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anonymous

  • one year ago

position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

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  1. geerky42
    • one year ago
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    \[\large v(t) = \dfrac{\mathrm d }{\mathrm dt}~s(t)\]Right?

  2. geerky42
    • one year ago
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    You just need to evaluate \(v(2)\)

  3. anonymous
    • one year ago
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    If I evaluate the formula at t=-2 I just get -14

  4. anonymous
    • one year ago
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    *2

  5. geerky42
    • one year ago
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    How did you get -14? Can you find derivative of -2 - 6t ?

  6. anonymous
    • one year ago
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    I don't think I know how to find derivatives properly.

  7. anonymous
    • one year ago
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    When it says derivative, I think derived from t equaling 2 and the formula -2 -6t. -2-6*(2)=-14

  8. anonymous
    • one year ago
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    I don't think that's how it works though.

  9. geerky42
    • one year ago
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    You can just take derivative of each term individually. For each term, use power rule: \(\dfrac{\mathrm d }{\mathrm dx}x^n = nx^{n-1}\) For constant, you have \(\dfrac{\mathrm d }{\mathrm dx}c = 0\)

  10. geerky42
    • one year ago
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    So here, you have \(\dfrac{\mathrm d }{\mathrm dt}(-2-6t) ~~~=~~~ \dfrac{\mathrm d }{\mathrm dt}(-2) + \dfrac{\mathrm d }{\mathrm dt}(-6t) = ~?\)

  11. anonymous
    • one year ago
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    (d/-2d)-(d/-2d)=0

  12. anonymous
    • one year ago
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    ?

  13. geerky42
    • one year ago
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    Seem you are not really familiar with basic derivative rules. Can you tell me what you learned recently (what teacher tried to teach you, etc)?

  14. anonymous
    • one year ago
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    I don't have a teacher

  15. anonymous
    • one year ago
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    Can you tell me what you got, then we can work it out.

  16. geerky42
    • one year ago
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    So you are self-taught?

  17. anonymous
    • one year ago
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    Probably online school :) ^^

  18. anonymous
    • one year ago
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    yes

  19. anonymous
    • one year ago
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    If someone could offer me the answer to this question, I would be glad to discuss how it was found

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