hard kinematics question
You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0s. At the beginning of that 2.0s, the police officer begins braking suddenly at 5.0 m/s^2.
a) What is the separation between the two cars when you attention finally returns? Suppose that you take another 0.40s to realize your danger and begin braking.
b) If you too brake at 5.0 m/s^2, what is your speed when you hit the police car?
i converted the 110km/h to about 30.56m/s before i started the problem
for part a), i just used the equation x = x0 + v0t + at^2 / 2. a = 5, t = 2, and v0 = 0 because the the 2 cars are going in the same direction so the relative velocity of the police car is just 0. plugging those values in, i got x  x0 = 10. since the displacement is 10, and the policeman was applying the brakes, he goes back 10m so the distance between the 2 cars is 2510 = 15m now. this was my book's answer.
for part b), i can't seem to get the right answer. i used the equations v = v0 + at and xx0 = v0t + at^2 /2. i have xx0 = 25, a = 5, and v0 = 0 since at time t=0, the two cars were traveling at the same speed in the same direction so i subtracted the two and got 0. when i solved, i got t = +/3.1. so i plugged that into the first equation v = v0 + at and i got v = 30.56 + (5)(3.1) and i got around 15 m/s. however the answer given by the book is 94 km/h. how did they get that answer?
