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Velocity problem # 2

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.06 m/s2.

    (a) How long does such a car, initially traveling at 23.6 m/s, take to stop?
    5.813 s This answer is correct.

    (b) How far does it travel in this time?
    I've tried 59.47 m and 137.1868 but Webassign says both are incorrect. I have no clue as to what I am doing wrong.

    2. Relevant equations

    x(t)= initial position + final velocity * time

    v(t)= (acceleration * time) + initial velocity

    x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

    x= initial position * (average velocity * time)

    average velocity= (final velocity - initial velocity) / (2)

    (final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position

    3. The attempt at a solution

    I got 59.47 meters by pluging in part a, 5.813 seconds, into the position function.
    Then I tried 137.19 meters after figuring that in 5.813 seconds at 23.6 m/s (I multiplied the two) it could travel 137.19.

    Any ideas on what I am doing wrong?
  2. jcsd
  3. Feb 5, 2009 #2


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    x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time)
    = .5*(-4.06)*...............
    The acceleration is negative because it is slowing the car. Decelerating.
  4. Feb 5, 2009 #3

    68.59 meters. Thanks! I really REALLY appreciate it. I'm seeing the little things I'm doing wrong now. :smile:
  5. Feb 5, 2009 #4


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