question_answer

A body covers first one third distance with speed 1 m/ s, the second one third distance with speed 2m/s and the last one third distance with speed 3m/s. Find its average speed.

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 \[{{V}_{av}}=\frac{{{S}_{1}}+{{S}_{2}}+{{S}_{3}}}{{}^{{{S}_{1}}}/{}_{{{V}_{1}}}+{}^{{{S}_{2}}}/{}_{{{V}_{2}}}+{}^{{{S}_{3}}}/{}_{{{V}_{3}}}}\] \[=\frac{3d}{{}^{d}/{}_{1}+{}^{d}/{}_{2}+{}^{d}/{}_{3}}=\frac{18d}{11d}=1.64m/s\]