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

$$\begin{gathered} \mathrm{Let} {\mathrm{v}}_{\mathrm{m}} \mathrm{be} \mathrm{the} \mathrm{maximum} \mathrm{speed}, \\ {\mathrm{v}}_{\mathrm{m}}={\mathrm{a}}_1{\mathrm{t}}_1 \mathrm{and} {\mathrm{v}}_{\mathrm{m}}=\sqrt{2{\mathrm{a}}_1\mathrm{l}} \\ {\mathrm{t}}_2= \frac{2\mathrm{l}}{{\mathrm{v}}_{\mathrm{m}}} \mathrm{and} {\mathrm{v}}_{\mathrm{m}}={\mathrm{a}}_2{\mathrm{t}}_3 =\sqrt{2{\mathrm{a}}_2\left(3\mathrm{l}\right)} \\ \mathrm{Now}, \mathrm{average} \mathrm{speed} {\mathrm{v}}_{\mathrm{av}}=\frac{\mathrm{l}+2\mathrm{l}+3\mathrm{l}}{{\mathrm{t}}_1+ {\mathrm{t}}_2+{\mathrm{t}}_3} \\ {\mathrm{v}}_{\mathrm{av}}=\frac{6\mathrm{l}}{\left({\mathrm{v}}_{\mathrm{m}}/{\mathrm{a}}_1\right)+\left(2\mathrm{l}/{\mathrm{v}}_{\mathrm{m}}\right)+\left({\mathrm{v}}_{\mathrm{m}}/{\mathrm{a}}_2\right)} \\ =\frac{6\mathrm{l}}{\left({\displaystyle \frac{{\mathrm{v}}_{\mathrm{m}}}{{\mathrm{v}}_{\mathrm{m}}^2/2\mathrm{l}}}\right)+\left({\displaystyle \frac{2\mathrm{l}}{{\mathrm{v}}_{\mathrm{m}}}}\right)+\left(\frac{{\mathrm{v}}_{\mathrm{m}}}{{\mathrm{v}}_{\mathrm{m}}^2/6\mathrm{l}}\right)} \\ =\frac{6\mathrm{l}}{\left(10\mathrm{l}/{\mathrm{v}}_{\mathrm{m}}\right)}=\frac{3{\mathrm{v}}_{\mathrm{m}}}{5} \\ \frac{{\mathrm{v}}_{\mathrm{av}}}{{\mathrm{v}}_{\mathrm{m}}}=\frac{3}{5} \end{gathered}$$