$\begin{array}{l}
{\rm{Distance,}}\,{\rm{x = }}{\left( {t + 5} \right)^{ – 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
Velocity,\,v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}{\left( {t + 5} \right)^{ – 1}} = {\left( {t + 5} \right)^{ – 2}}\,\,\,\,\,\,….\left( {ii} \right)\\
Acceleration\\
\,\,\,\,\,\,\,\,\,\,\,\,a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ { – {{\left( {t + 5} \right)}^{ – 2}}} \right] = 2{\left( {t + 5} \right)^{ – 3}}\,\,…\left( {iii} \right)\\
From\,equation\,\left( {ii} \right),\,we\,get\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v^{3/2}} =  – {\left( {t + 5} \right)^{ – 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {iv} \right)\\
Substituting\,this\,in\,equation\,\left( {iii} \right)\,we\,get\\
Acceleration,\,a =  – 2{v^{3/2}}\,\,or\,\,a \propto {\left( {velocity} \right)^{3/2}}\\
From\,equation\,\left( i \right),\,we\,get\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} = {\left( {t + 5} \right)^{ – 3}}\\
Substituting\,this\,in\,equation\,\left( {iii} \right),\,we\,get\\
Acceleration,\,\,\,\,\,\,\,\,\,\,\,a = 2{x^3}\,\,or\,\,a \propto {\left( {{\rm{distance}}} \right)^3}\\
Hence\,option\left( a \right)\,is\,correct.
\end{array}$