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Kinematics, driving question

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    you're driving down the highway late one night at 20ms when a deep steps onto the road 35m infront of you. your reaction time before steeping on the brakes is 0.5s and the max deceleration of ur car is 10m/s^2

    what is the max speed you could have and still not hit the deer?

    deltaX=35m V=?
    a=-10m/s^2 reaction time = 0.5s


    2. Relevant equations

    Xf= Xi+ViT+1/2aT^2
    Vf-Vi=aT
    Vf^2=Vi^2+1/2a(deltaX)

    3. The attempt at a solution
    so this is what i tried

    i pluged the numbers into the first equation to get

    35=0+ViT+1/2(-10)T^2

    and i have 2 missing variables so i tried to get it from the other equation

    Vf-Vi=aT
    0-Vi=(-10)T then that didnt work so i tried the other equation

    Vf^2=Vi^2+1/2a(deltaX)
    0=Vi^2+1/2(-10)35
    Vi= 13.23m/s

    and i know that Vi can't be right because i did the question before it and it had Vi of 20m/s and it was 5m from hitting the deer...i dont know what i did wrong can someone help please?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In order to JUST miss hitting the deer, it must take all 35 meters to slow to 0 m/s. So you have TWO of the equations you just gave: 35=0+ViT+1/2(-10)T^2 for the 35 m and 0-Vi=(-10)T for the time to slow to 0. Now you have two equations in the two "unknown" numbers Vi and T. 0-Vi= -10T tells you that Vi= 10T. Replace Vi by that in the first equation to get a simple equation for T alone. Once you have found T, put that into Vi= 10T to find Vi.

     
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