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Relative motion of a ship

  1. Jun 16, 2003 #1
    A ship cruises forward at vs = 5 m/s relative to the water. On deck, a man walks diagonally toward the bow such that his path forms an angle theta = 19 degrees with a line perpendicular to
    the boat's direction of motion. He walks at vm = 3 m/s relative to the boat.

    I found the speed at which he walks in relative to the water. the speed is 6.61566 m/s. the next thing i need is at what angle to his intended path does the
    man walk with respect to the water? Answer in units of degrees.

    here's what i have done: i have taken the inverse tangent of the x and y components of his speed, but that doesn't work. any help would be appreciated. thanks
  2. jcsd
  3. Jun 16, 2003 #2
    It sounds like you have the right idea but made a mistake somewhere. I get a different result for the man's velocity relative to the water (7.897 m/s). How are you getting yours?
  4. Jun 18, 2003 #3


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    gnome: The original post says that the man walks at an angle of 19 degrees to a line PERPENDICULAR to the boats line of motion. I get your answer for the speed if I take 19 degrees to be the angle with respect to the centerline of the boat rather than pependicular.

    The way I did this problem was: draw a vertical line of length 5 (representing the boats motion). Draw a horizontal line at the tip of that and draw another line at angle 19 degrees to that with length 3(representing the mans walk). Draw a vertical line from the tip of that to the horizontal line. You now have a right triangle with hypotenuse length 3 and angle 19 degrees. The vertical leg of that right triangle has length 3 sin(19)= 0.977 and the horizontal leg has length 3 cos(19)= 2.83.

    Thinking of these in terms of vectors (taking components parallel to and perpendicular to the motion of the boat), the velocity of the boat is the vector (5, 0) and the motion of the man is (0.977, 2.83).
    The sum of those two vectors is (0.977, 2.83). If you draw a line from the base of the line representing the motion of the boat (the first line drawn above) to the tip of the line representing the man's walk, you get a line representing the joint motion (the motion of the man relative to the water) which is the vector sum:
    (5,0)+ (0.977, 2.83)= (5.977, 2.83). If you draw the horizontal and vertical lines forming a right triangle with this last line, the components 5.977 and 2.83 are the lengths of the legs of that triangle. The length of the hypotenuse (and the speed of the man relative to the water) is, by the Pythagorean theorem,
    [sqrt]((5.977)2+ 2.832)= 6.61 mph.

    The tangent of the angle at the base of that triangle is given by the ratio of the legs: 2.83/5.977= 0.475. The angle is the arctan of that: tan-1(0.475)= 25.4 degrees.

    The man is walking at 6.61 mph relative to the water and at an angle (again, relative to the water) of 25.4 degrees to the line of motion of the boat.
  5. Jun 19, 2003 #4


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    A typo in the above: it should be, of course
    "The sum of those two vectors is (5.977, 2.83)"
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