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Help with Kinematics

  1. Jul 26, 2009 #1
    An object is moving in 3D space with position vector vectot r=(x)i+y(j)+z(k) and velocity vector v=a(i)+b(j)+c(k)
    i,j,k are unit vectors along x,y,z axes
    If the object is always moving towards origin,
    then
    (a) a/x =b/y = c/z
    (b) (a/x)(b/y)(c/z)<0
    (c)ax+by+cz<0
    (d)ab/xy >0
    x,y,z,a,b,c are functions of time
    question may have one or more correct answer(s)
    MY APPROACH
    integrating velocity eqn,
    r=at(i)+bt(y)+ct(z)
    which implies,
    x=at
    y=bt
    z=ct
    I dont get the correct ans if I proceed this way
    wats wrong?
     
  2. jcsd
  3. Jul 26, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Re: kinematics

    Angle between position vector and velocity vector is π.
    So the dot product of r and v i.e. r.v = - ( xa + yb + zc) < 0
     
  4. Jul 26, 2009 #3
    Re: kinematics

    plz explain me,how the angle is pi
    what does the statement the object is always moving towards origin
    imply?
     
  5. Jul 26, 2009 #4
    Re: kinematics

    For the sake of argument, let's simplify it to a one dimensional problem.
    The x axis is positive to the right of the origin, and negative to the left of the origin.
    The object starts with an initial displacement [tex]x_0 > 0[/tex]
    If the object is to move towards the origin, is its velocity negative or positive?

    The dot product is the scalar product of two vectors.

    [tex]\vec i\cdot \vec j \equiv |i|*|j|*\cos{\theta}[/tex] where [tex]\theta[/tex] is the angle between the two vectors.

    So for your velocity and displacement vectors, make an analogy from the one-dimensional case.

    The velocity runs opposite the displacement vector for each of the axes. Can you see why the dot product is necessarily negative, now?
     
    Last edited: Jul 26, 2009
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