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anonymous

  • 5 years ago

If the velocity, v, in ft/sec, of an object is given by v=sqrt(144-12t), where t is the time in seconds, how far does it travel coming to a stop?

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  1. anonymous
    • 5 years ago
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    okay lets do this together. what's the velocity of an object when it comes to a stop?

  2. anonymous
    • 5 years ago
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    0

  3. anonymous
    • 5 years ago
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    when t=12...velocity is zero. But this doesn't tell you how far you've gone. You need the position function. Integrate that thing.

  4. anonymous
    • 5 years ago
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    so, as maph said, if you equate v to 0, you will get t = 12 seconds. So, in 12 seconds, how many feet will the object travel given its velocity is v?

  5. anonymous
    • 5 years ago
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    distance = speed * time. now, since the speed is a function of time, you have to integrate the vtdt with limits 0 and 12

  6. anonymous
    • 5 years ago
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    don't i deserve a fan :(

  7. anonymous
    • 5 years ago
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    yea ill be ur fan but i'm new on here

  8. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{12}v(t)dt\]

  9. anonymous
    • 5 years ago
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    how do i become ur fan i clicked on ur named and nothing comes up about being ur fan.

  10. anonymous
    • 5 years ago
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    click "become a fan" next to the user's name :P

  11. anonymous
    • 5 years ago
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    o wow ha

  12. anonymous
    • 5 years ago
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    Omg every one is super smart i feel like i am just using every one to gat answers coz im too dumb to do it myself.=)

  13. anonymous
    • 5 years ago
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    I'm just lazy

  14. anonymous
    • 5 years ago
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    we are here to help you understand. You should always look to do your problems on your own. Its better to learn how to do problems than to post individual problems here and get a solution for them.

  15. anonymous
    • 5 years ago
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    Well, if you are lazy and uninterested to learn, I will not be helping you, sorry.

  16. anonymous
    • 5 years ago
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    Have a pleasant day.

  17. anonymous
    • 5 years ago
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    go for it up bub I appreciate what help you provided

  18. anonymous
    • 5 years ago
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    You are welcome.

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