Answer:
Let AB, BC and CD be three phases of the travelling train. And \[{{S}_{1}},{{S}_{2}}\] and \[{{S}_{3}}\] be the distances travelled in three phases respectively.
\[{{V}_{1}}=60kmph\,\,{{V}_{2}}=30kmph\,\,{{V}_{3}}=70kmph\]
\[{{v}_{1}}=60kmph\,\,{{v}_{2}}=30kmph\,\,{{v}_{3}}=70kmph\]
\[{{t}_{1}}=0.52\,\,hrs\,\,{{t}_{2}}=0.24hrs\,\,{{t}_{3}}=0.71hrs\]
\[{{S}_{1}}=?\,\,{{S}_{2}}=?\,\,{{S}_{3}}=?\]
Average speed of train \[({{V}_{av}})=?\]
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\[{{S}_{1}},{{S}_{2}}\]and\[{{S}_{3}}.\]
We know that:\[S=vt\]
Applying the above for we get,
\[{{S}_{1}}=60\times 0.52=31.2km\]
\[{{S}_{2}}=30\times 0.24=7.2km\]
\[{{S}_{3}}=70\times 0.71=49.7km\]
\[{{V}_{av}}=\frac{31.2+7.2+49.7}{0.52+0.24+0.71}=\frac{88.1}{1.47}=59.9kmph\]
\[\approx 60\,\,kmph\]
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