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Whatazarian
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[SOLVED] Motion in one dimension
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].
h[tex]^{2}[/tex](x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-3/2}[/tex]V[tex]^{2}_{boy}[/tex]=a
x(x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-1/2}[/tex]V[tex]^{1}_{boy}[/tex]=V
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.
Homework Statement
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].
Homework Equations
h[tex]^{2}[/tex](x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-3/2}[/tex]V[tex]^{2}_{boy}[/tex]=a
x(x[tex]^{2}[/tex]+h[tex]^{2}[/tex])[tex]^{-1/2}[/tex]V[tex]^{1}_{boy}[/tex]=V
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.
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