A car moving with a speed of 50km/h can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100km/h, the minimum stopping distance is 

Using work-energy theorem, $W_{total} = \Delta K$
i.e, work done by stopping forces should be equal to change in the kinetic energy of the car.
 $-F_{stopping} x_{stopping}$distance=$(k_{final}-k_{initial})$
  $-F_{stopping}$ x S= $\left(0-\frac{1}{2}m{v^2}_{initial}\right)$
$\Rightarrow -F_{stopping}$ x S =$\frac{1}{2}m{v^2}_{initial}$  
As stopping force is same for both the cars, hence
$S\alpha v^2{}_{initial}$=>Hence ,  $\frac{S_1}{S_2}=\frac{{\left({v^2}_{initial}\right)}_1}{{\left({v^2}_{initial}\right)}_2}$
Or  $\frac{6}{S_2}=\frac{{\left({v^2}_{initial}\right)}_1}{{\left({v^2}_{initial}\right)}_2}={\left(\frac{50}{100}\right)}^2\Rightarrow S_2=24m$