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Question

# Between the two stations, a train uniformly accelerates from rest at first, then moves with constant velocity and finally decelerates uniformly to come to rest. If the ratio of the time taken is 1:8:1 and the maximum speed attained is 60 km/h, then what is the average speed over the whole journey?

A
48 km/h
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B
52 km/h
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C
54 km/h
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D
56 km/h
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Solution

## The correct option is C 54 km/hLet us divide total journey between two stations into P, Q and R. Where P is journey when train uniformly accelerates, Q is journey when train moves with constant speed and R is journey when train decelerates uniformly. Also assume the time taken by journey's P, Q and R be tP, tQ and tR respectively. As, tP:tQ:tR=1:8:1. We can say, tP=t, tQ=8t and tR=t. As the maximum speed attained is 60 km/hr, we can plot the v−t graph: Now, average velocity is given by ¯v=total displacementtotal time Total displacement is given by area under v−t curve. ∴total displacement=12×60×(8t+10t)total displacement=30×18t ¯v=30×18t/10t/⇒¯v=54 km/hr

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