A particle of charge $$q$$ and mass $$m$$ moves rectilinearly under the action of electric field $$E = A - Bx,$$   where $$A$$ and $$B$$ are positive constants and $$x$$ is distance from the point where particle was initially at rest then the distance traveled by the particle before coming to rest and acceleration of particle at that moment are respectively :

Solution :
$$\eqalign{ & F = qE = q\left( {A - Bx} \right) \cr & ma = q\left( {A - Bx} \right) \Rightarrow a = \frac{q}{m}\left( {A - Bx} \right)\,......\left( {\text{i}} \right) \cr & \frac{{vdv}}{{dx}} = \frac{q}{m}\left( {A - Bx} \right);vdv = \frac{q}{m}\left( {A - Bx} \right)dx \cr & \int\limits_0^0 {vdv} = \frac{q}{m}\int\limits_0^x {\left( {A - Bx} \right)dx} ;Ax - \frac{{B{x^2}}}{2} = 0 \cr & x = 0,x = \frac{{2A}}{B}\,......\left( {{\text{ii}}} \right) \cr} $$
From eqs. (i) and (ii)
$$\frac{q}{m}\left( {A = Bx} \right) = \frac{q}{m}\left( {A - B \times \frac{{2A}}{B}} \right) = \frac{{ - qA}}{m}.$$