6. The brakes of a car allow it to stand on an inclined asphalt surface with an angle at the base of no more than $30^{\circ}$. Determine the minimum braking distance of this car when moving at a speed of $30 \, \text{m/s}$ on a flat horizontal road with the same surface. The acceleration due to gravity $g=10 \, \text{m/s}^2, \cos 30^{\circ} \approx 0.866, \sin 30^{\circ}=0.5 .(15$ points)
Answer: 78 m

Solution. For the situation when the car is on an inclined plane:
$F_{mp}=m g \sin \alpha$ (2 points), $\mu m g \cos \alpha=m g \sin \alpha$ (2 points). Therefore, the coefficient of friction: $\mu=\frac{\sin 30^{\circ}}{\cos 30^{\circ}} \approx 0.577$ (3 points). For the situation when the car is moving on a horizontal surface: $F_{mp}=m a, \mu m g=m a$ (2 points), $\mu g=a=5.77$ (3 points).

The braking distance of the car: $S=\frac{v^{2}}{2 a}=\frac{30^{2}}{2 \cdot 5.77} \approx 78 m$ (3 points).