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Relative velocity question

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A river is flowing 4.0 m/s to the east. A boater on the south shore plans to reach a dock on the north shore 30.0 Degrees downriver by heading directly across the river. What should be the boat's speed relative to the water? Also, what's the boat's speed relative to the dock?


    2. Relevant equations

    Not sure.

    3. The attempt at a solution

    I have no idea whatsoever how to do this.
     
    Last edited: Feb 11, 2012
  2. jcsd
  3. Feb 11, 2012 #2

    cepheid

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    Your velocity relative to the bank is a vector sum of your velocity relative to the water (which is northward, i.e. across the river) and the velocity of the water relative to the bank (which is eastward). Since these two vectors are perpendicular, they form a right triangle, with the resultant (sum) as the hypotenuse. The angle of this diagonal path across the river is going to depend on the relative lengths of the two legs of the right triangle. In other words, it's going to depend on how much the magnitude of the "downstream" velocity compares to the magnitude of the of the "across the river" velocity. The former is given, while the latter is what you're trying to solve for.

    So basically you have a right triangle with one of the two perpendicular sides, and one angle. Solving for the other perpendicular side should be trivial.
     
  4. Feb 11, 2012 #3
    How would you solve for it?

    Would it be sin30 = 4/V. V = 8?
     
  5. Feb 11, 2012 #4

    cepheid

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    Well, first of all, I'm assuming that the 30 degrees is measured from due north (i.e. the bearing of your path ends up being 30 degrees east of north).

    So the side of the triangle that is opposite the angle is the eastward vector with magnitude 4 m/s.

    The side of the triangle that is adjacent to the angle is the northward vector whose magnitude you are trying to solve for.

    Which trigonometric ratio is the ratio of these two sides? Hint: it's not sine. Sine involves the opposite side and the hypotenuse, which you're not trying to solve for.

    This is basic trig, so if you're feeling shaky about it, I would brush up.
     
  6. Feb 12, 2012 #5
    I think I understand now. Would it be tan30 = (4.0m/s)/(Vb/w) Vb/w = 6.93 m/s?

    Also, how would you answer the second part?
     
  7. Feb 12, 2012 #6

    cepheid

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    Seems ok. For the second part, as I mentioned before, the speed of the boat relative to the shore is the speed of the boat relative to the water + the speed of the water relative to the shore. So for the second part, you want to compute the magnitude of the vector sum of the velocities.
     
  8. Feb 12, 2012 #7
    So Vb/d (where dock is d) = Vb/w + Vw/d

    Vb/d = 6.93m/s + 4.0m/s

    Vb/d = 10.93 m/s

    Correct?
     
  9. Feb 12, 2012 #8

    cepheid

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    NO, not at all. Did you forget about the fact that we're doing a vector sum and that we therefore have a right triangle? The 6.93 m/s and 4.0 m/s velocity vectors are perpendicular to each other, so their magnitudes don't add directly the way scalars do.

    Which side of the triangle represents the magnitude of the vector sum of the northward and eastward velocities? Given the geometry, how would you compute the length of that side?

    A further hint: the boat's velocity relative to the bank is the "actual" direction of the boat across the river that would be seen by an observer sitting on the bank. It is neither straight north nor straight east.
     
  10. Feb 12, 2012 #9
    Ah, the magnitude of the vector sum is the hypotenuse given by the formula Vb/d[itex]^{2}[/itex] = Vb/w[itex]^{2}[/itex] + Vw/d[itex]^{2}[/itex].

    Vb/d = √(4.0m/s)[itex]^{2}[/itex] + (6.93m/s)[itex]^{2}[/itex]

    Vb/d = 8.0 m/s

    Correct?

    PS: Thanks for all your help, I really really do appreciate it.
     
  11. Feb 12, 2012 #10

    cepheid

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    Yeah, that looks okay to me. Glad to be of help.
     
  12. May 4, 2012 #11
    Thanks. The answer was right!
     
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