question_answer

A train starts from rest and moves with a constant acceleration of\[\text{2}.0\text{m}/{{\text{s}}^{\text{2}}}\] for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

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 \[{{S}_{s}}=\frac{{{v}^{2}}-{{u}^{2}}}{2a}=\frac{{{(30)}^{2}}-0}{2\times 2}=\frac{900}{4}=225m\]