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apple_pi Group Title

If a car travelling at vm/s takes X meters to stop, what would the stopping distance be if it was travelling at 3vm/s (assume stopping force is equal)

  • 2 years ago
  • 2 years ago

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  1. apple_pi Group Title
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    I remember doing this in class, but forgot which equation to use

    • 2 years ago
  2. lgbasallote Group Title
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    hmm could it be \[V_2 ^2 = V_1 + 2as\] where s is the distance

    • 2 years ago
  3. lgbasallote Group Title
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    correction.. \[V_2 ^2 = V_1^2 + 2as\]

    • 2 years ago
  4. apple_pi Group Title
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    V is velocity? a is acceleration? s is displacement?

    • 2 years ago
  5. lgbasallote Group Title
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    yup

    • 2 years ago
  6. apple_pi Group Title
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    Oh... yeah so that's the same one I have here as v^2=u^2+2as

    • 2 years ago
  7. lgbasallote Group Title
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    yep same shiz

    • 2 years ago
  8. lgbasallote Group Title
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    so first find the acceleration using v = 0 u = v s = X

    • 2 years ago
  9. apple_pi Group Title
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    0=v^2+2aX... what is a?

    • 2 years ago
  10. lgbasallote Group Title
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    you solve for it

    • 2 years ago
  11. lgbasallote Group Title
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    isolate a

    • 2 years ago
  12. apple_pi Group Title
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    a=-v^2/2X

    • 2 years ago
  13. lgbasallote Group Title
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    right

    • 2 years ago
  14. lgbasallote Group Title
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    now use v^2 = u^2 + 2as again this time use: v = 0 u = 3v a = -v^2/2X

    • 2 years ago
  15. lgbasallote Group Title
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    then solve for s

    • 2 years ago
  16. apple_pi Group Title
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    starting with 0=9v^2+2aX -9v^2/2a=X -9v^2/2(-v^2/2X)=X 9X=X now what?

    • 2 years ago
  17. lgbasallote Group Title
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    no...

    • 2 years ago
  18. lgbasallote Group Title
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    it's not 2aX

    • 2 years ago
  19. lgbasallote Group Title
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    it's 2as

    • 2 years ago
  20. lgbasallote Group Title
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    X is just for the first situation

    • 2 years ago
  21. apple_pi Group Title
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    oh i see

    • 2 years ago
  22. lgbasallote Group Title
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    \[0 = 9v^2 + 2as\] \[-9v^2 = 2as\] \[-\frac{9v^2}{2a} = s\] \[-\frac{9v^2}{2(-\frac{v^2}{2X})} = s\] \[\implies -\frac{9v^2}{-\frac{v^2}{X}} = s\] \[\implies 9X= s\]

    • 2 years ago
  23. apple_pi Group Title
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    Oh. THank You!

    • 2 years ago
  24. lgbasallote Group Title
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    welcome

    • 2 years ago
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