Physics - Kinematics Question with Solution | TestHub

Let$t_1$,$t_2$and$t_3$be the time taken by the particles to cover the distance$2x$,$4x$and$6x$respectively. Let$v$be the velocity of the particle at$\mathrm{B}$i.e, maximum velocity. The particle moves with uniform acceleration from$\mathrm{A}$to$\mathrm{B}$.

For motion from$\mathrm{A}$to$\mathrm{B}$.

Average velocity$=\frac{0+v}{2}=\frac{v}{2}$

Time taken,$t_1=\frac{2x}{{\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Particle moves with uniform retardation from$\mathrm{C}$TO$\mathrm{D}$,$V_{avg}=\frac{0+v}{2}=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Time taken ,$t_3=\frac{6x}{\left(\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

Total time =$t_1+t_2+t_3$

$t=\frac{2x}{\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{4x}{v}+\frac{6x}{\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{20x}{v}$

Average Velocity over=$\frac{6x+2x+4x}{\raisebox{1ex}{$20x$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}=\frac{3}{5}v$
Required Ratio =$\frac{\mathrm{Average}Speed}{v}=\frac{3}{5}$