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 {tex}\triangle{/tex}OAB
= ({tex}\frac{1}{2}{/tex})  {tex}\times{/tex} OB {tex}\times{/tex} OA
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 5s {tex}\times{/tex} 52 kmh^-1
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 5 {tex}\times{/tex} (52  {tex}\times{/tex} 1000)/3600)m
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 5 {tex}\times{/tex} ({tex}\frac{130}{9}{/tex} )m
= {tex}\frac{325}{9}{/tex}
= 36.11 m
Distance Travelled by second car before coming to rest =Area of {tex}\triangle{/tex}OCD
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} OD {tex}\times{/tex} OA
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 10 s {tex}\times{/tex} 3 kmh^-1 
=({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 10 {tex}\times{/tex} ({tex}\frac{3 \times 1000}{3600}{/tex}) m 
= ({tex}\frac{1}{2}{/tex}) {tex}\times{/tex} 10 {tex}\times{/tex} ({tex}\frac{5}{6}{/tex}) m
= 5 {tex}\times{/tex}({tex}\frac{5}{6}{/tex}) m 
= {tex}\frac{25}{6}{/tex} m
= 4.16 m
{tex}\therefore{/tex} Clearly the first car will travel farther(36.11 m)than the first car(4.16 m).