Explanation :
Speed of car, [tex]v_1=50\ km/h=13.8\ m/s[/tex]
Kinetic energy of the car, [tex]KE_1=400\ J[/tex]
Speed of car, [tex]v_2=100\ km/h=27.7\ m/s[/tex]
Let [tex]KE_2[/tex] is the kinetic energy of the car when it is moving with 100 km/h.
[tex]\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}[/tex]
[tex]\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}[/tex]
[tex]KE_2=KE_1(\dfrac{v_2}{v_1})^2[/tex]
[tex]KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2[/tex]
[tex]KE_2=1611.6\ J[/tex]
Initial velocity of both cars are 0. Using third equation of motion :
So, [tex]S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}[/tex]
and [tex]S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}[/tex]
t is same. So,
[tex]\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}[/tex]
[tex]\dfrac{S_1}{S_2}=\dfrac{1}{2}[/tex]
[tex]S_2=2\ S_1[/tex]
So, the braking distance at the faster speed is twice the braking distance at the slower speed.
