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Q.

A train takes t sec to perform a journey. it travels for t/n sec with uniform acceleration then for
$(\frac{n - 3}{n}) t$ 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

$(3 n - 2) \frac{v}{2 n}$

b

$(2 n - 3) \frac{v}{2 n}$

c

$(3 n - 2) \frac{v}{3 n}$

d

$(2 n - 3) \frac{v}{3 n}$

answer is B.

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Detailed Solution

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Total time of journey = t
First interval = t/n
Second interval = $\frac{n - 3}{n} t$

Third interval = 2t/n

$A v e r a g e s p e e d = \frac{A r e a u n d e r v - t g r a p h}{T i m e}$

$\frac{\frac{1}{2} \frac{t}{n} V + V . \frac{n - 3}{n} t + \frac{1}{2} \frac{2 t}{n} V}{t}$

$\Rightarrow$ $\frac{2 n - 3}{2 n} v$

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