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anonymous

  • one year ago

A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1. The time interval(s) when the particle is moving right The time interval(s) when the particle is going faster slowing down

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  1. nayef
    • one year ago
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    sorry wrong

  2. nayef
    • one year ago
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    I will try to solve

  3. anonymous
    • one year ago
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    i got 4.5 to 8 also

  4. nayef
    • one year ago
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    The time interval(s) when the particle is moving right: (0,2.5) & (6,8)

  5. IrishBoy123
    • one year ago
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    \(v(t) = t^2 - 9t + 18\) the first thing you should do is plot this: ie \(v(t) vs \ t\). then you have a visual for what is to follow. eg it is moving right if \(v(t) > 0\)

  6. nayef
    • one year ago
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    going faster: (3,4) & (6,8)

  7. nayef
    • one year ago
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    sorry going faster: (3,4.5) & (6,8)

  8. nayef
    • one year ago
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    Slowing down: (0,3) & (4.5,6)

  9. nayef
    • one year ago
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    All that can be understood from drawing V(t) and a(t). where a(t) is the acceleration , we can get it from diffrentating V(t)

  10. nayef
    • one year ago
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    If V(t) positive , Then the particle going right , vise versa.

  11. nayef
    • one year ago
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    If V(t) and a(t) with different signs , then the particle decelerating, vise versa. Hope you understand.

  12. anonymous
    • one year ago
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    thank you it makes much sense

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