Car Motion using Tire Friction I am a bit confused regarding how car moves. Following is the explanation which I've come up with after reading but I would like if someone verifies and corrects it.

If a car is moving, the rotation of it's tyre will exert a backward force on the road. Due to newton's third law, the road also exerts a force on the tyre. The force will be in the forward direction. This force causes the tire and hence the car to move forward.

 A: your explanation is correct, but I would add that the tyre that is rotating in your example is being forced to do so by the engine. 
A: 
I think that this figure can answer your question
The tire force $f_R$ cause that the vehicle move
$M\frac{dv}{dt}=f_R$.
The tire force is proportional to the slip between the tire and the road. No tire force without friction
Edit:
The equations of motion
\begin{align*}
\text{Vehicle}\\
M\dot{v}&=-f_R&(1)\\
\text{Wheel}\\
\Theta\ddot{\varphi}&=f_R\,r+ \tau_E\,b+\tau_B &(2)\\
&\text{with:}\\
&f_R\quad \text{Constraint force}\\
&\tau_E\quad \text{Engine torque}\\
&\tau_B\quad \text{Brake  torque}\\
&r\quad \text{Wheel radius}\\
&b\quad \text{Transmission between engine and wheel}
\end{align*}
I) Wheel rolling condition:
\begin{align*}
  &\text{The equation for rolling wheel is:}\\
  v&=\dot{\varphi}\,r\,,\quad \dot{v}=\ddot{\varphi}\,r&(3)\\
  &\text{with equations (1) (2) and (3) we get}\\\\
  &\left(\Theta+M\,r^2\right)\dot{v}=\left(
  \tau_{E}\,b+\tau_{B}\right)\,r
\end{align*}
II) Wheel slipping condition:
\begin{align*}
 &\text{The  wheel slipping equation is:}\\
 &s_L=\frac{v-r\,\dot{\varphi}}{|v|}\\
 &\text{for this case is the constraint force (tire force) $f_R$ is:}\\
 &f_R=C_s\,s_L
\end{align*}
