Time & Velocity Homework: Swimming Across a River

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Homework Help Overview

The problem involves a swimmer crossing a river that is 1.0 mile wide, with the swimmer's speed being 2.0 mi/hr perpendicular to the river flow, which also flows at 2.0 mi/hr. The objective is to determine the time taken to reach the opposite bank.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the decomposition of the swimmer's motion into perpendicular components, questioning the setup of the right triangle used to analyze the problem. There are attempts to calculate the resultant vector and time taken using different methods, leading to some confusion regarding units and calculations.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some guidance has been offered regarding the calculation of time and the need to consider units, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

Participants are navigating potential misunderstandings related to the geometry of the problem and the implications of the swimmer's speed relative to the river's flow. There is also mention of unit conversion that may affect the interpretation of the results.

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Homework Statement


A person who can swim at 2.0 mi/hr is swimming perpendicularly to the bank of a river (directly across the direction of river flow) which is flowing at 2.0 mi/hr. If the river is 1.0 mi wide, how long does it take to reach the other side?


Homework Equations


t=x/v


The Attempt at a Solution


I drew a right triange: the first being 2 mi/hr (the direction of the river flow) and the second being 2mi/hr (the person swimming to the bank) and tried to solve for the hypotenuse through Pythagoreans theorem. Then used t=1/2.83 but it was incorrect. I think I'm drawing the figure wrong?
 
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The resultant motion of the swimmer can be decomposed into two perpendicular independent motions - one with the speed of the swimmer across the river and the other with the speed of the river along the direction of the river. When he has reached the opposite embankment the perpendicular component "covered" a distance 1.0 mi at a speed of 2.0 mi/hr.
 
Ok, I'm still a little confused. So now I have set up one of my legs on my right triangle as 2 mi/hr for the direction of the water flow, and the other leg as 1.0 mile for the distance he covered. Then I solved for the resultant vector: square root of 2^2 + 1^2, which equals 2.24. Then I plugged it into the time equation, x=1, v=2.24 and got 0.45. Am I on the right track?
 
I think you had it right the first way you did it. Is the answer given in minutes? Because the time you calculated (t=1/2.83=0.35) is in hours. Convert it to minutes to see if you have the correct answer.
 
You can approach this problem similar to how the motion of a projectile is analyzed. That is the resultant motion is described by two separate motions - one in the x- and another in the y-direction. For the swimmer it is in one across the river and another along the river. To solve this problem you need only look at the component across the river. The swimmers component in this direction covers one mile at a constant speed of 2 mi/hr when the actual swimmer crosses over to the other side.
 

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