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  1. MHB Master
    \(\displaystyle \int\)
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    #1

    OK, this can only be done by observation so since we have v(t) I chose e
    but the eq should have a minus sign.

    here the WIP version of the AP Calculus Exam PDF as created in Overleaf



    the goal is to have 365 problems that align basically where students are first year calculus
    always appreciate comments since it needs to be a group effort.
    Last edited by karush; January 7th, 2020 at 17:30.

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    #2
    Quote Originally Posted by karush View Post
    OK, this can only be done by observation so since we have v(t) I chose e
    but the eq should have a minus sign.
    choice (e) works as written only if acceleration remains constant over the indicated interval of time

    fyi, the avg value of a function of time over the interval of time $[a,b]$ is $ \displaystyle \dfrac{1}{b-a} \int_a^b f(t) \, dt$

    which of the other 4 choices matches up?

  4. MHB Master
    \(\displaystyle \int\)
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    #3 Thread Author
    by f(t) do you mean v(t)?

    assuming c with abs enclosure

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    #4
    in general, the average value of a function is ...

    $\displaystyle \overline{f(x)} = \dfrac{1}{b-a} \int_a^b f(x) \, dx$

    in general, if $f$ is any function of time over the time interval $[a,b]$ ...

    $\displaystyle \overline{f(t)} = \dfrac{1}{b-a} \int_a^b f(t) \, dt$


    so, specifically ...

    average acceleration, $\displaystyle \overline{a(t)} = \dfrac{1}{b-a} \int_a^b a(t) \, dt$

    average velocity, $\displaystyle \overline{v(t)} = \dfrac{1}{b-a} \int_a^b v(t) \, dt$

    average position, $\displaystyle \overline{x(t)} = \dfrac{1}{b-a} \int_a^b x(t) \, dt$

    choice (c) is average speed, not velocity.
    Last edited by skeeter; January 7th, 2020 at 19:07.

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  7. MHB Master
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    #6 Thread Author
    pl I think B is the answer due to absence of absolute value.

    sorry just noticed I never replied to this.

    that was a lot of help .... kinda confusing at first.

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    #7
    Quote Originally Posted by karush View Post
    pl I think B is the answer due to absence of absolute value.

    sorry just noticed I never replied to this.

    that was a lot of help .... kinda confusing at first.
    (B) is correct

    $\displaystyle \dfrac{1}{8} \int_0^8 v(t) \, dt = \dfrac{x(8)-x(0)}{8-0} = \dfrac{\Delta x}{\Delta t} = \bar{v}$

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