You have your limiting acceleration (traction) equation and you use that to get the allowed acceleration
$$ \dot{v} = \pm \frac{ \sqrt{ \alpha^2 \rho^2 -v^4}}{\rho} $$
The distance traveled with this acceleration, starting from $v_1$ (at point S_1) is $$ s = \int \frac{v}{\dot{v}}\,{\rm d}v = \int \limits_{v_1}^v \frac{\rho v}{\sqrt{v_f^4-v^4}}\,{\rm d}v$$ where $v_f = \sqrt{\alpha \rho}$
The solution is $$s = \pm \frac{\rho}{2} \left( \sin^{-1} \left( \frac{v^2} {v_f^2} \right)-\sin^{-1} \left( \frac{v_1^2} {v_f^2} \right) \right) $$
or
$$v(s) = v_f \sqrt{ \sin\left(\pm \frac{2s}{\rho}+ \sin^{-1} \left( \frac{v_1^2}{v_f^2} \right) \right) } $$
This means that from $v_1$ to $v_f$ the car spans the distance $$ s_f = \frac{\rho}{2} \, \cos^{-1} \left( \frac{v_1^2} {v_f^2} \right) $$
and for the remaining distance $s_2-s_f$ the car moves with $v_f$.
You can also look at the power to do so, with $P = m\, v\, \dot{v}$ which yields $$P = \frac{m v_f^3}{\rho} \sqrt{ \sin\left(2 \theta + \sin^{-1} \left( \frac{v_1^2}{v_f^2} \right) \right)}\,\cos\left( 2\theta + \sin^{-1} \left( \frac{v_1^2}{v_f^2} \right) \right) $$ where $s=\rho \theta$ the arc length of angle $\theta$.
Peak power is $P_{max} = \frac{12^{0.25} m v_f^3}{3 \rho} \approx 0.62 m \sqrt{\rho} \alpha^{1.5}$ is achieved only at distance $$s_{P_{max}} =\frac{\rho}{2} \left( \tan^{-1}\left( \frac{1}{\sqrt{2}} \right) - \sin^{-1} \left( \frac{v_1^2}{v_f^2} \right) \right)$$
If starting from a standstill this happens at about $ s \approx 0.308 \rho$
The viscous stress must come in equal and opposite pairs because of Newton's third law. The forces on an element are exerted by the elements above and below it much like the friction between surfaces. If the force on the top face of a given element is rightwards, then it means that the element above is exerting this force on the given element. This means that the bottom face of the element above experiences the same force leftwards.
So we see there is really only one force variable here, $F(y)$, for every value of $y$. In other words, at each $y$, there is a pair of equal and opposite forces of magnitude $F(y)$ which act respectively on the elements above and below. So for the fluid element from $y$ to $y+\Delta y$, the force on the top face is $F(y+\Delta y)$ to the right and the force on the bottom face is $F(y)$ to the left. In steady flow, the net force on the element is zero, so the external force on the element must equal $-(F(y+\Delta y) - F(y))$, i.e. it is the (negative) derivative of $F$. After finding $F(y)$, the velocity profile can be determined using the definition of $F/A = \eta \partial v/\partial y$.
For example, in the simple case of Couette flow, the external force on all elements is zero. Therefore, the velocity varies linearly with $y$. If the external force is uniform, such as a river flowing downhill, the velocity will be a quadratic function of $y$.
Best Answer
I think that they are neglecting that term because it is of second order, that is
$$ \mathrm{d}x \gg \mathrm{d}x \mathrm{d}t\, .$$
So, you shouldn't have an equality there but an approximately equal sign.