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Calculations in regards to a car coming to a stop, help is needed.

  1. May 23, 2012 #1
    I have been given a physics assignment which requires that I go over the physics of why 40km/h is a better speed to be followed in a school zone rather than 50km/h. (must include calculations for stopping distances etc.)

    I used the formula Vo ^ 2 / 2 x μ x g.
    It should give me the stopping distance if I am correct, Vo being the inital velocity, μ the friction coefficient and g for gravity. I first tried using 40km/h for Vo, .7 for μ and 9.8 for g.

    40^2/2x.7x9.8

    I use this and end up with 116.61...kilometres, this is evidently wrong and I am not sure what I am doing wrong as it should come out to 20-30 metres.
    Can somebody please help?
    Thanks.
    (P.S I am in grade 11, so go easy on me :frown:)
     
  2. jcsd
  3. May 23, 2012 #2

    PhanthomJay

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    You forgot to convert km/hr into m/s. It is essential that units are consistent.
     
  4. May 23, 2012 #3
    Oh that gives me an answer of 13.72, I feel stupid now. Thanks for the help!
     
  5. May 23, 2012 #4

    PhanthomJay

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    13.72 m? Show how you arrived at that answer.
     
  6. May 23, 2012 #5
    40000^2/2x.7x9.8
    Just realized this is wrong as well. I don't even know.
     
  7. May 23, 2012 #6
    Actualy, just made it 11.1 m/s which does equal 40km/h, I put it into the equation and ended up with approximately 9 metres.
     
  8. May 24, 2012 #7

    PhanthomJay

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    Now, you're talking! Looks Good!:smile:
    And welcome to PF!
     
  9. May 24, 2012 #8
    Your starting formula is incorrect. The μg should be in the denominator. That is the deceleration. The larger the deceleration, the shorter the stopping distance.

    Chet
     
  10. May 24, 2012 #9

    PhanthomJay

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    It is in the denominator. Parentheses would help, but I think it is in proper PEMDAS, per my recollect.
     
  11. May 25, 2012 #10
    Got it, thanks for the help guys!
     
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