A train accelerating uniformly from rest attains a maximum speed of 40 m/s in 20 sec. It travels at this speed for 20 sec, and is brought to rest with uniform retardation in 40 sec. The average velocity during this period is

(\frac{80}{3}) m / s

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30 m / s

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25 m / s

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40 m / s

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Solution

The correct option is C $25 m / s$
$\begin{matrix} S_{1} = (\frac{u + v}{2}) t = 20 \times 20 = 400 m S_{2} = 40 \times 20 = 800 m S_{3} = (\frac{u + v}{2}) t = 20 \times 40 = 800 m \end{matrix}$
Total displacement $= 2000$ m
Total time taken $= 80$ m
$V = \frac{2000}{80} = 25 m / s$
$∴$ Option $C$ is correct.

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