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Thread: maximum velocity of a particle with 3d vector equation

  1. #1
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    maximum velocity of a particle with 3d vector equation

    Find the maximum speed of a particle whose velocity v m/s, at any time t seconds is given by v$\displaystyle =2sin(t)$ i $\displaystyle + cos(t)$ j $\displaystyle + 3$ k.
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  2. #2
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    Quote Originally Posted by shawli View Post
    Find the maximum speed of a particle whose velocity v m/s, at any time t seconds is given by v$\displaystyle =2sin(t)$ i $\displaystyle + cos(t)$ j $\displaystyle + 3$ k.
    Dear shawli,

    $\displaystyle v=2\sin{t}\underline{i}+\cos{t}\underline{j}+3\und erline{k}$

    $\displaystyle Therefore,~speed~of~the~particle~(S)=\left|2\sin{t }\underline{i}+\cos{t}\underline{j}+3\underline{k} \right|$

    $\displaystyle S=\sqrt{(2sint)^2+cos^{2}t+9}$

    $\displaystyle S=\sqrt{3sin^{2}t+10}$

    $\displaystyle Since,~(sin^{2}t)_{max}=1$

    $\displaystyle S_{max}=\sqrt{13}\approx{3.61}$

    Hope this will help you.
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  3. #3
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    $\displaystyle
    \|\vec{v}\|^2=(2sin(t))^2+(cos(t))^2+9=3sin^2t+10\ leq 13 ~,~
    \|\vec{v}\|_{max}=\sqrt{13}
    $
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