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# 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$m{{s}^{-2}}$ respectively. For the rest of the journey, the train maintains uniform speed. Calculate the average speed of the train, given distance between two stations is 4 km.

Last updated date: 20th Jun 2024
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Hint: We need to understand the relation between the acceleration of the train while speeding up to a constant velocity and retarding to the halt with the average speed of the train between the given two points to solve the problem as required.

Complete Step-by-Step Solution:
We are given that a train which travels for 4 km from a station ‘A’ to the station ‘B’ accelerates and decelerates for a constant distance of 200m respectively at the start and end of the journey at 1 $m{{s}^{-2}}$. We can find the maximum velocity achieved for the constant motion of the train after its deceleration using the equations of motion as –
\begin{align} & {{v}^{2}}-{{u}^{2}}=2as \\ & \Rightarrow {{v}^{2}}-0=2\times 1m{{s}^{-2}}\times 200m \\ & \Rightarrow {{v}^{2}}=400 \\ & \therefore v=20m{{s}^{-1}} \\ \end{align}

This will be velocity at which the train travels constantly during its uniform motion. Now, we can find the time taken for the train to accelerate to attain this velocity again using the equations of motion as –
\begin{align} & v=u+at \\ & \Rightarrow t=\dfrac{v-u}{a} \\ & \Rightarrow t=\dfrac{20m{{s}^{-1}}-0}{1m{{s}^{-2}}} \\ & \therefore t=20s \\ \end{align}

We can understand that the time taken for accelerating to the velocity and decelerating to rest from the same velocity travelling for the same distance will be the same. So, the time taken for acceleration and retardation can be as –
\begin{align} & {{t}_{a}}={{t}_{acc}}+{{t}_{ret}} \\ & \therefore {{t}_{a}}=40s \\ \end{align}

Now, we can find the time taken for the travelling of the train for the rest of the motion at a constant motion for the rest of the distance can be given as –
\begin{align} & v=\dfrac{s}{t} \\ & \Rightarrow t=\dfrac{s}{v} \\ & \Rightarrow t=\dfrac{4000m-400m}{20m{{s}^{-1}}} \\ & \therefore t=180s \\ \end{align}

Now, we get the total time taken for the travel as –
\begin{align} & {{t}_{total}}={{t}_{a}}+t \\ & \Rightarrow {{t}_{total}}=40s+180s \\ & \therefore {{t}_{total}}=220s \\ \end{align}

Now, we can find the average speed of the journey as –
\begin{align} & {{v}_{avg}}=\dfrac{{{S}_{total}}}{{{t}_{total}}} \\ & \Rightarrow {{v}_{avg}}=\dfrac{4000m}{220s} \\ & \therefore {{v}_{avg}}=18\dfrac{2}{11}m{{s}^{-1}} \\ \end{align}

This is the required solution which gives the average velocity of the train.

Note:
The average speed is the speed at which the particle or the body could have moved in a constant speed from the initial point to the final point in the same given time. The value of the average speed needn’t be related to the actual speeds at which the body travelled.