Answer:
As the train starts from rest its initial velocity is zero.
the acceleration of the train\[=2.0\,m/{{s}^{2}}\]
time duration\[{{t}_{1}}=30\]seconds
So, the distance travelled by the train during this period,
\[{{S}_{1}}=u{{t}_{1}}+\frac{1}{2}at_{1}^{2}\]
\[{{S}_{1}}=0+\frac{1}{2}\times 2\times {{(30)}^{2}}=900m\]
velocity of the train, \[V=u+at\]
\[=0+2\times 30=60m/s\]
This is the maximum velocity, when breaks applied, the train comes to rest that means its final velocity is zero and maximum velocity 60 m/s becomes initial velocity when breaks are applied, its acceleration changes.
\[retardation=\frac{v-u}{t}=\frac{0-60}{60}=-m{{s}^{-2}}\]
So, the distance travelled by the train in the second case
\[{{S}_{2}}=\frac{{{v}^{2}}-{{u}^{2}}}{2a}=\frac{0-60\times 60}{-2\times 1}=\frac{3600}{2}=1800m\]
(a) total distance moved by the train
\[={{S}_{1}}+{{S}_{2}}=900+1800=2700m=2.7km\]
(b) maximum speed attained by the train
\[{{V}_{man}}=60m/s.\]
(c) the position of the train at half the maximum speed
\[{{S}_{s}}=\frac{{{v}^{2}}-{{u}^{2}}}{2a}=\frac{{{(30)}^{2}}-0}{2\times 2}=\frac{900}{4}=225m\]
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