component will reduce the northerly component of his crossing speed. and there for increase the time it takes to reach the north bank.

Darryl Dennis - 2 years ago

I hope you read my comment in full. I mentioned that we could do it under 79 minutes, so my answer must be 78+ minutes. I was marked wrong for that, but in your explanations you claimed an answer of 78+ minutes? Edit : oh, I thought asker get to change the answer at will anytime. Sorry, I'll help to get the staff's attention by adding another report.

Saya Suka - 5 months, 1 week ago

K T
Dec 30, 2020

Suppose John walks upstream during $t_1$ , over a distance $3t_1$ and then paddles during $t_2$ . During the paddling the water will drift over $1.4 t_2$ downstream. Using pythagoras, we find an expression for the distance $s_2$ he has to paddle relative to the water:

$s_2^2=2^2+ (1.4t_2-3t_1)^2$

Also, because of his paddling speed $s_2=2t_2$ Combining these we find $4t_2^2=4+(1.4t_2-3t_1)^2$

$3t_1=1.4t_2-2\sqrt{t^2-1}$

We want to minimize the total time, therefore we use $\frac{d (3t_1+3t_2)}{d t_2} = \frac{d 4.4t_2-2\sqrt{t^2-1}}{d t_2}= 4.4-\frac{2t_2}{\sqrt{t^2-1}}=0$

So that for the minimum total time we find $4.4^2(t_2^2-1)=2^2t_2^2$ $t_2=\frac{11}{24}\sqrt{6}$ Using the above formula for $3t_1$ we find that $t_1=\frac{3}{40}\sqrt{6}$ $60(t_1+t_2)=32\sqrt{6} \approx 78.38367..$

River Crossing

The answer is 78.45.

2 solutions

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