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Math Help - Instantaneous velocity

  1. #1
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    Instantaneous velocity

    Man, these types of problems are given me a run for my money

    t2 - 7t + 16

    Find the instantaneous velocity at t = 4

    I get this problem simplified down to

    \frac{x(x-7)-a(a-7)}{x-a}

    Which I know is right.

    But can i cancel the denominator by making the equation \frac{(x-7)(a-7)}{1} then plugging in 4 and getting 6
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    Man, these types of problems are given me a run for my money

    t2 - 7t + 16

    Find the instantaneous velocity at t = 4

    I get this problem simplified down to

    \frac{x(x-7)-a(a-7)}{x-a}

    Which I know is right.

    But can i cancel the denominator by making the equation \frac{(x-7)(a-7)}{1} then plugging in 4 and getting 6
    What's t^2 - 7t + 16 meant to be? Displacement? Acceleration? Velocity? Without knowing this, your questions can't be answered.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    What's t^2 - 7t + 16 meant to be? Displacement? Acceleration? Velocity? Without knowing this, your questions can't be answered.
    "The displacement (in meters) of a particle moving in a straight line is given by s = t2 - 7t + 16, where t is measured in seconds."
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  4. #4
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    Quote Originally Posted by silencecloak View Post
    "The displacement (in meters) of a particle moving in a straight line is given by s = t2 - 7t + 16, where t is measured in seconds."
    v = ds/dt. Then substitute t = 4.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    v = ds/dt. Then substitute t = 4.
    that just confuses me

    I have to use the definition

    <br />
\lim_{x \to a} \frac{f(x)-f(a)}{x-a}<br />

    which is how i got down to the equation i posted above
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  6. #6
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    Quote Originally Posted by silencecloak View Post
    that just confuses me

    I have to use the definition

    <br />
\lim_{x \to a} \frac{f(x)-f(a)}{x-a}<br />

    which is how i got down to the equation i posted above
    1. Drip feeding important information over a series of posts in a thread doesn't help your cause in getting fast, simple and effective help.

    2. Using symbols like x and a in a motion problem is just plain confusing when they do not represent displacement and acceleration.

    You need to develop understanding on how to use these formulae with notation appropriate to the context. You also need to understand why this is important.


    v(4) = \lim_{t \rightarrow 4} \frac{(t^2 - 7t + 16) - (4^2 - 7(4) + 16)}{t-4} = \lim_{t \rightarrow 4} \frac{t^2 - 7t - 12}{t-4} = \lim_{t \rightarrow 4} \frac{(t - 4)(t - 3)}{t-4} = \lim_{t \rightarrow 4} (t - 3) = 1.
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    Quote Originally Posted by mr fantastic View Post
    1. Drip feeding important information over a series of posts in a thread doesn't help your cause in getting fast, simple and effective help.

    2. Using symbols like x and a in a motion problem is just plain confusing when they do not represent displacement and acceleration.

    You need to develop understanding on how to use these formulae with notation appropriate to the context. You also need to understand why this is important.


    v(4) = \lim_{t \rightarrow 4} \frac{(t^2 - 7t + 16) - (4^2 - 7(4) + 16)}{t-4} = \lim_{t \rightarrow 4} \frac{t^2 - 7t - 12}{t-4} = \lim_{t \rightarrow 4} \frac{(t - 4)(t - 3)}{t-4} = \lim_{t \rightarrow 4} (t - 3) = 1.
    Sorry I thought it was implied, and I am just now learning this in school..so...sorry...
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  8. #8
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    isn't instantaneous velocity acceleration, so he has to take the derivative twice?
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  9. #9
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    Quote Originally Posted by hockey777 View Post
    isn't instantaneous velocity acceleration, so he has to take the derivative twice?
    NO!
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