c
\(\begin{array}{l}
{\rm{Using}}\,equation,\,a = \frac{{v - u}}{t}\,and\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s = ut + \frac{1}{2}a{t^2}\\
{\rm{Distance}}\,{\rm{travelled}}\,{\rm{by}}\,{\rm{car}}\,{\rm{in}}\,{\rm{15}}\,{\rm{sec}}\\
{\rm{ = }}\frac{1}{2}\frac{{\left( {45} \right)}}{{15}}{\left( {15} \right)^2}\\
 = \frac{{675}}{2}\,m\\
{\rm{Distance}}\,{\rm{traveled}}\,{\rm{by}}\,{\rm{scooter}}\,{\rm{in}}\,{\rm{15}}\,{\rm{seconds}}
\end{array}\)

\(\begin{array}{l}
{\rm{ = 30}} \times 15 = 450\left( {{\rm{distance}} = speed \times time} \right)\\
Difference\,between\,{\rm{distance}}\,travelled\,by\,car\\
and\,scooter\,in\,15\,\sec ,\,450 - 337.5 = 112.5\,m\\
Let\,car\,catches\,scooter\,in\,time\,t;\\
\frac{{675}}{2} + 45\left( {t - 15} \right) = 30t\\
337.5 + 45t - 675 = 30t\\
 \Rightarrow \,15t = 337.5\\
 \Rightarrow \,t = 22.5\,\sec 
\end{array}\)