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Math Help - Instantaneous Velocity

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    Instantaneous Velocity

    Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
    s(t)= square root of (t+1)
    t= 1

    This is the only problem I have left and I cannot make much headway in it. Can someone please help me out?
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    Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
    s(t)= square root of (t+1)
    t= 1

    since ur in pre-calc, u probably dont know chain rule, but here is how to solve it:

    s(t)= (t+1)^(.5)

    v(t)= (.5)((t+1)^(-.5)) , where v(t) represents instantaneous velocity.

    v(1)= 1/ 2*(2^.5)
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  3. #3
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    Quote Originally Posted by Purpledog100 View Post
    Instructions: The position of an object at time t is given by s(t). Find the instantaneous velocity at the indicated value of t.
    s(t)= square root of (t+1)
    t= 1

    This is the only problem I have left and I cannot make much headway in it. Can someone please help me out?
    from first principles ...

    v(a) = \lim_{t \to a} \frac{s(t) - s(a)}{t - a}

    v(1) = \lim_{t \to 1} \frac{s(t) - s(1)}{t - 1}

    v(1) = \lim_{t \to 1} \frac{\sqrt{t+1} - \sqrt{2}}{t - 1}<br />

    v(1) = \lim_{t \to 1} \frac{\sqrt{t+1} - \sqrt{2}}{t - 1} \cdot \frac{\sqrt{t+1} + \sqrt{2}}{\sqrt{t+1} + \sqrt{2}}

    v(1) = \lim_{t \to 1} \frac{(t+1) - 2}{(t - 1)(\sqrt{t+1} + \sqrt{2})}

    v(1) = \lim_{t \to 1} \frac{t - 1}{(t - 1)(\sqrt{t+1} + \sqrt{2})}<br />

    v(1) = \lim_{t \to 1} \frac{1}{\sqrt{t+1} + \sqrt{2}} = \frac{1}{2\sqrt{2}}
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    Thank you for the help!
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