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Position and velocity

  1. Oct 23, 2008 #1

    An automobile and train move together along
    parallel paths at 38.9 m/s.
    The automobile then undergoes a uniform
    acceleration of −4 m/s2 because of a red light
    and comes to rest. It remains at rest for 71.7 s,
    then accelerates back to a speed of 38.9 m/s
    at a rate of 3.52 m/s2.
    How far behind the train is the automobile
    when it reaches the speed of 38.9 m/s, as-
    suming that the train speed has remained at
    38.9 m/s? Answer in units of m.

    Ok i know that the total distance that the train covered in those 71.7s is 2789.13m now what else do i need to do to find out how far behind is the automobile? How can i tell the total distance traveled by the car? is it d = vi(t)+1/2a(t)^2?
    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 23, 2008 #2


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    You need to consider the car's motion as 3 separate calculations:

    1. Deceleration at -4 m/s^2
    2. At rest
    3. Acceleration at 3.52 m/s^2

    How much time does each of those take?
  4. Oct 23, 2008 #3
    Ok i got the times
    1. 9.725 sec
    2. 71.7s
    3. 11.05s
  5. Oct 23, 2008 #4


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    Okay, all that's left is to figure out how far the car and the train move during those times.

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