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happykiddo

  • one year ago

If the particle’s position on the x-axis is now given by the function x=(1t^2+1t+-2)m, where t is in s. To the nearest 0.1 m, what is the turning point (where the particle reverses direction of motion) of the particle's position?

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  1. happykiddo
    • one year ago
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    Could someone explain to me what a "turning point" is in relation to this problem?

  2. happykiddo
    • one year ago
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    The answer is 5, if that helps any.

  3. EmmaTassone
    • one year ago
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    the function is:\[x(t)=t ^{2}+t-2\] right?

  4. happykiddo
    • one year ago
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    So sorry the actual function is x=(-2t^2+0t+5) not the one stated above.

  5. happykiddo
    • one year ago
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    not the one in the question above. Function different overall question the same.

  6. happykiddo
    • one year ago
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    sorry for the confusion.

  7. EmmaTassone
    • one year ago
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    the turning point is a point at which the derivative changes sign

  8. EmmaTassone
    • one year ago
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    so you should derivate the function and see when the derivative becomes zero. And when you find that turning point then you will have to evaluate in your position function to know at what position was the particle when it reverse his direction of motion.

  9. happykiddo
    • one year ago
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    I understand what I must do now. Thank you very much! : )

  10. EmmaTassone
    • one year ago
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    no problem

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spraguer (Moderator)
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