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Antiderivatives problem help

  1. Jan 18, 2008 #1
    1. The problem statement, all variables and given/known data

    A car going 70km/h comes to a stop in 6 seconds, assume that the deceleration is constant, find the distance traveled using a graph; find the distance traveled using antiderivatives.

    3. The attempt at a solution

    If the deceleration is constant, I would assume that the slope (derivative) of the graph would also be constant, hence the graph should look like a straight line with a negative slope.

    Converting 70km/h into m/s, I get 19.4m/s. To calculate the distance, I simply took the area under the curve, which is a triangle:

    (19.4 x 6)/2 = 58.2m

    But when I'm using antiderivatives to solve this, I get a different answer...

    S(6) = intergral v(t)dt = 19.4 (6) = 116.4m

    It seems like this answer is twice as much as the one when calculated using a graph, and I'm sure the graph should be the right answer... could someone please tell me which part did I do wrong?
  2. jcsd
  3. Jan 18, 2008 #2


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    The velocity is not constant, but rather a function of time..

    [tex]\int kt dt = \frac{1}{2} k t^2+C[/tex]
  4. Jan 18, 2008 #3
    Thanks for your correction Nate. I'm still a bit confused though, if velocity at time 0 = 19.4, and that at time 6 is 0, wouldn't C always equal to 0?
  5. Jan 18, 2008 #4


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    Not "always". For this particular problem, yes.
  6. Jan 19, 2008 #5
    Thanks HallsofIvy, I attempted to solve using the corrected equation, but I still can't seem to get the right answer. I believe k is a constant in this formula, but how do I determine it? At time 0, the whole 1/2(kt^2) is equal to 0, am I supposed to use the distance at a different time (eg. t = 1s) to solve for k?
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