A particle starts from rest and traverses a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2 l$ and finally comes to rest after moving a further distance $3 l$ under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways is

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$\frac{2}{5}$

$\frac{3}{5}$

$\frac{1}{5}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

Let $v_{m}$ be the maximum speed,
$v_{m} = a_{1} t_{1} v_{m} = \sqrt{2 a_{1} l} t_{2} = \frac{2 l}{v_{m}} v_{m} = a_{2} t_{3} = \sqrt{2 a_{2} (3 l)}$
Now, average speed
$v_{a v} = \frac{l + 2 l + 3 l}{t_{1} + t_{2} + t_{3}} V_{a v} = \frac{6 l}{(v_{m} / a_{1}) + (2 l / v_{m}) + (v_{m} / a_{2})} = \frac{6 l}{(\frac{v_{m}}{v_{m}^{2} / 2 l}) + (\frac{2 l}{v_{m}}) + (\frac{v_{m}}{v_{m}^{2} / 6 l})} = \frac{6 /}{(10 / (y_{m})} = \frac{3 v_{m}}{5} \frac{ν_{aV}}{ν_{mo}} = \frac{3}{5}$