b
\(\begin{array}{l}
{\rm{Distance}}\,along\,a\,{\rm{ilne}}\,{\rm{i}}{\rm{.e}}{\rm{.,}}\,{\rm{displacement}}\left( s \right)\\
 = {t^3}\left( {\,s\, \propto \,{t^3}\,given} \right)\\
By\,double\,differentiation\,of\,displacement,\\
Wr\,get\,acceleration.\\
V = \frac{{ds}}{{dt}} = \frac{{d{t^3}}}{{dt}} = 3{t^2}\,and\\
a = \frac{{dv}}{{dt}} = \frac{{d3{t^2}}}{{dt}} = 6t\,\\
a = 6t\,or\,a \propto \,t\\
Hence\,graph\,(b)\,is\,correct.
\end{array}\)