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Stopping distance, Book is wrong?

  1. Dec 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 2 ft/s2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don’t drive!

    2. Relevant equations
    s=vt
    y2 = y2 + 2ac (s - s0)

    3. The attempt at a solution
    I already know how the book solves it, the problem is that I'm not sure if the book is correct. Here's the solution from the book (12-15):
    http://image.slidesharecdn.com/solutionmanualhibbelerengineeringmechanics12thedition-130719121523-phpapp02-131228081729-phpapp02/95/solutionmanualhibbelerengineeringmechanics12thedition-10-638.jpg?cb=1388219005 [Broken]
    But the problem is that I think they gave the distance d relative to some coordinate where s=0 at t=0 and not the d that is illustrated. I think that the proper solution for this is something like this:
    let the distance they got at time t=0.75s be d1=33ft and the solution they obtain is d2=517ft. Now this are relative to the origin s=0, t=0. So the distance that we want to find "d" is d=d2-d1=484ft. Same procedure for the 2nd case. Am I right?...I think the illustration is for the breaking distance, isn't it?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 18, 2015 #2

    haruspex

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    I'm not grasping your difficulty with the book solution.
    The idea is to calculate stopping distance, which is the sum of thinking distance and braking distance.
    The book applies the formula vf2-vi2=2asbrake. But it substitutes sbrake=sstop-sthink.
     
  4. Dec 18, 2015 #3
    Hm...ok I just reread the question and I think I get it now. So the illustration shows the distance at the time I see the pedestrian relative to the pedestrian...I.e. the distance from when t=0 to the distance when v=0. I thought that illustration was showing the distance at time t=0.75 to the distance when v=0, which is the braking time.
     
  5. Dec 18, 2015 #4
    sorry about this question.
     
  6. Dec 18, 2015 #5

    gneill

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    Staff: Mentor

    Please don't drink and derive o0):smile:
     
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