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anonymous

  • one year ago

The position of an object at time t is given by s(t) = -9 - 5t. Find the instantaneous velocity at t = 4 by finding the derivative.

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  1. IrishBoy123
    • one year ago
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    \[{ds(t) \over dt} = v(t) = { d\over dt}(-9 - 5t)\] \[{d \over dt} (-9) = ??\] \[{d \over dt} (-5t) = ??\]

  2. anonymous
    • one year ago
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    what? sorry that is confusing me

  3. IrishBoy123
    • one year ago
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    ok you are asked to find the derivative of: \(s(t) = -9 - 5t\) that will give you an equation for the velocity \(v(t)\) of the object. how do you propose to do that?

  4. anonymous
    • one year ago
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    v (t) = ds/dt?

  5. wmj259
    • one year ago
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    That is correct. The velocity is the derivative of the position function.

  6. wmj259
    • one year ago
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    If you plug in a specific time into the velocity function you find the INSTANTANEOUS SPEED.

  7. anonymous
    • one year ago
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    So if i put v (9) = ds/d(9)

  8. anonymous
    • one year ago
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    i mean 4

  9. IrishBoy123
    • one year ago
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    i strongly suggest you do the derivative first. then you will see :p

  10. anonymous
    • one year ago
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    Yeah see the problem is I dont know how to do any of this, like for this lesson i missed it in school, and im doing a review right now for the quiz tomorrow and i dont understand this at all

  11. wmj259
    • one year ago
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    @IrishBoy123, Yes that would make your night much easier. By taking the derivative of the position function you get the velocity function. But in this case its not much of a velocity FUNCTION then it is a velocity CONSTANT.

  12. anonymous
    • one year ago
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    sO IT WOULD BE a velocity constant? whats that

  13. anonymous
    • one year ago
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    @Loser66 @satellite73

  14. wmj259
    • one year ago
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    Yes, But if you take the derivative of the position function, what do you get?

  15. anonymous
    • one year ago
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    idk? how would i find that

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