A particle moves a distance x in time t accor | vedclass

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Question

$\begin{array}{l}
{\rm{Distance,}}\,{\rm{x = }}{\left( {t + 5} \right)^{ - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\

Vedclass · Accepted Answer

$\left( velocity \right)^{\frac{3}{2}}$

Vedclass · Answer

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

A particle moves a distance $x$ in time $t$ according to equation $x = (t + 5)^{-1}$ The acceleration of particle is proportional to

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