An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 130km/h. Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Homework rules require you to actually attempt the work first...

Tell us what you have done.. and what you have gotten first.

first I found Δx of speeder = 130km/h × (1/3.6)(t)= (36.1t)m
then Δx of police car at first = 95km/h × (1/3.6)(2.50)= 66.0m

Δx of police car at second = 66(t-2.50)+.5(2.20)(t-2.50)^2
Δx of police car total = 66.0 + 66(t-2.50)+.5(2.20)(t-2.50)^2 = 36.1t

then I found t = -4.82 or -17.4 (the problem is that t can't be negative :s) so please help