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Physics car kinematics question

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of .5 m/s^2.
    A green car arrives at the position of the stop-light 7 s after the light has turned green.
    What is the lapse time of the blue car when the green car catches it if the green car maintains the slowest constant speed necessary to catch up to the blue car? Answer in units of s.

    2. Relevant questions
    I'm having trouble grasping how to find "the slowest constant speed necessary" I'm assuming I need to set up two equations and set them equal to each other, but I'm given no information about the green car!

    3. The attempt at a solution

    Well, short of the green car I have it set up that for the blue car

    Vi = 0 m/s
    a = .5
    t = 7 seconds

    V = AT
    V = 3.5 m/s when the green car gets to the stop light and the blue car has gone
    x-xi = Vi * t + 1/2 A T^2 = 42.25 meters.

    so if the blue car is going 3.5 m/s with .5 m/s^2 and it has an initial x component of 42.25 from the green car after 7 seconds, I need to find the time that the green car will catch up with the slowest possible speed..

    I'm stuck I can't think of how to deal with this.

    Do I assume the accleration on the green car is 0? Since it talks only about speed? Then I need to find a constant speed which will probably be quite high relative to the blue car where they intersect each other? How do I go about doing this?
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2


    User Avatar
    Homework Helper

    Welcome to PF.

    I think you are almost there, except I think your equation is not quite right. When Green catches Blue the difference equation is going to be 0 which I think you've already gathered. But you must be careful in constructing the equations so as to account for the time properly.

    For instance if t in your equation is the time the Green car passes the intersection, - the time for the blue car has already advanced to 7. Hence Blue will be 1/2 (.5)(7)2 m ahead at the moment Green hits the intersection. And for the purposes of his progress he will be traveling at an initial speed of 7(.5). At that point you can construct your equation for the Blue car and subtract the speed*t of the Green car to calculate intercept.

    This equation will give you then a relationship between time and velocity. Think then about how you would go about finding the minimum of that equation as to V.
  4. Sep 14, 2008 #3
    Oh, I can then take the derivative and set it equal to zero of that equation to find the minimum? Thanks :D

    Of course with testing the max min points so as to not confuse it with a max.
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