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Piecewise does not translate properly

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data[/b]A car is driving at a constant velocity of 18 m/s. There is a school bus on the road with its stop sign extended. The car is 40 metres away from the bus when the driver sees the stop sign.
    There is a time delay of 0.75 secondsbetween the time the driver sees the sign and when the driver can begin to slow down. This is called the "driver reaction time". During this reaction time the distance d, in m, travelled by the car is given by the equation d=18t, where t is the time in seconds from when the driver sees the bus. When brakes are applied, after the 0.75 second reaction time, the distance d travelled by the car in time t is given by the equation 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 car stops 3.75 seconds after seeing the school bus.
    i) Write a piecewise-defined function to describe the distance travelled by the car until it stops.iv)how far does the car travel befor it stops?Explain how you found this.

    d=(18t, 0<=t<=0.75
    (-3t^2+22.5t-1.6875, 0.75<t<=3.75

    When I solve the equations individually and add them together, I get 13.5+38.8125=53.3125
    When I graph the piecewise on graphing calculator, I get a graph with an end coordinate of
    (3.75, 40.5). How is this possible and which one is correct?
  2. jcsd
  3. Feb 15, 2009 #2


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    I don't know anything about the graphing solution, but there is something odd about this problem. The formula d=18t says that the initial speed is 18 m/s. But the second term of d=-3t^2+22.5t-1.6875 says that the initial velocity is 22.5 m/s, a contradiction. The -1.6875 doesn't make any sense. If you use the acceleration of -3 from an initial speed of 22.5, you get a deceleration time of 7.5 s, not 3 s. The question has conflicts in it so it is not surprising that different methods result in different answers.
  4. Feb 16, 2009 #3
    perhaps if i translate the parabola up 13.5 units and right 0.75 units we can get a better result. Something is lost in the translation from 2 individual functions to a piecewise. Any ideas are welcome and appreciated.
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