What is the ratio of time for two twins to cross a river at different angles?

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Homework Help Overview

The problem involves two twins rowing across a river with a current. Twin A rows straight across while Twin B rows at an angle to compensate for the current, aiming to reach a point directly across from their starting position. The task is to calculate the ratio of the time taken by each twin to cross the river and to determine the angle of Twin A's motion relative to the downstream direction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss setting up position formulas to find the time ratio for crossing the river. There are attempts to clarify the vector components involved in the motion of both twins, particularly regarding the use of sine and cosine functions in their calculations.

Discussion Status

Participants are exploring different interpretations of the vector components related to the twins' motions. Some have offered hints regarding vector addition and the relationship between the angles and components, while others are questioning their assumptions about the direction of the river and the corresponding components of velocity.

Contextual Notes

There is some confusion regarding the orientation of the river's flow and the definitions of upstream and downstream, which affects the calculations of angles and components. Participants are also navigating the implications of their assumptions on the setup of the problem.

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Homework Statement


Two twins set out to row separately across a swiftly moving river. They have identical canoes, and can row at the same speed in still water. Twin A aims straight across the river but, due to the current, is carried downstream before reaching the opposite bank. Twin B aims her canoe upstream at an angle of 56.0 degrees relative to the riverbank, so as to arrive on the opposite side at a point that is directly across from her starting point.

Calculate the ratio of the time it takes twin A to cross the river to the time it takes twin B to cross.

Determine the direction of twin A's motion, expressed as an angle relative to the downstream direction.


Homework Equations





The Attempt at a Solution


In order to find a ratio of time, I set up update of position formulas.

Twin A:
[tex]\Delta[/tex]X=Vavg*t

In the case of the x direction I'll call the initial velocity V which should equal Twin B's initial velocity

[tex]\Delta[/tex]X=V*t

time for twin A:
t=[tex]\Delta[/tex]X/V

Twin B:
[tex]\Delta[/tex]X=Vcos34*t
t=[tex]\Delta[/tex]X/Vcos34

Ratio of A/B for t is cos34, I got this part right.

I don't know how to go about the second part. The drag of the water represents a Force vector, and Twin A start off with a velocity vector in the x direction.

What I do know:
Initial y position=0
Initial y velocity=0
 
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Part 2 is a vector addition problem, where the two vectors are perpendicular:
The velocity of A with respect to the ground = velocity of A with respect to the water + velocity of the water with respect to the ground​

Hint: Make use of the fact that B travels directly across the river.
 
Alright, so can I assume that the vector downstream is Vsin(56)? And the vector to the other side of the stream is V?

Therefore, angle relative to the upstream direction should be arctan(1/sin56)?

Then 180-arctan(1/sin56)= direction relative to downstream direction = 150

But that's wrong. What did I do wrong?
 
danielatha4 said:
Alright, so can I assume that the vector downstream is Vsin(56)?
Why sine?
 
I assumed sine because it was the y component of the Vector V for twin B and I assumed that since he traveled 0 Y displacement that his Y vector (Vsin56) equaled the vector downstream.
 
danielatha4 said:
I assumed sine because it was the y component of the Vector V for twin B and I assumed that since he traveled 0 Y displacement that his Y vector (Vsin56) equaled the vector downstream.
The stream travels along the x-direction, so why use the y-component?
 
Sorry, I assumed the river to be in the Y direction. Upstream being higher than downstream.

Doesn't that coincide with the way I solved the first part?

Sorry for the misunderstanding, I'm sure I'm the messed up one though.
 
Wait, I see my confusion now. The sin of 56 is the x component. The sin of 34 is the y component. Besides that, did I do the rest of the problem correctly?
 
My bad for not realizing that you have the river moving along the y-direction. In any case, the angle is given with respect to the riverbank (the y-axis) and you want the component parallel to the riverbank.
 
  • #10
danielatha4 said:
Wait, I see my confusion now. The sin of 56 is the x component. The sin of 34 is the y component. Besides that, did I do the rest of the problem correctly?
There you go. Other than that, I think it was OK.
 
  • #11
Ok so arctan(1/sin34)=60.8 degrees.

Is this not 60.8 degrees relative to the riverbank from the upstream direction?

So the downstream angle should be 119.2 degrees? but this is wrong...
 
  • #12
danielatha4 said:
Ok so arctan(1/sin34)=60.8 degrees.

Is this not 60.8 degrees relative to the riverbank from the upstream direction?
No, it's relative to the downstream direction. (Rivers flow downstream. :wink:)

So the downstream angle should be 119.2 degrees? but this is wrong...
You're just mixing up downstream and upstream. You had the correct answer above.
 

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