Answer:
Let the total distance be 3d.
\[{{v}_{1}}=1m/s\,\,{{v}_{2}}=2m/s\,\,{{v}_{3}}=3m/s\]
\[{{S}_{1}}=d\,\,{{S}_{2}}=d\,\,{{S}_{3}}=d\]
\[{{t}_{1}}=?\,\,{{t}_{2}}=?\,\,{{t}_{3}}=?\]
We know,
\[{{V}_{av}}=\frac{Total\,\,dis\tan ce\,\,\operatorname{cov}ered}{Total\,\,time\,\,taken}\]
\[\Rightarrow {{V}_{av}}=\frac{{{S}_{1}}+{{S}_{2}}+{{S}_{3}}}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}}\] ……(1)
We need to find\[{{t}_{1}},\,\,{{t}_{2}}\]and \[{{t}_{3}}.\]
We know that, \[t=\frac{S}{v}\]
Substituting the above in (1), we get
\[{{V}_{av}}=\frac{{{S}_{1}}+{{S}_{2}}+{{S}_{3}}}{{}^{{{S}_{1}}}/{}_{{{V}_{1}}}+{}^{{{S}_{2}}}/{}_{{{V}_{2}}}+{}^{{{S}_{3}}}/{}_{{{V}_{3}}}}\]
\[=\frac{3d}{{}^{d}/{}_{1}+{}^{d}/{}_{2}+{}^{d}/{}_{3}}=\frac{18d}{11d}=1.64m/s\]
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