Answer
$4.3~m/s^2$
Work Step by Step
Let's find the position as function of time equation for both speeder and police car. Assume the positive direction to be the direction the cars are traveling with the origin at the starting position of the police car ($x_{pol_0}=0$). The initial time is the moment the police car starts ($v_{pol_0}=0$). The speeder is also at the origin ($x_{spe_0}=0)$.
Speeder: constant speed ($v=15 m/s$), $a=0$ -> $x=x_0+vt$
$x_{spe}=0+(15 m/s)t=(15 m/s)t$
Police: constant acceleration ($a$), -> $x=x_0+v_0t+\frac{1}{2}at^2$
$x_{pol}=0+0t+\frac{1}{2}at^2=\frac{a}{2}t^2$
When the police car catches the speeder $t=7~s$ and $x_{spe}=x_{pol}$
$(15 m/s)(7~s)=\frac{a}{2}(7~s)^2$
$a=\frac{2(15~m)(7~s)}{(7~s)^2}=4.3~m/s^2$
