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    #1
    $\tiny{205.o8.14}$
    $\textsf{The position of a partical moving along a coordinate line is given }\\$
    $\textsf{what is the velocity when $t=3$}\\$

    \begin{align}
    \displaystyle
    {s(t)}&={\sqrt{4+4t}}\\
    &=2(1+t)^{1/2}\\
    s'(t)&=\frac{1}{\sqrt{1+t}}\\
    s'(3)&=\color{red}{\frac{1}{2}\frac{m}{s}}
    \end{align}

    $\textit{think this is ok, but naybe sugestions}$

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    #2
    Hi karush,

    That looks perfectly fine to me - no suggestions from my end.

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    #3
    Quote Originally Posted by karush View Post
    $\tiny{205.o8.14}$
    $\textsf{The position of a partical moving along a coordinate line is given }\\$
    $\textsf{what is the velocity when $t=3$}\\$

    \begin{align}
    \displaystyle
    {s(t)}&={\sqrt{4+4t}}\\
    &=2(1+t)^{1/2}\\
    s'(t)&=\frac{1}{\sqrt{1+t}}\\
    s'(3)&=\color{red}{\frac{1}{2}\frac{m}{s}}
    \end{align}

    $\textit{think this is ok, but naybe sugestions}$

    I think the value you have found is the speed. If the question really means velocity then your answer should be a vector.

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    #4 Thread Author
    is a vector a related rate?

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    #5
    Quote Originally Posted by karush View Post
    is a vector a related rate?
    Velocity in 1 dimension can "point" in only 1 of 2 directions, which is indicated by the sign. Since you find the requested velocity to be positive, you know the particle is moving to the right.

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