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Homework Help: Stopping distance w human reaction time

  1. Sep 19, 2010 #1

    afa

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    1. The problem statement, all variables and given/known data

    determine the stopping distance for a car with an intitial speed of 26.9 m/s and a human reaction time of 0.9s for an acceleration of -4m/s^2

    2. Relevant equations

    x=vt t=v/a x=x+vt+.5at^2

    3. The attempt at a solution

    I used the second equation to find total time by adding it to .9s then plugged that into equation 3 and added that to x of the first equation?? what am i doing wrong??
     
  2. jcsd
  3. Sep 19, 2010 #2
    Alright, you are going 26.9 m/s. When you stop, you take .9s before the deceleration takes place.

    So, it's .9 seconds plus however much time the deceleration takes.

    As such,
    Alright, you are going 26.9 m/s. When you stop, you take .9s before the deceleration takes place.

    So, it's .9 seconds plus however much time the deceleration takes.

    As such,
    (Original Velocity)/(Deceleration rate)=Total Deceleration Time
    (26.9m/s)/(4m/s^2)=Total Deceleration Time
    6.725 seconds= Total Deceleration Time

    Stopping Distance=(Average Velocity)(Total Deceleration Time)
    (.5)(26.9 m/s)(6.725 seconds)=90.45125 meters

    Now, you know how much distance it takes to stop.
    You have to add how much distance you covered before stopping.
    (reaction time)(velocity during reaction time)=Distance traversed during reaction time
    (.9 seconds)(26.9m/s)=24.21 meters

    Add the two to get your answer.
    24.21 meters + 90.45125 meters= 114.66125 meters

    And that's your answer.
     
  4. Sep 19, 2010 #3

    afa

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    thank you so much, you seem to be the most helpful, do you think you could help me out on some more?
     
  5. Sep 19, 2010 #4
    Certainly. How else would I postpone doing my chemistry work? And I haven't actually taken AP physics, so I suggest checking my answers that I give you. I just like math and am good at figuring stuff out.
     
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