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Math Help - Particle motion

  1. #1
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    Post Particle motion

    Particle moves along vertical axis, position at any time t greater than or equal to 0 is given by the function s(t) = \frac{1}{3}t^3-3t^2+8t-4,  s is centimenters and t is seconds.

    a. Find the displacement during the first 6 seconds

    b. Find the average velocity during the first 6 seconds

    c. Find expressions for velocity and acceleration at time t

    d. For what values of t is the particle moving downward?

    Here are my answers...are they correct? thanks.

    a. 12

    b. -10

    c. v(t) = t^2-6t+8
     <br />
a(t)=2t-6<br />

    d. t is less than 0? I'm not sure how to do this one..
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
    Particle moves along vertical axis, position at any time t greater than or equal to 0 is given by the function s(t) = \frac{1}{3}t^3-3t^2+8t-4,  s is centimenters and t is seconds.

    a. Find the displacement during the first 6 seconds

    b. Find the average velocity during the first 6 seconds

    c. Find expressions for velocity and acceleration at time t

    d. For what values of t is the particle moving downward?

    Here are my answers...are they correct? thanks.

    a. 12 ok

    b. -10 no ... avg v = displacement/time

    c. v(t) = t^2-6t+8 ok
     <br />
a(t)=2t-6<br />
ok

    d. t is less than 0? I'm not sure how to do this one..

    find the interval where v(t) < 0
    ...
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  3. #3
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    Quote Originally Posted by skeeter View Post
    ...

    so,
    b would be 2 sec/centimeter?

    and would d be all real numbers less than 0.64?
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  4. #4
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    Quote Originally Posted by live_laugh_luv27 View Post
    so,
    b would be 2 sec/centimeter? cm/sec

    and would d be all real numbers less than 0.64?
    t^2 - 6t + 8 = 0

    (t - 2)(t - 4) = 0

    v(t) < 0 on the interval 2 < t < 4
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