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Math Help - Vectors problem

  1. #1
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    Vectors problem

    If anyone could explain how the following problem is done, I'd be really grateful!


    Given the acceleration vector a(t), the initial position r(0) = r_0, and the initial velocity v(0)=v_0 of a particle moving in space, find the position vector r(t) at time t.

    a(t) = 6ti - 5j + 12t^2 k
    r_0 = 3i + 4j
    v_0 = 4j - 5k
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  2. #2
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    From the given we get
    v(t) = \left( {3t^2  + C_1 } \right)i + \left( { - 5t + C_2 } \right)j + \left( {4t^3  + C_3 } \right)k
    r(t) = \left( {t^3  + C_1 t + D_1 } \right)i + \left( {\frac{{ - 5t^2 }}{2} + C_2 t + D_2 } \right)j + \left( {t^4  + C_3 t + D_3 } \right)k

    Now use the initial values to find the C & D constants.
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  3. #3
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    Quote Originally Posted by Plato View Post
    From the given we get
    v(t) = \left( {3t^2  + C_1 } \right)i + \left( { - 5t + C_2 } \right)j + \left( {4t^3  + C_3 } \right)k
    r(t) = \left( {t^3  + C_1 t + D_1 } \right)i + \left( {\frac{{ - 5t^2 }}{2} + C_2 t + D_2 } \right)j + \left( {t^4  + C_3 t + D_3 } \right)k

    Now use the initial values to find the C & D constants.
    Sorry, I'm confused - how did you come up with those?
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  4. #4
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    Hello, faure72!

    Given: . \begin{array}{cccc}\text{the acceleration vector:} & {\bf a}(t) & =  & 6t{\bf i} - 5{\bf j} + 12t^2{\bf k} \\<br />
\text{the position vector:} & {\bf r}_0 & = & 3{\bf i} + 4{\bf j} \\<br />
\text{and the velocity vector:} & {\bf v}_0 & =  & 4{\bf j} - 5{\bf k}\end{array}

    and: . \begin{array}{cc}\text{initial position:} & {\bf r}(0) \:=\:{\bf r}_0 \\ \text{initial velocity:} & {\bf v}(0)\:=\:{\bf v}_0\end{array} . of a particle moving in space,

    find the position vector {\bf r}(t) at time t.

    Integrate: . {\bf v}(t) \;=\;\int\left(6t{\bf i} - 5{\bf j} + 12t^2{\bf k}\right)\,dt \;=\;3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf C}

    Since {\bf v}(0) = {\bf v}_0, we have: . 0^2\!\cdot\!{\bf i} - 0\!\cdot\!{\bf j} + 0\!\cdot\!{\bf k} + C \;=\;{\bf v}_0\quad\Rightarrow\quad {\bf C} \;=\;{\bf v}_0

    . . Hence: . {\bf v}(t)\;=\;3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf v}_0


    Integrate: . r(t) \;=\;\int\left(3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf v}_0\right)\,dt \;=\;t^3{\bf i} - \frac{5}{2}t^2{\bf j} + t^4{\bf k} + {\bf v}_ot + {\bf C}

    Since {\bf r}(0) = {\bf r}_0, we have: . 0\!\cdot\!{\bf i} - 0\!\cdot\!{\bf j} + 0\!\cdot\!{\bf k} + {\bf v}_0\!\cdot0 + {\bf C} \;=\;{\bf r}_0\quad\Rightarrow\quad {\bf C} \,=\,{\bf r}_0

    . . Therefore: . \boxed{{\bf r}(t) \;=\;t^3{\bf i} - \frac{5}{2}t^2{\bf j} + t^4{\bf k} + {\bf v}_0t + {\bf r}_0}

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