there are two kinetic types of energy that are involved in a moving car:
Linear and rotational kinetic energy, $$ K_{e \ linear} = \frac {1}{2}mv^2 \ ; K_{e \ rotatinal} = \frac {1}{2}I \omega^2 $$
The deceleration from V1 to V2 will reduce the energy. This energy is being saved in hybrid and electrical cars as regenerative energy.
The linear energy change is equal to:
$ K_{e \ linear \ chang} = \frac {1}{2}mv_{1}^2 - \frac {1}{2}mv_{2}^2 $
And rotational energy change is equal to:
$ K_{e \ rotatinal \ change} = \frac {1}{2}I_{1} \omega^2 - \frac {1}{2}I_{2} \omega^2 $
And $ \ V = r \omega = d \omega/2 \ and\ \omega= V/r $ with r being the radius of the wheel, r = d/2.
However, because of small "I" of the wheels compared to the mass of the car, the contribution of the wheel's change of energy is not significant.