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A train takes t sec to perform a journey. it travels for t/n sec with uniform acceleration then for (n-3n)t sec with uniform speed v and finally it comes to rest with uniform retardation. Then average speed of train is

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a

(3n-2)v2n

b

(2n-3)v2n

c

(3n-2)v3n

d

(2n-3)v3n

answer is B.

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Detailed Solution

Total time of journey = tFirst interval = t/nSecond interval = n-3ntThird interval = 2t/n Average speed= Area under v-t graphTime 12tnV +V.n-3nt + 122tnVt ⇒2n-32nv

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