Answer:
[tex]v_{p1}[/tex] = 19.32 m/s  and  [tex]v_{p2}[/tex]  = 27.32 m/s
Explanation:
The kinetic energy has as formula
    K = ½ m v²
Where m is the mass and v the speed of the body.
If the policeman calculates the kinetic energy, let's clear the speed
    v = √ 2K/m
Let's reduce the units to the SI system
    K = 16.32 KJ (1000 J / kJ) = 16.32 10³ J
Let's calculate
   v = RA (2 16.32 10 3/60)
   v = 23.32 m / s
The rider at 14.4 km/h, reduce
   [tex]v_{h}[/tex] = 14.4 km / h (1000m / 1km) (1h / 3600s) = 4.00 m / s
We already have the relative speed of the two (rider and police car) which is 23.32, we also have the rider speed 4.0 m / s, let's calculate the possible police speeds
These two speeds come from going in the same direction or in opposite directions
    [tex]v_{r}[/tex] = [tex]v_{h}[/tex] ± [tex]v_{p}[/tex]
    [tex]v_{p}[/tex]  = [tex]v_{r}[/tex] - vj
    [tex]v_{p1}[/tex]  = 23.32 - 4
    [tex]v_{p1}[/tex] = 19.32 m/s
    [tex]v_{p2}[/tex]  = [tex]v_{r}[/tex] + vj
    [tex]v_{p2}[/tex] = 23.32 +4
    [tex]v_{p2}[/tex]  = 27.32 m/s