Answer
$53~m$
Work Step by Step
Let's find the position as function of time equation for both red car and blue car. Assume the positive direction to be the direction the cars are traveling with the origin at the starting position of the blue car ($x_{blue_0}=0$). The initial time is the moment the police car (blue car) starts ($v_{blue_0}=0$). The red car is at $x_{red_0}$ (head-start distance).
Red car (speeder): constant speed ($v=25~m/s$), $a=0$ -> $x=x_0+vt$
$x_{red}=x_{red_0}+(25~m/s)t$
Blue car (police): constant acceleration ($3.8~m/s^2)$, -> $x=x_0+v_0t+\frac{1}{2}at^2$
$x_{blue}=0+0t+\frac{1}{2}(3.8~m/s^2)t^2=(1.9~m/s^2)t^2$
When the police car catches the speeder $t=15~s$ and $x_{red}=x_{blue}$
$x_{red_0}+(25~m/s)(15~s)=(1.9~m/s^2)(15~s)^2$
$x_{red_0}=(1.9~m/s^2)(15~s)^2-(25~m/s)(15~s)=53~m$
