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The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3}$ $ms^{-2}$ in the third second is
$\frac{{10}}{3}\,m$
$\;\frac{{19}}{3}\,m$
$6\,m$
$4\,m$
Solution
$\begin{array}{l}
{\rm{Distance}}\,travelled\,in\,the\,{3^{rd}}\,{\rm{second}}\\
= {\rm{Distance}}\,travelled\,in\,3\,s\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, – \,d{\rm{istance}}\,{\rm{travelled}}\,{\rm{in}}\,\,{\rm{2}}\,\,{\rm{s}}{\rm{.}}\\
As,\,u = 0,\\
{S_{\left( {{3^{rd}}s} \right)}} = \frac{1}{2}a{.3^2} – \frac{1}{2}a \cdot {2^2} = \frac{1}{2}a \cdot 5\\
Give\,a\frac{4}{5}m{s^{ – 2}};\,\,\,\therefore \,\,\,{S_{\left( {{3^{rd}}s} \right)}} = \frac{1}{2} \times \frac{4}{3} \times 5 = \frac{{10}}{3}\,m
\end{array}$