A particle starts from rest and travels a distance s with uniform acceleration, then it travels a distance 2s with uniform speed finally it travels a distance 3s with uniform retardation and comes to rest. If the complete motion of the particle is a straight line then the ratio of its average velocity to maximum velocity is

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answer is C.

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Detailed Solution

Maximum velocity $v_{max} = \sqrt{2 as}$
$\begin{array}{l} v_{avg} = \frac{6 s}{T} T = t_{1} + t_{2} + t_{3} \\ t_{1} = \sqrt{\frac{2 s}{a}}, t_{2} = \frac{2 s}{v_{max}} = \sqrt{\frac{2 s}{a}} \\ Retardation \Rightarrow B \\ β = \frac{v_{max}^{2}}{6 s} \\ β = \frac{2 a \times s}{6 \times s} = \frac{a}{3} \\ t_{3} = \frac{v_{max}}{β} = 3 \sqrt{\frac{2 a \times s}{a}} = 3 \sqrt{\frac{2 s}{a}} \\ T = 5 \sqrt{\frac{2 s}{a}} \begin{array}{l} v_{avg} = \frac{6}{5} \frac{5}{\sqrt{\frac{2 s}{a}}} \\ v_{avg} = \frac{6}{5} \sqrt{\frac{as}{2}} \\ \frac{v_{avg}}{v_{max}} = \frac{6}{5} \sqrt{\frac{as}{2}} \times \frac{1}{\sqrt{2 as}} = \frac{3}{5} \end{array} \end{array}$