1.You are arguing over a cell phone while trailing an unmarked police car by 29 m. Both your car and the police car are traveling at 105 km/h. Your argument diverts your attention from the police car for 2.0 s. At the beginning of that 2.0 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? 2. v=v0+at x-x0=v0t+(1/2)at2 v2=v02+2a(x-x0) x-x0=1/2(v0+v)t x-x0=vt-1/2at2 3. Part (a) gave me no trouble. 105 km/h=29.2 m/s To find the position of the police car, I used equation 2 (x0=29 m) to find the position of the police car at t=2.0 and t=2.4 (the time at which you start braking). At t=2.0, I found the separation to be 19 m (correct). At t=2.4, I found the separation to be 14.6 m (used eqn. 2 to find x of the police car and your car and subtracted). After 2.4s of braking, I found the velocity of the police car to be 17.2 m/s, so v(t)=17.2-5t (according to equation 1). Your velocity when you begin braking is v(t)=29.2-5t. I tried using equation 3 to find v when you hit the police car: x-x0=14, substituting the values above I found it to be approximately 26.6 m/s or 96.1 km/h. However, this is not correct. I feel like it has something to do with a system of equations (to account for the different velocities) but I've been at this for close to two hours now and can't seem to wrap my head around it. Any and all help is greatly appreciated.