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Hungry lioness

  1. Aug 6, 2010 #1
    A hungry lioness runs after a gazelle. [tex]v_{L}>v_{G}[/tex] are the constant magnitudes of the velocities of the two animals. The lioness always directs it's velocity towards the gazelle, the latter always runs in a straight line. Calculate when will the lioness eat the gazelle, as a function of the initial positions of the two animals in the savannah (the horizontal xy plane) and of the direction of the gazelle's speed.
     
  2. jcsd
  3. Aug 6, 2010 #2

    arildno

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    Sorry, I can only solve a similar problem where a cat chases a mouse.

    You have to help me out here:

    SHOW ME YOUR HOMEWORK SO FAR!!!! :grumpy:
     
  4. Aug 6, 2010 #3
    ...it's about 10 years I don't have to do homework, luckily...:rofl:

    Anyway...
    ...in the rest frame of the gazelle, or the mouse, or the thief...
    ...in plane polar coordinates [tex]r,\alpha[/tex]...
    ...I found...

    [tex]\dot r=-v_L-v_G\cos\alpha[/tex]

    [tex]r\dot\alpha=-v_G\sin\alpha[/tex]

    Theese should be the equations of motion.
    The first one can be replaced by

    [tex]v_G\ddot\alpha\sin\alpha+v_L\dot\alpha^2=0[/tex]

    But I can't go any further!
     
  5. Aug 6, 2010 #4

    arildno

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    Now, I am unsure as to why polar coordinates would be your salvation here.

    I suggest that you first try the following special case:

    The direction the gazelle will run in is orthogonal to the line connecting the initial positions of the gazelle and lioness.

    Try and set up a set up equations of motion in the ground's rest frame; let for example the gazelle run up along the y-axis.

    I'll help you out if you get stuck.
     
  6. Aug 6, 2010 #5
    In the rest frame it should be [tex](x_G,y_G)=(0,v_Gt)[/tex] and

    [tex]\dot x=\frac{-vx}{[x^2+(v_Gt-y)^2]^{1/2}}[/tex]

    [tex]\dot y=\frac{v(v_Gt-y)}{[x^2+(v_Gt-y)^2]^{1/2}}[/tex]

    (I suppressed the index L for readability).
    I thought this looked awful, that's why I used polar coordinates. But maybe you have got the clue here?

    EDIT:

    Oh yeah and you can write

    [tex]\frac{\dot x}{x}=\frac{\dot y}{y-v_Gt}[/tex]

    that looks better, but I still can't solve!
     
    Last edited: Aug 6, 2010
  7. Aug 6, 2010 #6
  8. Aug 6, 2010 #7

    arildno

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    1. This can be written as:
    [tex]x\frac{\dot{y}}{\dot{x}}=(y-v_{G}t)(1)[/tex]

    2. Now we have:
    [tex]\frac{\dot{y}}{\dot{x}}=\frac{dy}{dx}(2)[/tex]

    Now, we need to eliminate "t" from (1)!
    Remember that:
    [tex]\int_{0}^{t}\sqrt{\dot{x}^{2}+\dot{y}^{2}}dt=v_{L}t[/tex]
    The integral, can, however, be rewritten, in terms of an arbitrary x as:
    [tex]t=\frac{1}{v_{L}}\int_{x_{0}}^{x}\sqrt{1+(\frac{dy}{dx})^{2}}dx[/tex]
    whereby we get:
    [tex]\frac{dt}{dx}=\frac{\sqrt{1+(\frac{dy}{dx})^{2}}}{v_{L}}[/tex]

    Now, let us differentiate (1) with respect to x:
    [tex]\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}-\frac{v_{G}}{v_{L}}\sqrt{1+(\frac{dy}{dx})^{2}}[/tex]
    which is, at first a separable equation for the variable Y=dy/dx:
    [tex]x\frac{dY}{dx}=-\frac{v_{G}}{v_{L}}\sqrt{1+Y^{2}}[/tex]
     
  9. Aug 6, 2010 #8

    arildno

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