Who Crosses the River Faster, Boy A or Boy B?

[daysrunaway]

Homework Statement

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 v_R. Boy A aims upstream at such an angle that he actually travels at right angles to v_R. Boy B aims at right angles to the bank, but is carried downstream. Which boy crosses the river in less time?

Homework Equations

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

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 v_R.

Here I got confused: v_B > v_A, 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?

 

[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.

 

[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?

 

[Villyer]

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?

I arrive at the same conclusion.

 

[asteriatic]

I would approach this problem by first clarifying the information and assumptions given. It is important to note that the problem states that both boys can paddle at the same speed in still water, but does not mention any specific speed for the kayaks or the river's velocity (vR). Without this information, it is difficult to accurately determine which boy crosses the river in less time.

Additionally, the cosine and sine laws provided may not be necessary for solving this problem. It seems that the key factor here is the angle at which each boy paddles, rather than the distance they travel. Therefore, I would suggest using basic trigonometry to determine the components of each boy's velocity in the direction perpendicular to the bank.

Overall, it is important to carefully consider the information provided and use appropriate equations and assumptions to solve the problem accurately. It may also be helpful to discuss the problem with your teacher or a peer to gain further insight.