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Math Help - Rates of Change help, a simple problem.

  1. #1
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    Rates of Change help, a simple problem.

    A construction worker drops a bolt while working on a high rise building 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where s(t) = 320 - 5t^2, 0 < t < 8.

    a. Find the avg. velocity during the first, third, and eighth seconds.
    b. Find the avg. velocity for the interval 3 < t < 8.
    c. Find the velocity at t = 2.

    How do I do this problem?
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  2. #2
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    Hello, Jeavus!

    There is a typo in the problem . . .


    A construction worker drops a bolt while working on a 320-m building.
    After t seconds, the bolt has fallen a distance of s metres,
    where: . {\color{blue}s(t) \:= \:320 - 5t^2},\quad0 < t < 8 . . . . This is incorrect

    a. Find the avg. velocity during the first, third, and eighth seconds.
    b. Find the avg. velocity for the interval 3 < t < 8.
    c. Find the velocity at t = 2.
    The height of the bolt at time t is: . s(t) \:=\:320 - 5t^2


    a) At t = 0:\;s(0) \:= \:320 - 5(0^2) \:=\:320\;m
    . . At t = 1:\;s(1) \:=\:320 -5(1^2) \:=\:315\text{ m}
    In the first second, it has fallen: . 320 - 315 \:=\: 5\text{ m}
    . . Its average velocity is: . \frac{5\text{ m}}{1\text{ sec}} \:=\:5\text{ m/sec}

    At t = 2:\;s(2) \:=\:320-5(2^2) \:=\:300\text{ m}
    At t = 3:\;s(3) \:=\:320 5(3^2) \:=\:275\text{ m}
    In the third second, it has fallen: . 300 - 275 \:=\:25\text{ m}
    . . Its average velocity is: . \frac{25\text{ m}}{1\text{ sec}} \:=\:25\text{ m/sec}

    At t = 7:\;s(7) \:=\:320 - 5(7^2) \:=\:75\text{ m}
    At t = 8:\;s(8) \:=\:320 = 5(8^2) \:=\:0\text{ m}
    In the eighth second, it has fallen: . 75 - 0 \:=\: 75\text{ m}
    . . Its average velocity is: . \frac{75\text{ m}}{1\text{ sec}} \:=\:75\text{ m/sec}


    b) \;s(3) \:=\:275\;m,\quad s(8) \:= \:0
    The bolt fell: . 275 - 0 \:=\:275\text{ m} in 5 seconds.
    . . Its average velocity is: . \frac{275\text{ m}}{5\text{ sec}} \:=\:55\text{ m/sec}


    c) For velocity: . v(t) \:=\:s'(t) \:=\:-10t
    . . At t=2\!:\;v(2) \:=\:-10(2) \:=\:-20\text{ m/sec}

    (The minus indicates a downward velocity.)
    .
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