A particle starts from point A moves along a straight line path with an acceleration given by a = p - qx where p,q are constants and x is distance from point A. The particle stops at point B. The maximum velocity of the particle is

Given = a = P − qx

Vdv/dx​ = P − qx

${\int }_0^{v}vdv={\int }_0^x(p-qx)dx$

v²/2 ​= px − qx²/2​                  ...(1)

because body come to fast , hence velocity will be maximum at point where 

a = 0 or dv/dx ​= 0 ,p − qx = 0 

For Vmax

x = p/q​

V_max² = 2p(p/q​)

V_max² ​= 2p(p/q​)−q(p/q​)² = p²/q​

V_max ​= $\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$\sqrt{q}$}\right.$