Question

A train leaves station 'A' to station 'B'. The train travels straight without any halts between the stations. During the first and last $200$ m of its journey, the train has uniform acceleration and retardation both equal to $1{ms}^{-1}$ respectively. For the rest of the journey, the train maintains uniform speed Calculate the average speed of the train, given the distance between two stations is 4 km.

• A. $18\frac{2}{11}{ms}^{-1}$
• B. $9\frac{2}{11}{ms}^{-1}$
• C. $\frac{100}{11}{ms}^{-1}$
• D. $17\frac{6}{5}{ms}^{-1}$

Answer 1

The correct option is A $18\frac{2}{11}{ms}^{-1}$

For the First 200 $m$,

Initial velocity $u=0$, Final Velocity $v$, acceleration $a$= 1 $m{s}^{-2}$,

Using $v^2=u^2+2as$ $\Rightarrow v^2=0+2\times 1\times 200$ $\Rightarrow v=20m{s}^{-1}$,

Let $t_1$ is time taken to travel first 200 $m$.

Using $v=u+at$ $\Rightarrow 20=0+\times t_1$ $\Rightarrow t_1=20s$,

For next 3600 $m$, Speed is constant $v=20m{s}^{-1}$ ,

Let $t_2$ is time taken to travel 3600 $m$.

Using $s=ut$ $\Rightarrow t_2=\frac{3600}{20}=180s$,

For the last 200 $m$ , initial velocity $u=20m{s}^{-1}$, acceleration $a$=-1 $m{s}^{-2}$, Final Velocity $v=0$,

Let $t_3$ is time taken to travel last 200 $m$

Using $v=u+at$ $\Rightarrow 0=20-1\times t_3$ $\Rightarrow t_3=20s$,

Average speed =$\frac{\text{Total Distance}}{\text{Total Time}}$

$\Rightarrow$ Average speed= $\frac{4000}{t_1+t_2+t_3}=\frac{4000}{20+180+20}$= $18\frac{2}{11}m{s}^{-1}$