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A relative velocity problem

  1. Apr 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Rain is falling down vertically. To a man walking on the road, velocity of rain appears to be 1.5 times his velocity. Then to protect himself from rain, he holds his umbrella at an angle (theta) to the vertical such that tan (theta) =


    2. Relevant equations

    The relative velocity of rain to man is R-M where R is rain velocity and M is man's velocity.


    3. The attempt at a solution

    Assume the rain velocity vector is R. We can think it is -rJ. J is a unit vector along y axis. r is the magnitude of rain velocity. The negative sign comes because of the direction of rain - down.

    Similarly M = mI where I is unit vector along x axis. The relative velocity of rain to man is R-M = -rJ-mI.

    The magnitude of R-M is 1.5m (given).

    1.5m = sqrt(r^2+m^2) implies m/r = 2/sqrt(5) = tan(theta).

    But the answer stated in the book is the reciprocal i.e. sqrt(5)/2. Who is right?


    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2008 #2

    Doc Al

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    Staff: Mentor

    The book is correct. Note that they ask for the angle with respect to the vertical. :wink:
     
  4. Apr 6, 2008 #3
    I considered angle with the vertical only. For angle with the vertical the tan would be

    side an x axix/side on y axis

    which in our case is m/r which was derived equal to 2/sqrt(5).

    Please let me know where I am going wrong if I am.
     
  5. Apr 6, 2008 #4

    Doc Al

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    Staff: Mentor

    Oops! My bad. You were correct. The book gave the angle w.r.t. the horizontal, not the vertical. (Sorry about that! :redface:)
     
  6. Apr 6, 2008 #5
    Another related question. Rain is falling in NW direction. Will the vector be in the 2nd quadrant or the 4th quadrant? Why I am asking this is that it is a subtle but importnat difference. The angle will be different in the two cases event thoufh raising out of the same event. Generally we draw all vectors starting from origin. Here if we draw the vector starting from origin (parallel shifts of vectors are equal vectors) it will be in the 4th quadrant. If we draw the vector ending at the origin it will be in the second quadrant.
     
  7. Apr 6, 2008 #6

    Doc Al

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    Staff: Mentor

    I don't understand your question. How are you defining your coordinates?
     
  8. Apr 8, 2008 #7
    OK, let me put the question in other words. The rain falling from NW direction will it be making 135 degrees with the EAST (x axis) and fall in the second quadrant or will it be making 315 degrees with the EAST (x axis) and fall in the 4th quardrant)? Direciton of the rain is NW, so if we draw a vector originating from the origin ( as we do for all position vectors) then this will be in the fourth quadrant, the vector makes an angle of 315 degrees. But we generally think NW makes an angle of 135 with the EAST. Hence my question? Thanks.
     
  9. Apr 8, 2008 #8

    Doc Al

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    Staff: Mentor

    from the NW or to the NW?

    If the direction of the rain is from the NW, it will make an angle of 315 degrees with the East. If you drew the velocity vector of the rain from the origin, it would lie in the 4th quadrant.

    On the other hand, if the direction of the rain is to the NW, it will make an angle of 135 degrees with the East and lie in the 2nd quadrant.

    Take your pick!
     
  10. Apr 9, 2008 #9
    Thanks, that is what I too thought and that appears logical. Even though NW, in the normal usage means 135 degrees with the east when we say from NW, the vector should be 315 degrees with the East.
     
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