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anonymous

  • 5 years ago

How do I solve for #12? I've attached the problem.

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  1. anonymous
    • 5 years ago
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  2. amistre64
    • 5 years ago
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    t = 8 is the maximum acceleration

  3. amistre64
    • 5 years ago
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    maybe :)

  4. anonymous
    • 5 years ago
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    haha thanks do you know how to do #12?

  5. amistre64
    • 5 years ago
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    acceleration is any change in velocity; where the velocity is level, there is no acceleration.

  6. amistre64
    • 5 years ago
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    we have the velocity changing at a rate of 1 in the first part. zero in the next segment; 6/3 in the next = 2 and 4/1 in the last part. so maybe between 8 and 9

  7. anonymous
    • 5 years ago
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    the answer to 12 is D if that helps

  8. anonymous
    • 5 years ago
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    i think your looking at #10

  9. amistre64
    • 5 years ago
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    lol....without knowing what D stands for, it aint much help ;)

  10. amistre64
    • 5 years ago
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    oh.... theres more if you scroll it lol

  11. amistre64
    • 5 years ago
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    at t=8 the objects position was at x=10 ?? how do they figure that?

  12. dumbcow
    • 5 years ago
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    ok for #12 find equation for line from t=5 and t=8 slope is -2 v= 2 when t=5 2=-2*5 + b b = 12 v(t) = -2t +12 for 5<t<8 now we need position function x(t) so integrate v(t) x(t) = -t^2+12t +c we know x(8) = 10 -8^2 +12*8 +c = 10 c = -22 find x(5) -5^2 +12*5 -22 = 13

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