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apple_pi

  • 3 years ago

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)

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  1. apple_pi
    • 3 years ago
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    I remember doing this in class, but forgot which equation to use

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

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

  4. apple_pi
    • 3 years ago
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    V is velocity? a is acceleration? s is displacement?

  5. lgbasallote
    • 3 years ago
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    yup

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

  7. lgbasallote
    • 3 years ago
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    yep same shiz

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

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

  10. lgbasallote
    • 3 years ago
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    you solve for it

  11. lgbasallote
    • 3 years ago
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    isolate a

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

  13. lgbasallote
    • 3 years ago
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    right

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

  15. lgbasallote
    • 3 years ago
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    then solve for s

  16. apple_pi
    • 3 years ago
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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?

  17. lgbasallote
    • 3 years ago
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    no...

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

  19. lgbasallote
    • 3 years ago
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    it's 2as

  20. lgbasallote
    • 3 years ago
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    X is just for the first situation

  21. apple_pi
    • 3 years ago
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    oh i see

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

  23. apple_pi
    • 3 years ago
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    Oh. THank You!

  24. lgbasallote
    • 3 years ago
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    welcome

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