A train stops at two stations A and B. It accelerates from rest from station A to a speed of 144 k m / h − 1 in 3 minutes and maintains this speed for 10 minutes. It then decelerates for 2 minutes and comes to rest at station B. Find the total distance between A and B.

Question

A train stops at two stations A and B. It accelerates from rest from station A to a speed of   144  k  m      /        h          −      1       in 3 minutes and maintains this speed for 10 minutes. It then decelerates for 2 minutes and comes to rest at station B. Find the total distance between A and B.

afermavas4h · Accepted Answer

Answer:The total distance between A and B is 30 kmExplanation:The given information are;The time duration of the acceleration of the train from station A = 3 minutes = 0.05 hoursThe speed attained by the train after the acceleration = 144 km/hThe time duration the train maintains the speed = 10 minutes = 0.1 hoursThe time duration in which the train decelerates to station B = 2 minutes = 0.0 hoursThe equation of motion required are;The initial acceleration,   a  =            (      144      −      0      )        0.05    =  2  ,  880  k  m      /        h    2  The distance covered,   s  =  u  ⋅  t  +      1    2    ⋅  a  ⋅      t    2  Where;u The initial velocity = 0  ∴      s    1    =  0  ×  0.05  +      1    2    ×  2880  ×      0.05    2    =  3.6  k  m      s    1    =  3.6  k  mThe distance, s₂ the train covers at the constant speed 144 km/h for 10 minutes (1/6 hours) is given as follows;      s    2    =   Velocity Time   =  144  ×      1    6    =  24  k  m      s    2    =  24  k  mThe deceleration, a₂ that brings the train to a stop in 2 minutes (1/30 hours) is given as follows;      a    2    =            (      0      −      144      )              (              1        30            )        =  −  4320  k  m      /        h    2  The distance covered, s₃ by the train as it decelerates to rest from the initial constant speed is given as follows;      s    3    =  u  ⋅  t  +      1    2    ⋅  a  ⋅      t    2  Where;u = The initial velocity =144 km/hWe have;      s    3    =  144  ×      1    30    −      1    2    ×  4320  ×  (      1    30        )    2    =  2.4      s    3    =  2.4  k  mThe total distance between A and B,   s  =      s    1    +      s    2    +      s    3    =  3.6  +  24  +  2.4  =  30  k  mThe total distance between A and B = 30 km.

A train stops at two stations A and B. It accelerates from rest from station A to a speed of 144 km/h^(-1) in 3 minutes and maintains this speed for 10 minutes. It then decelerates for 2 minutes and comes to rest at station B. Find the total distance between A and B.

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