question_answer

A train starting from rest and moving with a uniform acceleration attains a speed of 90 km/h in 5 minutes. Find the distance travelled by the train.

Answer:

Given, u = 0 \[t=5\,\,\min utes=5\times 60s=300s\] \[v=90km/h=90\times \frac{5}{18}=25m{{s}^{-1}}.\] We know, \[v=u+at\] \[a=\frac{v-u}{t}=\frac{25-0}{300a}=\frac{1}{12}m{{s}^{-2}}\] and \[s=ut+\frac{1}{2}a{{t}^{2}}=0+\frac{1}{2}\times \frac{1}{12}\times {{(300)}^{2}}\] \[=3750m=3.75km\] Hence the acceleration of the train is \[\frac{1}{12}m{{s}^{-2}}\]and it covers a distance of 3.75 km.