Under constant power, the force on the car due to the engine is indeed $$F_{\rm motor} = \frac{P}{v}$$
But you want to account for air resistance
$$ F_{\rm drag} = m \beta\, v^2 $$
where $beta$ is some constant related to the geometry of the car and the density of the air.
And rolling resistance (friction)
$$ F_{\rm roll} = m a_{\rm f} $$
where $a_{\rm f}$ is constant deceleration due to the rolling resistance of the tires and drivetrain.
Combine the above and Newton's 2nd law to get
$$F_{\rm motor} - F_{\rm drag} - F_{\rm roll} = m a $$
which is solved for the acceleration
$$ \boxed{ a = \frac{P}{m v} - a_{\rm f} - \beta v^2 } $$
The above has a direct solution, but is rather complex to write it out here. To find it use the following relationships
$$ \begin{aligned} x - x_1 &= \int _{v_1}^v \frac{ v}{a}\,{\rm d}v \\ t - t_1 &= \int _{v_1}^v \frac{ 1}{a}\,{\rm d}v \\ \end{aligned}$$
where $t_1$ is the initial time, $x_1$ the initial distance and $v_1$ the initial speed.
In a simulation setting, you can use the above as an integrator, given a acceleration is a function of speed only $a = f(v)$ then at each time-frame you have
$$\begin{aligned} \Delta x &= \frac{v}{f(v)} \Delta v \\ \Delta t &= \frac{1}{f(v)} \Delta v \end{aligned}$$
The integrator finds the distance and time needed to go from speed $v$ to speed $v+\Delta v$ when $\Delta v$ is a finite but small change in speed.