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Mechanics Problem with I/J vectors

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Two particles P and Q move in an x-y plane. At time t seconds the position vector of P is [itex](t^2\hat{i}+4t\hat{j})m[/itex] and Q is [itex](2t\hat{i}+(t+1)\hat{j})m[/itex].

    A: Prove the particle never collide
    B: Show that the velocity of Q is constant, and calcualte the magnitude and direction
    C: Find the value for t when P and Q have paralell velocities and find the distance between them at this point.


    2. Relevant equations
    ?


    3. The attempt at a solution
    Its part C that is confusing me but incase I messed the others up I will post them as well.

    Part A:
    For the particles to collide their position vectors will equal each other at the same time so setting their I an J components to equal eachother should produce any times they are equal. First their I components.
    [tex]
    2t=t^2 \\
    0=t^2 - 2t \\
    0=t(t-2)
    [/tex]
    Therefore when t=0 and t=2 the I components are the same. And now the J components.
    [tex]
    t+1=4t \\
    1=3t \\
    t=\frac{1}{3}
    [/tex]
    Therefore their J components are only equal when t=1/3, therefore they never collide as the I and J's are never the same at the same time.

    Part B:
    The velocity of Q will be constant if neither of the components are functions of time so differentiation will find me the velocity vector
    [tex]
    \dot{r_Q}=2\hat{i}+1\hat{j}
    [/tex]
    Therefore the velocity is constant as neither are functions of time. To find the magnitude next
    [tex]
    \sqrt{2^2+1^2}=\sqrt{5}m/s
    [/tex]
    And the direction
    [tex]
    tan^{-1}(\frac{1}{2})=26.57°
    [/tex]

    Part C:
    To find the time when P and Q have parallel velocities first P's velocity vector needs to be found
    [tex]
    \dot{r_P}=2t\hat{i}+4\hat{j}
    [/tex]
    It can be seen that both P and Q have constant J components therefore they will be parallel when the I components are the same.
    [tex]
    2t=2 \\
    t=1
    [/tex]

    However the answer to part C is apparently 4 not 1. Any help is appreciated :)
     
  2. jcsd
  3. May 22, 2013 #2
    Hello FaraDazed,
    For part C you are asked to find t for parallel vectors.What you are doing is making them equal (by equating both components ).When two vectors are parallel their components are proportional. I think this should give you the correct answer.
    If you use t=1 as you got in C
    v(p)=[itex](2\hat{i}+4\hat{j})[/itex]
    and v(q)=[itex](2\hat{i}+\hat{j})[/itex]
    They are neither equal nor parallel.
    Just try making the velocity components proportional.
    Regards
    Yukoel
     
  4. May 22, 2013 #3

    CAF123

    User Avatar
    Gold Member

    If two vectors are parallel, then one can be expressed as a scalar multiple of the other. i.e we can express this as $$\begin{pmatrix} {2}\\{1} \end{pmatrix} = \alpha \begin{pmatrix} {2t}\\{4} \end{pmatrix}$$ Solve this system for ##\alpha## and hence for ##t##.
     
  5. May 22, 2013 #4
    Thank you guys I get it now, thanks :)
     
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