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## Homework Statement

You are arguing over a cell phone while trailing an unmarked police car by 25.0 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell,“I won't do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.00 m/s2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.400 s to realize your danger and begin braking. (b) If you too brake at 5.00 m/s2, what is your speed (in km/h) when you hit the police car?

## Homework Equations

[tex] v^2 =u^2 +2as, s=ut +\frac{1}{2} at^2 [/tex]

## The Attempt at a Solution

I solved the first part. I get the wrong results for part b) because I think I'm making calculation mistakes. The way how I tried to solve it is by making the distances traveled equal.

From [tex] s=ut +\frac{1}{2} at^2 [/tex] where [tex] a=\frac{-5m}{s^2} [/tex]$ we have that since I start breaking [tex]2.4s[/tex] later than the police car, I must travel a distance [tex]s=u(t-2.4) +\frac{1}{2}at^2 [/tex] whilst the police car travels a distance [tex]s'=ut +\frac{1}{2}at^2 [/tex]. Since we crash, we must have [tex] s=s' [/tex]... this way finding [tex] t [/tex] and [tex] s [/tex], and from there using [tex] v^2 =u^2 +2as [/tex] we find [tex] v [/tex]

P.S. The LaTeX looks weird.