A particle is moving such that its position c | vedclass

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Question

$\begin{array}{l}
\,\,\,At\,time\,t = 0,\,the\,position\,vector\,of\
the\,particle\,is\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bar r}_1} = 2\hat i + 3\

Vedclass · Accepted Answer

$\;\frac{{11}}{5}\left( {\hat i + \hat j} \right)$

Vedclass · Answer

$\begin{array}{l}
\,\,\,At\,time\,t = 0,\,the\,position\,vector\,of\
the\,particle\,is\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bar r}_1} = 2\hat i + 3\hat j\
At\,time\,t = 5\,s,\,the\,position\,vector\,of\
the\,particle\,is\,\
\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bar r}_2} = 13\hat i + 14\hat j\
Displacement\,from\,\,\,{{\bar r}_1}\,to\,\,{{\bar r}_2}\,is
\end{array}$

$\begin{array}{l}
Displacement\,from\,\,\,{{\bar r}_1}\,to\,\,{{\bar r}_2}\,is\
\Delta \bar r = {{\bar r}_2} - {{\bar r}_1} = \left( {13\hat i + 14\hat j} ight) - \left( {2\hat i + 3\hat j} ight)\
 = 11\hat i + 11\hat j\
	herefore \,\,\,Average\,velocity,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\bar v}_{av}} = \frac{{\Delta {\bar {r}} }}{{\Delta t}} = \frac{{11\hat i + 11\hat j}}{{5 - 0}} = \frac{{11}}{5}\left( {\hat i + \hat j} ight)
\end{array}$

A particle is moving such that its position coordinates $(x, y)$ are

$(2\, m, 3 \,m)$ at time $t = 0$

$(6\,m,7\,m) ,$ at time $ t=2 \,s$ and

$ (13\,m,14\,m),$ at $ t=5\,s$.

Average velocity vector $\vec v_{av}$ from $ t=0$ to $t= 5 $ is

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