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!