A car is travelling at a speed of 21.6 km/hr when it reaches a 12.8...

Initial speed of the car, $u = 21.6 \text{ km/hr} = 21.6 \times \dfrac{5}{18} \text{ m/s} = 6 \text{ m/s}$.

To reach the bottom with the minimum possible speed, the driver must apply the brakes as hard as possible, which means the maximum frictional force will act opposite to the direction of motion (up the slope).

The forces acting on the car along the slope are the component of gravity downwards and the frictional force upwards. The net acceleration $a$ down the slope is given by:
$a = g \sin \theta - \mu g \cos \theta$

Given \theta = 30^{}^{{}^{\circ}}, $\mu = \dfrac{1}{2\sqrt{3}}$, and $g = 10 \text{ m/s}^2$:
a = 10 \left( \sin30^{}^{{}^{\circ}} -\dfrac{1}{2\sqrt{3}} \cos30^{}^{{}^{\circ}} \right)
$a = 10 \left( \dfrac{1}{2} - \dfrac{1}{2\sqrt{3}} \times \dfrac{\sqrt{3}}{2} \right)<br>$a = 10 \left( \dfrac{1}{2} - \dfrac{1}{4} \right) =$10$\times \dfrac{1}{4} =$2.5$\text{ m/s}^2$<br><br>Using the third equation of motion to find the final velocity$v$at the bottom of the slope of length$s = 12.8 $\text{ m}:<br>$v^2 = u² + 2as$<br>$v^2 = (6)^2 + 2(2.5)(12.8)$<br>$v^2 = 36 + 5 \times12.8$<br>$v^2 = 36 + 64 = 100$<br>$v = 10 $\text{ m/s}<br><br>Converting the final speed back to km/hr:<br>$v = 10 $\times \dfrac{18}{5} \text{ km/hr} =$36 $\text{ km/hr}$