Multiple choice

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 $\displaystyle 1\quad { ms }^{ -1 }$ 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.

  1. $\displaystyle 18\dfrac { 2 }{ 11 } { ms }^{ -1 }$
  2. $\displaystyle 9\dfrac { 2 }{ 11 } { ms }^{ -1 }$
  3. $\displaystyle \dfrac { 100 }{ 11 } { ms }^{ -1 }$
  4. $\displaystyle 17\dfrac { 6 }{ 5 } { ms }^{ -1 }$
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A Correct answer
Explanation

The train accelerates from 0 to v over 200m, travels at v for the middle distance, and decelerates from v to 0 over 200m. Using v^2 = 2as, v^2 = 2 * 1 * 200 = 400, so v = 20 m/s. Time for acceleration/deceleration is t = v/a = 20/1 = 20s each. Distance covered during these phases is 200m + 200m = 400m. Remaining distance is 4000m - 400m = 3600m. Time for constant speed is 3600/20 = 180s. Total time = 20 + 180 + 20 = 220s. Average speed = 4000m / 220s = 400/22 = 200/11 = 18 and 2/11 m/s.

AI explanation

Using the kinematic equation v squared equals u squared plus 2as, the maximum speed is the square root of 2 times 1 times 200, which equals 20 m/s. The time for acceleration and retardation is 20 seconds each, and the time for the remaining 3600 m at uniform speed is 180 seconds, giving an average speed of 4000 m divided by 220 seconds, or 18 and 2/11 m/s.

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