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A train starts from rest from station A with constant acceleration of 2 m/s2. After some time it starts moving with constant retardation of 3 m/s2 and finally stops at station B. It the separation between the stations is 1215 m, the maximum velocity attained by the train is

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54 m/s

48 m/s

60 m/s

24 m/s

answer is A.

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Detailed Solution

$x_{1} + x_{2} = 1215 \Rightarrow \frac{{(v_{m a x})}^{2}}{2 \times 2} + \frac{{(v_{m a x})}^{2}}{2 \times 3} = 1215 \Rightarrow v_{m a x} = 54 m / s$

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