[beren lper]

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)?

[smoothy]

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

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

[beren lper]

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