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Motion in one dimension

  1. Feb 3, 2008 #1
    [SOLVED] Motion in one dimension

    1. The problem statement, all variables and given/known data
    To protect his food from hungry bears, a boy scout raises his food pack with a rope that is thrown over a tree limb at height h above his hands. He walks away from the vertical rope with constant velocity V[tex]^{1}_{boy}[/tex], holding the free end of the rope with his hand.

    (a) Show that the speed V of the food pack is given by x(x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-1/2}[/tex]V[tex]^{1}_{boy}[/tex] where x is the distance he has walked away from the vertical rope.

    (b) Show that the acceleration a of the food pack is h[tex]^{2}[/tex](x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-3/2}[/tex]V[tex]^{2}_{boy}[/tex].

    2. Relevant equations


    3. The attempt at a solution

    First of all I would advise you all to look at the attached diagram of the situation (as expressed in the question).

    First, I concluded that V[tex]^{1}_{boy}[/tex]= x/t, since the boy travels a distance of x in some time t. Then, knowing that the boy started at x = 0, I calculated the distance the pack covered in this time would be (x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{1/2}[/tex] by Pythagoras' Theorem. Hence, V= ((x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{1/2}[/tex])/t, and since t =x/V[tex]^{1}_{boy}[/tex] we can conclude that V = V[tex]^{1}_{boy}[/tex]((x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{1/2}[/tex])/x by substitution.

    Am I right or wrong? Obviously if you switch the distances around, the equation would come out as expected, but, I cannot fathom why and how they would be the other way around.

    b) Taking V from a), and dividing by t to give acceleration, a = V[tex]^{2}_{boy}[/tex]((x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{1/2}[/tex])/x[tex]^{2}[/tex]
    or a = (V[tex]^{2}_{boy}[/tex]x)/((x[tex]^{2}[/tex]+h[tex]^{2}[/tex]))

    From there I don't know what to do since the acceleration is changing, so my method is probably wrong.

    Any help would be appreciated.
    Last edited: Feb 3, 2008
  2. jcsd
  3. Feb 3, 2008 #2
    Adding the attachment.


    Attached Files:

    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    host the image at http://imageshack.us
    and post link to image
  5. Feb 3, 2008 #4
    I think the formula in the problem statement is correct. You are right to assume that the pack is moving at the same rate as the rope is, but the rope is not moving at the same rate as the boy. The only time this would be true is if the rope and the boy were in a straight line (as if he were dragging it along the ground).

    If I could make a suggestion: [tex] dx/dt = v_{boy} [/tex] and you can get the Pythagorean theorem states that [tex] l^2 = x^2 + h^2 [/tex] where l is the length of the hypotenuse. Perhaps you would consider differentiating the Pythagorean theorem as a first step? This would also give you dl/dt, which is the rate of change of the hypotenuse, or, in other words, the speed of the rope.

    I couldn't open your attachment, but I'm assuming the rope makes a right triangle, with the food pack on the perpendicular to the direction of the boy.

    Last edited: Feb 3, 2008
  6. Feb 3, 2008 #5
    I think you are correct, thank you. I was under the misconception that It was an average velocity, but, when I treat it as an instantaneous velocity the answer falls through easily and also makes more sense.

    I can now also solve part b) with the new information.

    Cheers :D
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