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

a )( 2 / 5 ) b )( 3 / 5 ) c )( 4 / 5 ) d ) (6 / 7 )

The correct answer is (b): 3/5

Let for the first half of the journey, the time taken by the body is $t_1$, as the body starts from rest, therefore using equation of motion we have,
 $$\begin{gathered} x=\frac{1}{2}a{t_1}^2 \\ {t}_1=\sqrt{\frac{2x}{a}} \end{gathered}$$
Now, the maximum  speed of the body can also be calculated using equation of motion as, 
 $$\begin{gathered} v_{max}=0+a{t}_1 \\ {v}_{max}=a{t}_1 \end{gathered}$$
Now, let us assume that the time taken by the body in its second journey (uniform speed) be,​
$$\begin{gathered} t_2=\frac{displacement}{speed} \\ {t}_2=\frac{2x}{v_{max}} \\ or, \\ {t}_2=\frac{2x}{a{t}_1} \\ or, \\ {t}_2=\frac{2x}{a\left(\sqrt{{\displaystyle \frac{2x}{a}}}\right)} \\ {t}_2=\sqrt{\frac{2x}{a}} \end{gathered}$$
Now, let us assume that the time taken by the body in its third journey (uniform retardation $-a_2$) be $t_3$. The retardation can be found out by using equation of motion as,
 $$\begin{gathered} 0={v_{max}}^2-2a_2\times 3x \\ {a}_2=\frac{v_{max}}{6x} \\ or, \\ {a}_2=\frac{a^2{t_1}^2}{6x} \end{gathered}$$
Now, the time for third journey can be calculated as,
 $$\begin{gathered} 0=v_{max}-a_2{t}_3 \\ {t}_3=\frac{v_{max}}{a_2} \\ or, \\ {t}_3=\frac{a{t}_1\times 6x}{{\left(a{t}_1\right)}^2} \\ {t}_3=\frac{6x}{a{t}_1} \\ or, \\ {t}_3=\frac{6x}{a\left(\sqrt{{\displaystyle \frac{2x}{a}}}\right)} \\ {t}_3=2\sqrt{{\displaystyle \frac{2x}{a}}} \end{gathered}$$
Now, the average speed can be calculated as,
 $$\begin{gathered} v_{av}=\frac{x+2x+3x}{t_1+t_2+t_3} \\ or, \\ {v}_{av}=\frac{6x}{\sqrt{{\displaystyle \frac{2x}{a}}}+\sqrt{{\displaystyle \frac{2x}{a}}}+3\sqrt{{\displaystyle \frac{2x}{a}}}} \\ {v}_{av}=\frac{6x}{5\sqrt{{\displaystyle \frac{2x}{a}}}} \end{gathered}$$
Now, the maximum speed is given as,
 $$\begin{gathered} V_{max}=t_1\times a \\ or, \\ {V}_{max}=\sqrt{\frac{2x}{a}}\times a \end{gathered}$$
Therefore on taking ratio we get,
 $$\begin{gathered} \frac{v_{av}}{v_{max}}=\frac{\frac{6x}{5\sqrt{{\displaystyle \frac{2x}{a}}}}}{\sqrt{{\displaystyle \frac{2x}{a}}}\times a} \\ \frac{v_{av}}{v_{max}}=\frac{\frac{6x}{5\sqrt{{\displaystyle \frac{2x}{a}}}}}{\sqrt{{\displaystyle \frac{2x}{a}}}\times \sqrt{2ax}} \\ \frac{v_{av}}{v_{max}}=\frac{3}{5} \end{gathered}$$