The problem asks for the “fastest” process to turn the car $90^\circ$ (from perpendicular to parallel) to avoid a wall. This implies we need to minimize the distance or time required to effect this change in velocity direction.

The maximum force available to the car is the limiting friction $f_{max} = \mu mg$. Consequently, the maximum acceleration magnitude available is: $$ a_{max} = \frac{f_{max}}{m} = \mu g $$ Given $\mu = 0.6$ and $g = 10 \, \text{m/s}^2$, we have $a_{max} = 6 \, \text{m/s}^2$.

To achieve the turn in the minimum distance while avoiding the wall, the driver should apply this maximum acceleration in a constant direction such that the car decelerates in the initial direction ($x$) and accelerates in the final direction ($y$). This creates a Parabolic Path.

To stop motion in the $x$-direction and generate velocity in the $y$-direction simultaneously, the acceleration vector $\vec{a}$ must be directed at $135^\circ$ to the initial velocity (backwards and inwards).

Components of acceleration ($a = \mu g$): $$ a_x = -a \cos(45^\circ) = -\frac{\mu g}{\sqrt{2}} $$ $$ a_y = a \sin(45^\circ) = \frac{\mu g}{\sqrt{2}} $$

Time to turn ($T$): The car completes the turn when $v_x = 0$. $$ v_x = u + a_x t \implies 0 = u – \frac{\mu g}{\sqrt{2}} T $$ $$ T = \frac{u\sqrt{2}}{\mu g} $$

The distance covered in the x-direction before the car is moving parallel to the wall is: $$ x = uT + \frac{1}{2} a_x T^2 $$ Substituting $T$: $$ x = u\left(\frac{u\sqrt{2}}{\mu g}\right) – \frac{1}{2} \left(\frac{\mu g}{\sqrt{2}}\right) \left(\frac{2u^2}{\mu^2 g^2}\right) $$ $$ x = \frac{\sqrt{2}u^2}{\mu g} – \frac{u^2}{\sqrt{2}\mu g} = \frac{u^2}{\sqrt{2}\mu g} $$

Substitute values: $u = 30 \, \text{m/s}$ ($108 \, \text{km/h}$), $\mu = 0.6$, $g=10$. $$ x = \frac{900}{\sqrt{2}(6)} = \frac{150}{\sqrt{2}} = 75\sqrt{2} \, \text{m} $$ Thus, the initial distance must be greater than $75\sqrt{2}$ m. Statement (d) is correct.

The radius of curvature $\rho$ is given by $\rho = \frac{v^2}{a_n}$, where $a_n$ is the normal component of acceleration. The minimum radius occurs at the vertex of the parabola (at $t = T/2$), where the velocity is minimum and perpendicular to the acceleration is maximal relative to speed.

At $t = T/2$, $v_x = u/2$ and $v_y = u/2$. The speed is $v_{mid} = \frac{u}{\sqrt{2}}$. At this point, velocity is at $45^\circ$ and acceleration is at $135^\circ$. The angle between them is $90^\circ$, so the total acceleration acts as normal acceleration ($a_n = \mu g$). $$ \rho_{min} = \frac{v_{mid}^2}{a_{total}} = \frac{(u/\sqrt{2})^2}{\mu g} = \frac{u^2}{2\mu g} $$ $$ \rho_{min} = \frac{900}{2(6)} = \frac{900}{12} = 75 \, \text{m} $$ Statement (c) is correct.

Since the acceleration vector is constant in both magnitude and direction throughout the turn, the resulting trajectory is a parabola. Statement (b) is correct.