question_answer

A train travels at a speed of 60 km/h for 0.5 h, at 30 km/h for the next 0.2 h and then at 70 km/h for the next 0.7 h. What is the average speed of the train?

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, \[{{S}_{1}}=60\times 0.52=31.2km\] \[{{S}_{2}}=30\times 0.24=7.2km\] \[{{S}_{3}}=70\times 0.71=49.7km\] \[{{V}_{av}}=\frac{31.2+7.2+49.7}{0.52+0.24+0.71}=\frac{88.1}{1.47}=59.9kmph\] \[\approx 60\,\,kmph\]