The distance travelled by a body moving along | vedclass

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Question

$\begin{array}{l}
{m{Distance}}\,along\,a\,{m{ilne}}\,{m{i}}{m{.e}}{m{.,}}\,{m{displacement}}\left( s ight)\
 = {t^3}\left( {\,s

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Vedclass · Answer

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

The distance travelled by a body moving along a line in time $t$ is proportional to $t^3$. The acceleration-time $(a, t)$ graph for the motion of the body will be

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