Answer :


a. Here initial velocity, u=0


Let v be the final velocity


Acceleration, a=2 m/s2


Now, at t=30 s


We know,








When breaks are applied u1=60 m/s


v1=0


t=60 s




Total distance moved is


b. Maximum speed attained by the train is v= 60 m/s


c. Half the maximum speed=60/2=30 m/s


When acceleration of the train, a=2 m/s2



When the train is decelerating with, a1=-2 m/s2



Position from the starting point is 900+1350=2250 m=2.25 km


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