A driver of a car travelling …

As given in the figure below AB (in red line) and CD(in red line) are the Speed-time graph for given two cars with initial speeds 52kmh^-1 and 3 km h^-1 respectively.
Distance Travelled by first car before coming to rest =Area of $\triangle$OAB
= ($\frac{1}{2}$)  $\times$ OB $\times$ OA
= ($\frac{1}{2}$) $\times$ 5s $\times$ 52 kmh^-1
= ($\frac{1}{2}$) $\times$ 5 $\times$ (52  $\times$ 1000)/3600)m
= ($\frac{1}{2}$) $\times$ 5 $\times$ ($\frac{130}{9}$ )m
= $\frac{325}{9}$
= 36.11 m
Distance Travelled by second car before coming to rest =Area of $\triangle$OCD
= ($\frac{1}{2}$) $\times$ OD $\times$ OA
= ($\frac{1}{2}$) $\times$ 10 s $\times$ 3 kmh^-1 
=($\frac{1}{2}$) $\times$ 10 $\times$ ($\frac{3 \times 1000}{3600}$) m 
= ($\frac{1}{2}$) $\times$ 10 $\times$ ($\frac{5}{6}$) m
= 5 $\times$($\frac{5}{6}$) m 
= $\frac{25}{6}$ m
= 4.16 m
$\therefore$ Clearly the first car will travel farther(36.11 m)than the first car(4.16 m).