A train starts from rest and accelerate uniformly at the rate of 2 m/s2 for 10 s. It then maintains a constant speed for 200 seconds. The brakes are then applied on the train is uniformly retarded and comes to rest in 50 seconds. Find:
1) The maximum velocity reached
2) The retardation in the last 50 seconds.
3) The total distance travelled.
4)The average velocity of the train

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Solution

Train accelerate from start for 10 s
so velocity reached at 10s is a×t= 2×10=20m/s
1) 20m/s is max velocity as after this the train runs on constant speed
2) the train is retarded for 50s from this speed to rest so retardation is velocity /time=20/50=
-0.4m/s2
3) distance travelled is total of distance travelled with accelaration+distance with constant speed+ distance with retardation
distance with acceleration is ut+1/2at2
=1/2210*10=100m
distance with constant speed is s×t=20×200=4000m
distance with retardation is ut+1/2at2
20×50-1/2×.4×50×50
=1000-500
=500m
total distance =100+4000+500=4600m
4) average velocity= total distance/total time
= 4600/260
=17.69m/s

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