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Relative Speeds with vectors

  • Thread starter Klymene15
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Homework Statement



A swimmer heads directly across a river, swimming at 1.4 m/s relative to the water. She arrives at a point 48 m downstream from the point directly across the river, which is 74 m wide. What is the speed of the river current? What is the swimmer's speed relative to the shore? In what direction should the swimmer head so as to arrive at the point directly opposite her starting point?


Homework Equations


I'm guessing it's a mixture of x=v*t and the Pythagorean theorem.


The Attempt at a Solution



So, Displacement=83.2 m, which I figured out by making a triangle with the vertical and horizontal displacement. There are so many moving parts I've gotten so confused! I know I have to use vectors somehow to make a triangle. 1.4 meters per second is the hypotenuse of the triangle based on velocity (I made two. One based on displacement and another on the velocity. I figured they couldn't be mixed, and I couldn't find a way to translate either one to make them the same.)
 

Answers and Replies

  • #2
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The river current makes her moved 48m down stream.
Within that time too, she swims across for 74m.
 
  • #3
cepheid
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The velocity of the swimmer (s) relative to the water (w) is 1.4 m/s, (call this vsw, and this is in the vertical direction). The velocity of the water relative to the ground (g) is unknown, and this is in the horizontal direction (call it vwg). The velocity of the swimmer relative to the ground is just the vector sum of these two. I.e.:$$\vec{v}_{sg} = \vec{v}_{sw} + \vec{v}_{wg}$$Make sense? So the thing on the left hand side (the resultant velocity) is the hypotenuse of your right triangle, and the two things on the right hand side are the two perpendicular arms of the triangle.

Now, you don't know vwg, that's what you're trying to solve for. But you can figure out vsg, the resultant velocity, because you know the displacement of the swimmer (although you should re-check your calculation of this), and you can figure out the time interval (i.e. how long it took her to swim across the river).
 
  • #4
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Okay. Thanks, cephid. I'm kind of getting it, but not quite. Oh. and displacement is actually 88.2 Thanks for the tip.

So. I get the vector addition, but I'm not understanding how I can find the time. I would use the 1.4 m/s times 48 but the velocity's relative to the moving water. Is there an equation I should use?
 
  • #5
cepheid
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Okay. Thanks, cephid. I'm kind of getting it, but not quite. Oh. and displacement is actually 88.2 Thanks for the tip.

So. I get the vector addition, but I'm not understanding how I can find the time. I would use the 1.4 m/s times 48 but the velocity's relative to the moving water. Is there an equation I should use?
The horizontal (parallel to the bank) motion of the water doesn't affect the vertical (perpendicular to the bank) motion of the swimmer. You have the vertical displacement and the vertical speed, so it should be a simple matter to find the time required to traverse that vertical distance at that vertical speed (hint: you don't multiply the 1.4 and the 48, but you do do something with those numbers). According to an observer on the bank, the swimmer is moving along a diagonal path, that's true, but you've resolved that path into horizontal and vertical components, and now you're just dealing with the vertical component of the motion.

If you still don't get it conceptually, think about it from another frame of reference. The swimmer can, with equal validity, claim that he and the river are stationary, and that it is the bank and the rest of the world that is flowing past them in the opposite direction. So if the swimmer starts out stationary at one side of the river and then swims directly across its 74 m width at 1.4 m/s, how long will it take him to get to the other side? Notice that this is completely independent of what the rest of the world is doing.
 
  • #6
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Wait, just to clarify, 1.4 m/s represents what? The vertical speed (the speed to get across the river) or horizontal speed (the speed along the bank)
 
  • #7
cepheid
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Wait, just to clarify, 1.4 m/s represents what? The vertical speed (the speed to get across the river) or horizontal speed (the speed along the bank)
The vertical speed. It says in the problem that 1.4 m/s is the speed with which the swimmer moves across the river.
 
  • #8
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But the problem says "1.4 m/s relative to the water". I figured, since the water is only going horizontally, 1.4 m/s would represent the horizontal component of the swimmer's velocity vector
 
  • #9
cepheid
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Nope. Look at the first sentence. "A swimmer heads directly across the water."

Have a look at what I said above about how the trajectory is straight across the river *in the frame of reference of the river.* See if that makes sense to you.
 
  • #10
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Oh! So the horizontal component of the swimmer's velocity is the velocity of the river, and vertical component is the number listed. But how does that connect with the displacement?

I figure....

Velocity of Swimmer(horizontal)^2+velocity of river^2=total displacement of swimmer(88.2)/time^2

But where would I find the time, again? Is it 34 seconds? 48/1.4? I feel like it should be 74/1.4 instead...
 
  • #11
cepheid
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Oh! So the horizontal component of the swimmer's velocity is the velocity of the river, and vertical component is the number listed. But how does that connect with the displacement?

I figure....

Velocity of Swimmer(horizontal)^2+velocity of river^2=total displacement of swimmer(88.2)/time^2

But where would I find the time, again? Is it 34 seconds? 48/1.4? I feel like it should be 74/1.4 instead...
Yeah, it's 74 m / 1.4 m/s. The time taken to get across the river is the *vertical* displacement divided by the *vertical* speed.
 

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