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Relative Motion in Two Dimensions

  1. Sep 20, 2006 #1
    A 340-m-wide river has a uniform flow speed of 0.84 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 3.8 m/s with respect to the water. There is a clearing on the north bank 65 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?


    I figured that I would take the velocity of the powerboat, which is 3.8 m/s and resolve it into the x and y components. I let theta be the angle that (a) is trying to find, and I resolved it into 3.8 cos(theta). Taking into consideration the velocity of the river, the final velocity in the x direction is 3.8 cos(theta) + 0.84. I do the same thing for y, giving 3.8 sin(theta).

    I now figure that the boat must travel the x distance and the y distance in the same amount of time. So I get the two equations:

    t[3.8cos(theta)+0.84] = 340
    t[3.8sin(theta)] = 65

    I solve for t and theta and get 109.9s and 53.67 degrees. However, that's not the right answer. What's wrong with my reasoning?
  2. jcsd
  3. Sep 21, 2006 #2


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    The 0.84 m/s should be added to the eastward and not to the northward motion.
  4. Sep 21, 2006 #3
    I did add it to the eastward motion, didn't I?
  5. Sep 21, 2006 #4


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    You are right. I was considering the angle measured relative to north.
    What you have changed are the distances. You should have:
    [tex]t[3.8cos(\theta)+0.84] = 65[/tex] (The clearings are 65 m apart in the east direction)
    [tex]t3.8sin(\theta) = 340[/tex] (The width of the river)
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