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Question

# 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 1msâˆ’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.

A
18211ms1
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B
9211ms1
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C
10011ms1
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D
1765ms1
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Solution

## The correct option is A 18211ms−1For the First 200 m, Initial velocity u=0, Final Velocity v, acceleration a= 1 ms−2, Using v2=u2+2as ⇒v2=0+2×1×200 ⇒v=20ms−1, Let t1 is time taken to travel first 200 m.Using v=u+at ⇒20=0+×t1 ⇒t1=20s, For next 3600 m, Speed is constant v=20ms−1 , Let t2 is time taken to travel 3600 m.Using s=ut ⇒t2=360020=180s, For the last 200 m , initial velocity u=20ms−1, acceleration a=-1 ms−2, Final Velocity v=0,Let t3 is time taken to travel last 200 mUsing v=u+at ⇒0=20−1×t3 ⇒t3=20s, Average speed =Total DistanceTotal Time⇒ Average speed= 4000t1+t2+t3=400020+180+20= 18211ms−1

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