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Kinematic Question -- Car braking to avoid hitting a pedestrian

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Driving along a steady speed of 26m/s and suddenly see a child 150m from you. Breaks can produce acceleration of -2.5m/s² but it takes time to get the foot from the gas to the brake pedal. How much time do you have, if to avoid hitting the child?
    Known:
    d=150m
    a=-2.5m/s²
    Vi=26m/s
    t=?
    2. Relevant equations
    d=Vit+½at²
    3. The attempt at a solution
    150=26(t)+½(-2.5)t²
    150=26t+(-1.25)t²
    150=26t-1.25t²
    1.25t²-26t+150=0
    Quadratic formula:
    qform01.gif
    x=[26±√(-26)²-4(1.25)(150)]/2(1.25)
    x=[26±√676-750]/2.5
    x=[26±√-74]/2.5
    x=[26±8.6i]/2.5
    Here I got stuck. How can I solve it if I have an imagery number? I guess my way is wrong?
     
  2. jcsd
  3. Mar 6, 2016 #2
    i think you are trying to solve without realising the physical situation.
    try to ask - what should be car;s deceleration such that it comes to velocity zero after traversing the separation distance ? using other kinematical equation relating initial velocity , final velocity , distance and acceleration you can get info of required acceleration -compare this value with the maximum one can have -then relate the time taken ....
     
  4. Mar 6, 2016 #3

    PhanthomJay

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    The problem is asking about your "reaction" time, that is , the time it takes for you to hit the brakes after first seeing the child. You are traveling at 26 m/s constant speed during this reaction time. You should first calculate how far you will travel to a stop after hitting the brakes , by using another kinematic equation. Then you can calc your max reaction time based on what distance you have left.
     
  5. Mar 6, 2016 #4
    Kinda confused.. what should I solve for then? Vf?
     
  6. Mar 6, 2016 #5

    PhanthomJay

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    No, you should solve for the distance it takes to go from Vo (26 m/s ) to Vf (0 m/s ) when you hit the brakes.
     
  7. Mar 6, 2016 #6
    Got 135.2, now I sub this as my distance?
     
  8. Mar 6, 2016 #7

    haruspex

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    135.2m is the distance for which your motion will be ... what? What will your motion be prior to that?
     
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