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gandalfwiz Group Title

Anybody here good at physics and curious to know how long it takes a speeding Moped to reach Ke$ha?

  • 2 years ago
  • 2 years ago

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  1. gandalfwiz Group Title
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    The moped accelerates uniformly over a time of 5.21 seconds for a distance of 110 meters. I have to a) find the average velocity, b) find the velocity at 5.21 seconds, and c) determine acceleration of the Moped.

    • 2 years ago
  2. gandalfwiz Group Title
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    I'm pretty sure that the average velocity is 21.11 m/s (distance/time), but I'm unsure how to solve the others. Help?

    • 2 years ago
  3. Broken_Fixer Group Title
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    does the moped start from rest?

    • 2 years ago
  4. gandalfwiz Group Title
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    I'm assuming so

    • 2 years ago
  5. Broken_Fixer Group Title
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    My formula chart sez \[\Delta d = v_i \Delta t + \frac{1}{2}a (\Delta t)^2\] and you know d, v_initial, and t, so it's easy to solve for "a" ...

    • 2 years ago
  6. ravi623 Group Title
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    average velocity= (u+v)\2=21.21 m\s v=u+at u+v=42.22 m\s

    • 2 years ago
  7. Broken_Fixer Group Title
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    final velocity is pretty fast, can we hope that Ke$ha won't be able to get out of the way of the speeding moped?

    • 2 years ago
  8. ravi623 Group Title
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    42.22-u=u+at 2u=42.22-5.21 a

    • 2 years ago
  9. gandalfwiz Group Title
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    lol @ fixer... but your formula is way too intense for me :) I think that there's a way to figure it out without one other than the basics...

    • 2 years ago
  10. gandalfwiz Group Title
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    ravi: what do u and v stand for?

    • 2 years ago
  11. Broken_Fixer Group Title
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    i think ravi is saying that final velocity is twice average velocity (since initial velocity is zero), and than acceleration is (final velocity) / (time).

    • 2 years ago
  12. ravi623 Group Title
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    a=(42.22-2u)\2 u=initial velocity v=final velocity a=acceleration

    • 2 years ago
  13. gandalfwiz Group Title
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    hold on... let me process that information...

    • 2 years ago
  14. gandalfwiz Group Title
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    ravi- how'd you get 21.21? My calculations give 21.11

    • 2 years ago
  15. ravi623 Group Title
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    sir, it was written by mistake

    • 2 years ago
  16. gandalfwiz Group Title
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    okay, so it is 21.11? and then from there you said that velocity=initial velocity + acceleration * time?

    • 2 years ago
  17. ravi623 Group Title
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    yeah

    • 2 years ago
  18. gandalfwiz Group Title
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    and after that was initial velocity + final velocity= 42.22 m/s. Where did 42.22 come from?

    • 2 years ago
  19. Tomas.A Group Title
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    everyone whos good at physics are at physics group

    • 2 years ago
  20. gandalfwiz Group Title
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    no one in the physics group :P

    • 2 years ago
  21. ravi623 Group Title
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    distance traveled=(average velocity). time=110

    • 2 years ago
  22. sunsetlove Group Title
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    im glad some one could help u

    • 2 years ago
  23. Tomas.A Group Title
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    20ppl there

    • 2 years ago
  24. gandalfwiz Group Title
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    Ravi? Where's u go :(

    • 2 years ago
  25. ravi623 Group Title
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    i am here

    • 2 years ago
  26. ravi623 Group Title
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    u+v=42.22 v-u=5.21 a so adding these equation, we get 2v=42.22+5.21a

    • 2 years ago
  27. ravi623 Group Title
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    v=(42.22+5.11a)\2

    • 2 years ago
  28. ravi623 Group Title
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    a=(v-u)\t if moped started from rest then a=v\t=v\5.11

    • 2 years ago
  29. ravi623 Group Title
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    now we have final equation for v v=( 42.22+5.11*(v\5.11) ) \ 2 solving this equation for v, we can find the value of final velocity .

    • 2 years ago
  30. ravi623 Group Title
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    here average velocity= (u+v)\2

    • 2 years ago
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