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Chasing Cars Problem!

  1. Sep 8, 2006 #1
    A blue car pulls away from a red stop-light just after it has turned green w/ a constant acceleration of .6 m/(s^2).

    A green car arrives at the position of the stop-light 4.5s after the light had turned green.

    What is the slowest constant speed which the green car can maintain and still catch up to the blue car?

    Answer in units of m/s.

    Formulas: Kinetmatics equations:
    f=final, i=initial, x=position, v=velocity, a=acceleration, t=time
    1. Vf = Vi + at
    2. Xf = Xi + Vi(t) + .5a(t^2)
    3. (Vf ^2) = (Vi ^2) + 2a(Xf - Xi)

    may also need general formula : V=at

    Vi of BC = 0 m/s
    a of BC = .6 m/(s^2)
    Xi of BC = 0

    using formula 2, I have found out that Xf of BC @ 4.5 seconds = 6.075 meters.

    Need to Find Vf for GC!! (remember for green car Vf = Vi stated in the problem, constant speed)

    Thanks for your help!
  2. jcsd
  3. Sep 8, 2006 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    If you plot speed vs. time on a graph for both cars (distance being the area under the graph) you will easily solve this problem . In order for the green car to catch up, the area under its vt graph at some positive time t has to be equal to or greater than the area under the vt graph of the accelerating blue car.

    The distance that the green car catches up on the blue car is the area between the green and blue graphs. Green will catch up only if this distance is greater than the initial deficit (ie the area under the blue graph when the green car reaches the origin, at t=4.5 sec).

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