Relative velocity of a swimmer

  • #1
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Homework Statement


A person decides to swim across a river 84m wide that has a current moving with a velocity of 0.40m/s[E]. The person swims 0.70m/s[N] relative to the water. In what direction should she swim if she lands at a point directly north of her starting position?

Note: This is part d of the question- in earlier parts we solved for the velocity of the person with respect to the earth: 0.80m/s[N 30° E], time it takes to cross the river: 1.2x10^2s, how far downstream the person will land: 48m.

Homework Equations


v=d/t?

The Attempt at a Solution


I assumed the speed of the person will remain the same: 0.70m/s and the speed and direction of the current will remain constant. I know the person will have to swim in the western direction to so that when the stream pushes them east, they'll land north relative to the starting position. I drew a diagram of this, where the original x and y components are the same(x component is the current's velocity and direction and y component is the swimmer's velocity and direction relative to the water, but the hypotenuse of the right triangle formed by these 2 vectors can be described as having some velocity, with the direction [N θ W], θ being its angle. I tried solving for θ using the tan ratio: tan-1(0.40/0.70)=29.74°. However, the answer in the textbook is [N 35° W]. What did I do wrong?
 

Answers and Replies

  • #2
billy_joule
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It sounds like you've drawn the swimmers vector triangle wrong. The hypotenuse is the swimmers velocity relative to the water, (in a NW direction), it's x component is equal and opposite to the waters velocity, and the Y component is of no interest to you (swimmers velocity relative to earth).
 
  • #3
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It sounds like you've drawn the swimmers vector triangle wrong. The hypotenuse is the swimmers velocity relative to the water, (in a NW direction), it's x component is equal and opposite to the waters velocity, and the Y component is of no interest to you (swimmers velocity relative to earth).
Shouldn't the x component be the current? And how do we solve for the angle if we don;t know the y component?
 
  • #4
billy_joule
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Shouldn't the x component be the current?
No, for the swimmer to head due north their net (ie relative to earth)x velocity must be zero, so, relative to the water, the swimmers x velocity needs to cancel the waters x velocity so should be equal and opposite.
Like running on a treadmill, you must run in the opposite direction to the belt to remain stationary relative to earth.

And how do we solve for the angle if we don;t know the y component?
We know two sides (opposite & hypotenuse) so can solve for theta.
 
  • #5
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This does work to match the answer in the text but why is the swimmers speed only .7 m/s on the hypotenuse when before they could swim that fast due north? This means they are now only swimming .6 m/s north.
 
  • #6
haruspex
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This does work to match the answer in the text but why is the swimmers speed only .7 m/s on the hypotenuse when before they could swim that fast due north? This means they are now only swimming .6 m/s north.
The swimmer's speed is always .7m/s relative to the water, no matter in which direction she heads.
It was initially given as .7m/s N relative to the water because in some earlier part of the question that was how she chose to swim. But in that part her velocity relative to the bank was somewhat to the E because of the current.
She now wishes to swim in such a direction relative to the water that her velocity relative to the bank is due N. Since some of her velocity will go into opposing the current, her northerly speed is reduced.
 

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