Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1:8:1 and the maximum speed attained be 60 km h^-1, then what is the average speed over the whole journey?

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$48 {kmh}^{- 1}$

$56 {kmh}^{- 1}$

$52 {kmh}^{- 1}$

$54 {kmh}^{- 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} S_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} (a t) t = \frac{60 t}{2} = 30 t \\ S_{2} = 60 \times 8 t = 480 t, S_{3} = S_{1} = 30 t \end{array}$

$v_{av} = \frac{S_{1} + S_{2} + S_{3}}{t + 8 t + t} = 54 {kmh}^{- 1}$

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