Terminal Velocity of a Car When explaining how a car reaches its terminal velocity, is it solely down to how aerodynamic the car is? My thoughts are that if you have a fixed mass $m$, then the contact frictional force will be constant,$ F=\mu{R}$, and because the thrust due to the engine will be much greater than this value, then surely only air resistance will increase the total value of drag? And this is why as the speed increase the acceleration begins to tend to 0 as Drag ∝$v^2$.
 A: It is just a simplification to say that total drag is $\propto v^2$ and to ignore all other forms of friction. This applies because the power to overcome aerodynamic drag is $\gg$ power loss on the drivetrain and rolling resistance. If you included all the other terms the final speed will decrease only very slightly. 
So a simplified model of acceleration (as a function of speed $v$) is
$$ a = \frac{P(v)}{m v} - \beta v^2 $$
Where $P(v)$ is the engine power at speed $v$, $m$ is the mass and $\beta$ is some drag constant.
Top speed occurs when $a=0$ and hence $$v_{top} \approx \sqrt[3]{ \frac{P_{max}}{m \beta} } $$
This applies only the the car is drag limits (and not gear limited) and that it is designed with getting peak power at the rpm corresponding to top speed. Some cars with 6 speeds, may only achieve top speed in 5th gear, and 6th gear is just an "overdrive".
A: The car reaches terminal velocity when power in = power out. The power that an engine can deliver depends on the gear ratio and the rpm.
The car has several sources of power dissipation - rolling friction and air drag are the two most important ones. The power these take is a function of velocity - so more power (but a smaller fraction) goes into rolling friction as the car goes faster.
Simple way of formulating this:
$$\begin{align}P_{engine}&=f(v,...)\\
F_{air}&=\frac12\rho v^2 A\cdot C_D\\
F_{roll}&=\mu_r F_n\\
P_{resistance}&=(F_{air}+F_{roll})v\end{align}$$
A: Answer is "it depends" (on speed).
Forces acting against engine force $F_e$ are:
Air friction: $ F_a = - \frac 1 2 \rho C_D A v^2$
Rolling friction: $ F_r = -m g \mu_{rr} $
Equation for terminal velocity (balance of forces):
$F_e - F_a - F_r = 0$
$F_e = F_a + F_r$
Expressing them in terms of power (P = F*v):
$P_e = F_a * v + F_r * v$
$P_e =  \frac 1 2 \rho C_D A v^3 + m g \mu_{rr} v$
If you plot separately the two factors, you can see that for speeds lower than $V_{threshold}$ the $m g \mu_{rr} v$  factor will prevail, hence at low speed the rolling friction prevails on air friction:

Y = Power [W]
X = Speed [m/s]

*

*mass = 1000 kg

*$\mu_{rr} = 0.01$

*$\rho = 1.225 kg/m3$

*$g = 9.81 \frac m{s^2}$

*Cd = 0.3

*$A = 2.2 m^2$
Link to interactive plot
This means that for low speed you can approximate calculations by neglecting air drag. But terminal velocity of a car is high, hence you cannot neglect it.
Calculating trend of speed in time is not a linear problem because it involves derivatives. Expressing the equation again in terms of force tather than power:
$F_e - F_a - F_w = m * \frac {dv}{dt}$
$\frac {P_e} v - \frac 1 2 \rho C_D A v^2 - m g \mu_{rr} = m * \frac {dv}{dt} $
$m  \frac {dv}{dt} = \frac {K_1} v +  K_2 v^2 + K_3$
I think the solution involves hyperbolic tangent ( tanh() ).
