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Math Help - word problem

  1. #1
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    word problem

    the young folk decide to ride the (the Chaperones opt to watch from the comfort of the ground). At its maximum height, the ride reaches 65 m above the ground. When one boy reaches the top of the tower, his gum falls out of his mouth. The height of the gum can be given by the mathematical model: h=65-4.9t^2
    where t is in seconds and h is measured in metres.



    a. Find the average velocity of the gum on the intervals 2 < or equal t < or equal 3 and 2 < or equal t < 2.1
    b. Find the instantaneous velocity when t=2
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  2. #2
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    Quote Originally Posted by william View Post
    the young folk decide to ride the (the Chaperones opt to watch from the comfort of the ground). At its maximum height, the ride reaches 65 m above the ground. When one boy reaches the top of the tower, his gum falls out of his mouth. The height of the gum can be given by the mathematical model: h=65-4.9t^2
    where t is in seconds and h is measured in metres.



    a. Find the average velocity of the gum on the intervals 2 < or equal t < or equal 3 and 2 < or equal t < 2.1
    b. Find the instantaneous velocity when t=2
    average velocity between two times t_1 and t_2 ...

    \bar{v} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}


    instantaneous velocity at a specific time t = a

    v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    average velocity between two times t_1 and t_2 ...

    \bar{v} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}


    instantaneous velocity at a specific time t = a

    v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}
    thanks, although i'm still not quite getting it can you give me a step further
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  4. #4
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    On a sidenote:


    For the first part of the problem, think a second for what "average" actually means.
    Skeeter is saying (and correctly so) that he calculates the distance that the gum has fallen within the interval and then divide it with the difference in time.

    I'd assume the solution to b would be the derivative of the function given - the derivative of a distance/time graph gives you velocity.

    <br />
f(x) = 65 - 4.9t^2<br />
    <br />
f'(x) = 2 \cdot ( -4.9t ) = - 9.8t<br />
    <br />
f'(2) = -9.8 \cdot 2 = -19.6<br />
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  5. #5
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    Quote Originally Posted by skeeter View Post
    average velocity between two times t_1 and t_2 ...

    \bar{v} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}


    instantaneous velocity at a specific time t = a

    v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}
    for a i am getting both negative for the first -24.5 and the second -2.01,correct?

    i still can't get b
    Last edited by william; August 5th 2009 at 09:47 AM.
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  6. #6
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    Quote Originally Posted by william View Post
    for a i am getting both negative for the first -24.5 and the second -2.01,correct?
    avg velocity over the interval [2,3] is -24.5 m/s

    avg velocity over the interval [2,2.1] is -20.09 m/s


    i still can't get b
    v(2) = \lim_{t \to 2} \frac{(65 - 4.9t^2) - 45.4}{t-2}

    v(2) = \lim_{t \to 2} \frac{19.6 - 4.9t^2}{t-2}

    v(2) = \lim_{t \to 2} \frac{4.9(4 - t^2)}{t-2}

    v(2) = \lim_{t \to 2} \frac{4.9(2-t)(2+t)}{t-2}

    v(2) = \lim_{t \to 2} \left[-4.9(2+t)\right] = \, \, ?
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  7. #7
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    Quote Originally Posted by skeeter View Post
    avg velocity over the interval [2,3] is -24.5 m/s

    avg velocity over the interval [2,2.1] is -20.09 m/s


    v(2) = \lim_{t \to 2} \frac{(65 - 4.9t^2) - 45.4}{t-2}

    v(2) = \lim_{t \to 2} \frac{19.6 - 4.9t^2}{t-2}

    v(2) = \lim_{t \to 2} \frac{4.9(4 - t^2)}{t-2}

    v(2) = \lim_{t \to 2} \frac{4.9(2-t)(2+t)}{t-2}

    v(2) = \lim_{t \to 2} \left[-4.9(2+t)\right] = \, \, ?
    i appreciate your help! i get -24.5 but for the second with

    <br />
\frac{\delta h}{\delta t}=\frac{h(2.1)-h(2)}{2.1-2}=[65-4.9(2.1)^2]-[65-4.9(2)^2]=43.39-45.4=-2.01

    and for b/ i get 2
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  8. #8
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    Quote Originally Posted by william View Post
    i appreciate your help! i get -24.5 but for the second with

    <br />
\frac{\delta h}{\delta t}=\frac{h(2.1)-h(2)}{\textcolor{red}{2.1-2}}=[65-4.9(2.1)^2]-[65-4.9(2)^2]=43.39-45.4=-2.01

    and for b/ i get 2
    look at the denominator of the difference quotient ... you have to divide by 0.1
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  9. #9
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    Quote Originally Posted by skeeter View Post
    look at the denominator of the difference quotient ... you have to divide by 0.1
    i understand, how do i finish that limit? i have not yet done limits like that, i am learning mathematics on my own, please pardon my slowness
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  10. #10
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    Quote Originally Posted by william View Post
    i understand, how do i finish that limit? i have not yet done limits like that, i am learning mathematics on my own, please pardon my slowness
    then go and learn about limits ... find the limit when you are able.
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  11. #11
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    Quote Originally Posted by skeeter View Post
    then go and learn about limits ... find the limit when you are able.
    is the answer 19.6?
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  12. #12
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    Quote Originally Posted by william View Post
    is the answer 19.6?
    -19.6 m/s
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