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Piecewise Linear Functions!

  1. Jan 20, 2005 #1
    Ok... this is a long one, please bear with me!

    A car is driving at a constant velocity of 18 m/s. There is a school on the road with the stop sign extended. The car is 40 m away from bus when he sees the sign. There is a time dealy of 0.75 seconds bewtween the time the driver sees the sign ans when the driver can begin to slow down. During the reaction time the distance "d" in meters travelled by the car is given by the equation d= 18t, "t" is time in seconds from when the driver sees the bus.
    When the brakes are appled after the 0.75 second reaction time, the equation is: d= -3t^2 + 22.5t - 1.6875.
    After the brakes are applied it takes 3 seconds for the car to come to a stop. These 3 seconds plus the 0.75 second driver reaction time means the care stops 3.75 seconds after seeing the school bus.

    What is the piecewise-defined function to describe the distance travelled by the car until it stops
    How far does the car travel before stopping?
    Does the car hit the bus?


    I'd appreciate some tips on this one!

    Thanks guys!
    Erin :smile:
     
  2. jcsd
  3. Jan 20, 2005 #2

    HallsofIvy

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    Your title "piecewise linear functions" is misleading. For the first 0.75 seconds, the function is linear, in fact it is exactly the d= 18t you are given, but after that it is quadratic.

    All you need to do is "patch" those two formulas together. The first "piece" is simply 18t for 0< t< 0.75. The second piece is the second formula: for 0.75< t< 3,
    d= -3t2+ 22.5t- 1.6875. Write those in the standard for for "piecewise functions".
     
  4. Jan 21, 2005 #3
  5. Jan 23, 2005 #4

    HallsofIvy

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    The speed "when starting to apply the brakes" is not 22.5.

    The speed according to the second formula is given by -6t+ 22.5.
    At t= 0.75, that is -6(3/4)+ 22.5= -9/2+ 45/2= 36/2= 18 m/s.
    That is, of course, the speed given by the first formula.
     
  6. Jan 23, 2005 #5
    Ah, I see

    but what an unpractical way to express the distance travelled during braking, you will have to subtract -3(0.75)2 + 22.5*0.75 - 1.6875 from the equation to know how much distance is travelled during braking
     
  7. Apr 16, 2009 #6
    it takes 40.5 m for the car to stop and yes it hits the bus. If you attempt to use the formulas individually it does not work out. If you input the entire piecewise into your TI-83 it should give the right answer.
     
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