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Find position vector from given velocity

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle A of mass m is travelling on a horizontal surface with velocity 7i-4j and collides with a stationary particle of mass 3m at the origin of the co-ordinate system at time t=0

    Find an expression for the position vector of A at time t<0

    NB: There is more to this question but this is the part I don't understand. The question itself is confusing to me.

    2. Relevant equations

    3. The attempt at a solution

    I integrate the velocity, a constant, so taking the velocity of A as v I integrate and get vt + C

    So: converting back to component form I get 7ti-4tj but still with the '+ C'

    Clearly this cant be right.

    If you can point me in the right direction I'd appreciate it.
  2. jcsd
  3. Jul 8, 2009 #2
    The constants of integration are solved for with the initial conditions given in the problem. For example V=7i-4j when t=0 and subsequently when t< 0 V = ? So by knowing when t=0 one can solve for the constant, C, and then insert that value back into the original equation.
  4. Jul 8, 2009 #3
    Just to make work a bit more simple, let's assume that a unit vectors in the i and j directions are 1 meter in length each. I'd also prefer to name the two axes x and y, rather than i and j, just as to avoid confusion with terms like initial velocity.

    The velocity in the x direction is constant at 7 m/s
    The velocity in the y direction is constant at 4 m/s

    At [tex]t=0[/tex], you're told that the position vector is equal to 0.
    These are the kinematic equations describing motion at a constant velocity, they can be constructed using integration, or the more intuitive definitions of velocity and initial position.
    [tex]\vec x(t)=\vec v_{x}t + x_0[/tex]
    [tex]\vec y(t)=\vec v_{y}t+y_0[/tex]

    Now, you know exactly how those relations look like, since you know the constant velocities and the initial positions. Try and substitute [tex]t[/tex] for [tex]t*[/tex], and define [tex]t*=-t[/tex]

    See what sort of expression that nets you.

    Like GURU said, the integration constant in physics is often very closely associated with the initial conditions of the system. Once you reach more advanced subject material such as work and energy, or others, you'll often find integration leaves you with other constants that are dependent on the initial conditions of the system.

    One way to look at it is through dimensional considerations. You can only take the sum of two quantities that have the same dimensions. So the only constants you can add to an expression describing, displacement, for instance, are ones who themselves describe displacement.

    For an expression dependent on any sort of variable, [tex]a(b)=kb^n+c[/tex] just look at the 'initial' conditions, where [tex]b=0[/tex] and you'll find [tex]c[/tex].
    Last edited: Jul 8, 2009
  5. Jul 8, 2009 #4
    I'm trying to follow..

    So r(0)=0

    Therefore r(t)=vt + C must give C=0

    so r(t)=vt = 7ti-4tj

  6. Jul 8, 2009 #5
    Yes, that's right.

    Integration of the velocity gives the following displacement vector:
    [tex]\vec r(t)=\vec v t + C[/tex]
    [tex]\vec r(0)=0[/tex]
    [tex]\vec r(0)=\vec v*0 + C[/tex]
    And from there it immediately follows that:
    [tex]\vec r(t)=\vec v t[/tex]
    [tex]\vec r(t) = 7t \hat i -4t \hat j[/tex]
  7. Jul 8, 2009 #6
    Well that makes sense thanks... back to the books now

    Thanks again for the help CFDFEAGURU and RoyalCat
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