A train, starting from rest from station A, travels along a straight horizontal track until it stops at station B, which is 2400 m from A. Initially, the train accelerates at a uniform rate of 0.4...

A train, starting from rest from station A, travels along a straight horizontal track until it stops at station B, which is 2400 m from A. Initially, the train accelerates at a uniform rate of 0.4 ms–2 until it reaches a speed of 16 ms–1. It then maintains this speed of 16 ms–1 for T s, before decelerating uniformly to rest in 20 s.

  1. Calculate the time taken for accelerating.

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We can use the equation of motion here

`V=U+at`

`V = 16m/s`

`U = 0` since the train start from rest

`a = 0.4m/(s^2)`

`t = ???`

`16=0+0.4xxt`

`t=16/0.4 =160/4=40`

So the time taken for acceleration is 40s.

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