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Math Help - Help needed with a simple velocity problem

  1. #1
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    Help needed with a simple velocity problem

    Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

    A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

    Also find the instantaneous velocity when t=2.

    Thanks for your time.
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  2. #2
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    Quote Originally Posted by kep84 View Post
    Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

    A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

    Also find the instantaneous velocity when t=2.

    Thanks for your time.
    The average velocity between t_1 and t_2>t_1 is:

    <br />
\bar{v}=\frac{y(t_2)-y(t_1)}{t_2-t_1}<br />

    RonL
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  3. #3
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    Quote Originally Posted by kep84 View Post
    Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

    A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

    Also find the instantaneous velocity when t=2.

    Thanks for your time.
    The instantaneous velocity at t is:

    <br />
v(t)=\left. \frac{dy}{dt}\right|_t =40-32t<br />

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    The average velocity between t_1 and t_2>t_1 is:

    <br />
\bar{v}=\frac{y(t_2)-y(t_1)}{t_2-t_1}<br />

    RonL

    ummm could you go a bit further please? i don't know what i'm supposed to do with this.

    Quote Originally Posted by CaptainBlack View Post
    The instantaneous velocity at t is:

    <br />
v(t)=\left. \frac{dy}{dt}\right|_t =40-32t<br />

    RonL
    i've never seen this before...
    <br />
\left. \frac{dy}{dt}\right|_t<br />
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  5. #5
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    Quote Originally Posted by kep84 View Post
    i've never seen this before...
    <br />
\left. \frac{dy}{dt}\right|_t<br />
    It means you evaluate the derivative at t.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kep84 View Post
    ummm could you go a bit further please? i don't know what i'm supposed to do with this.



    i've never seen this before...
    <br />
\left. \frac{dy}{dt}\right|_t<br />
    to use other notations you may be familiar with.

    Let y = f(t)

    Average velocity from a point a to a point b is: \frac {f(b) - f(a)}{b - a}

    the "points" here are values of t

    The instantaneous velocity at time t is: f'(t)

    so you need to find f'(2) for the second part
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  7. #7
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    thanks for the help so far guys. can someone please just do the problem and show me how they are plugging in the numbers so i can just see exactly how it's done one time?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kep84 View Post
    Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

    A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.
    we want the average velocity between t = 2 and t = 2.05

    Let y = f(t)

    \Rightarrow \mbox {Average Velocity} = \frac {f(2.05) - f(2)}{2.05 - 2}

    = \frac {14.76 - 16}{0.05}

    = -24.8

    Also find the instantaneous velocity when t=2.
    f(t) = 40t - 16t^2

    \Rightarrow f'(t) = 40 - 32t

    Now, the instantaneous velocity when t = 2 is given by:

    f'(2) = 40 - 32(2) = -24
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    we want the average velocity between t = 2 and t = 2.05

    Let y = f(t)

    \Rightarrow \mbox {Average Velocity} = \frac {f(2.05) - f(2)}{2.05 - 2}

    = \frac {14.76 - 16}{0.05}

    = -24.8



    f(t) = 40t - 16t^2

    \Rightarrow f'(t) = 40 - 32t

    Now, the instantaneous velocity when t = 2 is given by:

    f'(2) = 40 - 32(2) = -24
    thx a bunch jhevon and the rest of you.
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