1. The problem statement, all variables and given/known data An unmarked police car traveling a constant 80 km/h is passed by a speeder traveling 125 km/h. Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s^2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)? 2. Relevant equations ΔX (speeder) = V(speeder)T → (125km/h) (1m/s / 3.6km/h) (T) = (34.72222T)m ΔX (police) = V (police)(3.00s) →(80km/h)(1m/s / 3.6km/h) (3.00s) = 66.666667m ΔX (police during accelerated motion) = V (t-3.00) + 1/2(2.40m/s^2)(t-3.00)^2 Total ΔX (police) = 66.666667 + 66.666667(t-3.00)+1/2(2.40)(t-3.00)^2 ΔX (speeder) = ΔX (police) 3. The attempt at a solution 4.0 4.1 2.5 0.64 Were all my answers but all are wrong. Please let me know what I did wrong. Thank you so much in advance!