Between two stations a train accelerates uniformly at first, then moves with a constant speed and finally retards uniformly. If the ratios of the time taken are 1:8:1 and the greatest speed attained by the train is $45 k m h^{- 1}$ and the average speed is v $m s^{- 1}$ over the whole journey then the value of ‘v’ is ________

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answer is 11.25.

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Detailed Solution

Since time to accelerate from zero to $60 k m h^{- 1}$ is equal to the time taken to decelerate from $60 k m h^{- 1}$ to zero. Hence, we can say both OA and BC are identical intervals. So, total distance travelled from O to A to B to C is
$l = l_{1} + l_{2} + l_{1} = \frac{1}{2} υ_{\max} t + (υ_{\max}) 8 t + \frac{1}{2} υ_{\max} t$
Since,
Average speed = $\frac{Total Distance Travelled}{T o t a l T i m e T a k e n}$
$\Rightarrow υ_{a υ} = \frac{l_{1} + l_{2} + l_{1}}{t + 8 t + t}$

$\Rightarrow υ_{a υ} = \frac{\frac{1}{2} υ_{\max} t + (υ_{\max}) 8 t + \frac{1}{2} υ_{\max} t}{10 t}$
$\Rightarrow υ_{a υ} = \frac{9}{10} υ_{\max} = \frac{9}{10} (45) \times \frac{5}{18} = 11.25 m s^{- 1}$