Results 1 to 2 of 2

Thread: I need help finding a position vector

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    Alaska
    Posts
    31

    I need help finding a position vector

    A particle starts at the initial position r(0)=<0,0,1> with the initial velocity v(0)=<1,0,0> and the accelerlation a(t<sin(t),e^(t),12t>. Find its position function r(t).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    Re: I need help finding a position vector

    Hello, allstar2!

    $\displaystyle \text{A particle starts at the initial position: }\:r(0)\:=\:\langle 0,0,1\rangle$
    $\displaystyle \text{with initial velocity: }\:v(0)\:=\:\langle1,0,0\rangle$
    $\displaystyle \text{and accelerlation: }\:a(t) \:=\:\langle \sin t,\,e^t,\,12t\rangle.$

    $\displaystyle \text{Find its position function, }r(t).$

    The velocity function is the integral of the acceleration function.

    $\displaystyle v(t) \;=\;\int\langle \sin t,\;e^t,\;12t\rangle\,dt \;=\;\langle \text{-}\cos t\!+\!c_1,\; e^t\!+\!c_2,\; 6t^2\!+\!c_3\rangle$


    We are told that: .$\displaystyle v(0) \:=\:\langle 1,\,0,\,0\rangle$

    So we have: .$\displaystyle \langle \text{-}\cos0\!+\!c_1,\:e^0\!+\!c_2,\:0\!+\!c_3\rangle \;=\;\langle 1,\,0,\,0\rangle$

    . . . . . . . . . . . . . .$\displaystyle \langle \text{-}1\!+\!c_1,\;1\!+\!c_2,\;c_3\rangle \;=\;\langle 1,\,0,\,0\rangle$

    Then we have: .$\displaystyle \begin{Bmatrix}\text{-}1+c_1 \:=\:1 & \Rightarrow & c_1 \:=\:2 \\ 1+c_2 \:=\:0 & \Rightarrow & c_2 \:=\:\text{-}1 \\ c_3 \:=\:0\end{Bmatrix}$

    Hence: .$\displaystyle v(t) \;=\;\langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\; 6t^2\rangle$



    The position function is the integral of the velocity function.

    $\displaystyle r(t) \;=\;\int \langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\;6t^2\rangle\,dt \;=\;\langle -\sin t\!+\!2t\!+\!k_1,\;e^t\!-\!t\!+\!k_2,\;2t^3\!+\!k_3\rangle $


    We are told: .$\displaystyle r(0) \:=\:\langle 0,0,1\rangle$

    So we have: .$\displaystyle \langle \text{-}\sin0\!+\!0\!+\!k_1,\;e^0\!-\!0\!+\!k_2,\;0\!+\!k_3\rangle \;=\;\langle 0,0,1\rangle$

    . . . . . . . . . . . . . . $\displaystyle \langle k_1,\;1\!+\!k_2,\;k_3\rangle \;=\;\langle 0,0,1\rangle$

    Then we have: .$\displaystyle \begin{Bmatrix}k_1 \:=\:0 \\ 1+k_2 \:=\:0 & \Rightarrow & k_2 \:=\:\text{-}1 \\ k_3\:=\:1 \end{Bmatrix}$


    Therefore: .$\displaystyle r(t) \;=\;\langle \text{-}\sin t\!+\!2t,\;e^t\!-\!t\!-\!1,\;2t^3\!+\!1 \rangle$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector Calculus (Position Vector)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Aug 23rd 2011, 01:43 PM
  2. Replies: 4
    Last Post: Jan 26th 2011, 02:11 PM
  3. Replies: 0
    Last Post: Jan 25th 2011, 02:01 PM
  4. [SOLVED] Find velocity vector from position vector
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 18th 2010, 12:10 PM
  5. Position vector...
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: Apr 22nd 2009, 10:59 PM

Search Tags


/mathhelpforum @mathhelpforum