You are arguing over a cell phone while trailing an unmarked police car by 36 m. Both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.5 s, the police officer begins emergency braking at 5 m/s2. a) What is the separation between the two cars when your attention finally returns? b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car? The following work is for part (a) which is correct. My work is shown below: distance between car and police= 36m speed of cars v=110km/hr =110(5/18) = 30.55m/s after police car breaks: initial v=20.55m/s, final v= 0m/s, a=-5m/s^2 from V^2-V^2-2as 0^2-30.55^2=2(-5)s s=933.64/10= 93.364 m velocity of police car after 2.5 secs of breaking v=v+at 30.55+(-5)2.5=18.05 m/s distance traveled in this time by police is S: (18.05)^2-(30.55)^2-2(-5)s =60.78m distance traveled by car in 2.5s=velocity x time: 30.55 x 2.5= 76.375m separation between 2 cars: (36+60.78)-76.375= 20.405m For part (b) I did the following but its still incorrect. Any ideas? v=u +at --> t=20.405/(30.55-18.05), t=1.6324s, from there I plugged the values into v=u+at 20.55-5(1.6324) and got 22.38775, rounded to 22.4 and then converted back to km/h and ended w/ 80.64 km/h which is still wrong.