Answer
The separation between the cars after $2.0~s$ is $~~15~m$
Work Step by Step
We can convert $110~km/h$ to units of $m/s$:
$(110~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 30.56~m/s$
We can find the distance we would travel without braking in a time of $2.0~s$:
$x = (30.56~m/s)(2.0~s) = 61.12~m$
We can find the distance the police officer travels after applying the brake:
$x = v_0~t+\frac{1}{2}at^2$
$x = (30.56~m/s)(2.0~s)+\frac{1}{2}(-5.0~m/s^2)(2.0~s)^2$
$x = 51.12~m$
The original separation was $25~m$. Since the police officer traveled 10 meters less, the separation between the cars after $2.0~s$ is reduced to $15~m$
