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.
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