Time & Velocity Homework: Swimming Across a River

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SUMMARY

The problem involves a swimmer crossing a river that is 1.0 mile wide, swimming at a speed of 2.0 mi/hr perpendicular to the river's flow, which is also 2.0 mi/hr. The swimmer's resultant velocity can be calculated using the Pythagorean theorem, resulting in a speed of approximately 2.24 mi/hr. The time taken to cross the river is determined using the formula t = x/v, yielding a time of 0.45 hours, which should be converted to minutes for clarity. The analysis emphasizes the decomposition of motion into two independent components: one across the river and one along the river.

PREREQUISITES
  • Understanding of basic physics concepts, particularly motion and velocity
  • Familiarity with the Pythagorean theorem
  • Knowledge of time calculation using the formula t = x/v
  • Ability to convert time units from hours to minutes
NEXT STEPS
  • Study vector decomposition in physics to analyze motion in two dimensions
  • Learn about projectile motion and its similarities to this swimming problem
  • Practice additional problems involving relative motion in flowing mediums
  • Explore the concept of resultant velocity in various physical scenarios
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Students studying physics, particularly those focusing on kinematics and motion analysis, as well as educators looking for practical examples of vector decomposition in real-world scenarios.

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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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