# River crossing problem

1. May 7, 2010

### daysrunaway

1. The problem statement, all variables and given/known data
Two boys can each paddle their kayaks at the same speed in still water. They paddle across a river which is flowing at a velocity of vR. Boy A aims upstream at such an angle that he actually travels at right angles to vR. Boy B aims at right angles to the bank, but is carried downstream. Which boy crosses the river in less time?

2. Relevant equations
Cosine law: a2 = b2 + c2 - 2bccosA
Sine law: a/sinA = b/sinB = c/sinC
velocity = distance/time

3. The attempt at a solution
I first drew a vector diagram. It consists of two right-angled triangles with a common leg. The hypotenuse of the first is the velocity of boy B and the hypotenuse of the second is the hypothetical path that boy A would travel without current. The shared leg is both the velocity of boy A and the hypothetical path that boy B would travel without current. The other two legs are both vR.

Here I got confused: vB > vA, but isn't boy B's distance also greater than boy A's? I emailed my teacher for help and he gave me this terse answer:

"Draw vector diagrams for boat A and boat B. Determine which boat has the larger component to it's course in the direction perpendicular to the bank. It will make it across first."

But they have the same component to their course perpendicular to the bank, don't they?

I don't understand what I'm missing. Could someone please help me?

Last edited: May 7, 2010
2. May 7, 2010

### danielatha4

"But they have the same component to their course perpendicular to the bank, don't they?"

Not quite, if Boy B aims completely perpendicular to the bank then the entire magnitude of his velocity is, well, perpendicular to the bank, whereas Boy A's velocity is aimed up at an angle (if I understood the question correctly, the wording is weird) so only an "x" component of the velocity will be perpendicular.

3. May 29, 2012

### Jimbo57

I would like to resurrect this question since I'm working on it myself. I'm pretty sure boy B would arrive first because, as danielatha4 said, his entire velocity magnitude is perpendicular to the bank and assuming that there is a downriver landing then it would take him the same time getting across the river as if it were still water (ignoring other variables). He would just be further down depending on the river velocity. Boy A would always be spending a percentage of his velocity maintaining his trajectory to the shore, although he would make it directly across and not downriver.

Can anyone else verify this?

4. May 29, 2012

### Villyer

I arrive at the same conclusion.