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YumYum247

  • one year ago

help please!!!!

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  1. YumYum247
    • one year ago
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  2. YumYum247
    • one year ago
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    @Elsa213 give it a shot!!!!

  3. Teddyiswatshecallsme
    • one year ago
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    I don't fluttering know. ;~;

  4. YumYum247
    • one year ago
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    awn :(

  5. Teddyiswatshecallsme
    • one year ago
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    Yeah. Sorry dude.

  6. YumYum247
    • one year ago
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    dude? how do u know my gender??????????

  7. YumYum247
    • one year ago
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    @oldrin.bataku plz help :)

  8. YumYum247
    • one year ago
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    @oldrin.bataku help plz

  9. arindameducationusc
    • one year ago
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    Let me try....

  10. arindameducationusc
    • one year ago
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    Do I have to check the steps?

  11. YumYum247
    • one year ago
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    sure i guess!!! O-o

  12. anonymous
    • one year ago
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    so he's going \(v_i=75\text{ km/h}=75000\text{ m/h}=1250\text{ m/s}\) and \(v_f=0\) m/s over a time span of \(3.4\) s

  13. YumYum247
    • one year ago
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    yesh

  14. anonymous
    • one year ago
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    oops, actually i forgot to divide by 60 again for 1 min = 60 s

  15. anonymous
    • one year ago
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    so \(v_i\approx20.83\) m/s

  16. YumYum247
    • one year ago
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    yah i just nticed that too :D

  17. anonymous
    • one year ago
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    supposing the brakign is a constant deceleration, the average velocity during breaking is \(\bar v=\frac12(v_i+v_f)\approx \frac12 (0+20.83)=10.42\) m/s

  18. anonymous
    • one year ago
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    so in \(3.4\) s the distance the car goes through is \(\bar v\cdot \Delta t\approx10.42\cdot 3.4=35.428\) m

  19. YumYum247
    • one year ago
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    so am i right?????????????? XD

  20. arindameducationusc
    • one year ago
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    I don't get it. why average velocity is the constant deacceleration? It is not necessary @oldrin.bataku

  21. arindameducationusc
    • one year ago
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    It should be.... x/t=v+u/2 ... for x

  22. anonymous
    • one year ago
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    @arindameducationusc if the braking is not a constant deceleration then we cannot reason the average velocity is \(\bar v=\frac12 (v_i+v_f)\), since this is the midpoint formula and only applies where the velocity is varying with a constant slope (i.e. \(v=v_0+at\) where acceleration \(a\) is constant)

  23. anonymous
    • one year ago
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    in fact, none of our kinematic equations we hope to use will work in the case the acceleration is not constant

  24. arindameducationusc
    • one year ago
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    That is only Yumyum did.... Ofcourse it wouldn't work. It should be given in question though....

  25. YumYum247
    • one year ago
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    i wrote down the question as is......

  26. arindameducationusc
    • one year ago
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    @oldrin.bataku What did you say in my question of Big O IO did not understand......

  27. YumYum247
    • one year ago
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    @arindameducationusc let's just postpone for now......

  28. arindameducationusc
    • one year ago
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    okay

  29. YumYum247
    • one year ago
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    i don't want my master to get his fingers tired XD

  30. YumYum247
    • one year ago
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    Thank you @oldrin.bataku :")

  31. anonymous
    • one year ago
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    @arindameducationusc we say that \(f(x)\in O(g(x))\) if we mean that eventually (i.e. for sufficiently big \(x\)) we have that \(|f|<C|g|\), so the magnitude is bounded above by some constant \(C\) times the magnitude of \(g\)

  32. anonymous
    • one year ago
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    which is why it's called asymptotic -- "in the long run"; we're looking at the comparative behavior of the functions for larger and larger \(x\), further and further along the x-axis

  33. YumYum247
    • one year ago
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    hey one last thing, so this is how you get the average velocity if you have the initial and final velocity????? v¯=1/2(vi+vf),

  34. anonymous
    • one year ago
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    if the acceleration is constant, yes, since then the velocity vs time graph is a straight line and that's just the midpoint

  35. YumYum247
    • one year ago
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    ok thanks :"|

  36. arindameducationusc
    • one year ago
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    @oldrin.bataku Nice... but can you explain in terms of graph? By making a graph in rough @oldrin.bataku

  37. arindameducationusc
    • one year ago
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    Okay thanks @oldrin.bataku

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