Starting from rest, a particle moves a distance x in a straight line with constant acceleration a. Then, the particle moves a further distance x along the same line at constant velocity. Finally, the particle, moving along the same line, decelerates uniformly at rate $\frac{a}{2}$ until it comes to rest. Let the maximum velocity of the particle during its motion be ${\text{v}}_{\text{max}}$ and let its average velocity over the entire journey be ${\text{v}}_{\text{AVG}}$. The ratio $\frac{{\text{v}}_{\text{MAX}}}{{\text{v}}_{\text{AVG}}}$ is equal to ________.

We know that           $v_{MAX}^2=2\left(a\right)\left(x\right)\Rightarrow v_{MAX}=\sqrt{2ax}$
Time taken during the first segment,               $T_1=\frac{v_{MAX}}{a}=\sqrt{\frac{2x}{a}}$
Time taken during the segment,                     $T_2=\frac{x}{v_{MAX}}=\sqrt{\frac{x}{2a}}$
Time taken during the third segment,              $T_3=\frac{v_{MAX}}{\left(\frac{a}{2}\right)}=\sqrt{\frac{8x}{a}}$
Total time,            $T=T_1+T_2+T_3=(\sqrt{2}+\frac{1}{\sqrt{2}}+2\sqrt{2})\sqrt{\frac{x}{a}}=\frac{7}{\sqrt{2}}\sqrt{\frac{x}{a}}$
Distance travelled during the third segment,           $X_3=\frac{v^{2}MAX}{2\left(\frac{a}{2}\right)}=2x$
Total distance,                 $X=x+x+2x=4x$
Therefore,                      $v_{AVG}=\frac{\text{Total\hspace{0.17em}distance}}{\text{Total\hspace{0.17em}Time}}=\frac{X}{T}=\frac{4}{7}\sqrt{2ax}$
So,                              $\frac{v_{MAX}}{v_{AVG}}=\frac{7}{4}=1.75$