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Will there be a collision?

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Speedy sue, driving at 30.0m/s enters a one lane tunnel. She then observes a slow-moving van 155m ahead traveling at 5.00m/s. Sue applied her breaks and accelerates at -2.00 m/s^2. Will there be a collision? If yes, at what distance and what time?


    2. Relevant equations
    initial v = 30 m/s
    final v = 0
    a = -2/s^2
    displacement = [(final velocity - initial velocity) / 2*acceleration)]

    3. The attempt at a solution
    By using the above equation, I found out that a crash does occur when Sue stops. However, I do not understand how to find out at exactly what displacement this crash occurs.
     
  2. jcsd
  3. Sep 19, 2007 #2

    Dick

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    The position of the truck at a time t is given by xtruck(t)=155m+t*(5m/sec). I've picked x=0 to be the tunnel entrance and t=0 to be the time when Sue enters the tunnel. Can you write an equation for Sue's position at time t? Then set them equal and solve for t.
     
  4. Sep 19, 2007 #3

    Doc Al

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    Start by expressing Sue's displacement as a function of time and the truck's displacement as a function of time. At some time they end up in the same spot at the same time. (Measure the displacement of both from Sue's initial position.)
     
  5. Sep 19, 2007 #4
    I'm sorry, I don't understand. How can I express sue's, or the truck's displacement as a function of time?
     
  6. Sep 19, 2007 #5

    Doc Al

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    Dick gave you the truck's position as a function of time. Now you find Sue's position as a function of time.
     
  7. Sep 19, 2007 #6
    I don't understand how Dick came up with that function :)
    I want to say that the function for Sue's car is car(t) = t*30m/s, but that doesn't work out because I don't know how to incorporate the acceleration. Could you please explain how a function of time is created?
     
    Last edited: Sep 19, 2007
  8. Sep 19, 2007 #7

    Dick

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    You must have some kinematical equations to use. The one I'm thinking of looks like:

    x(t)=x0+v0*t+(1/2)*a*t^2.

    Does that look familiar?
     
  9. Sep 19, 2007 #8

    Doc Al

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    You need to be familiar with the basic kinematic equations for constant speed and accelerated motion. Here's a summary that might prove helpful: Basic Equations of 1-D Kinematics
     
  10. Sep 19, 2007 #9
    aha! so the equation for the car will be car(t) = 30t + .5(-2 m/s^2)(t^2)?
     
  11. Sep 19, 2007 #10

    Doc Al

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    Exactly!
     
  12. Sep 19, 2007 #11

    Dick

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    Absolutely right.
     
  13. Sep 19, 2007 #12
    i love you all
     
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