Moving charged particle 2

Please refer to this image for the solution. A basic background of high school electric field is required to understand this in a snap. http://tinypic.com/r/aawymt/9

First, we have x=v0 t1=>2d=v0 t1=>t1=2d/v0, since we have no acceleration in x direction. F=m a1=qE => so acceleration in y direction is a1=qE/m, this means that the electric charge would travel DELTAy1=1/2 a1 (t1)^2 in y direction, since we have no initial speed in y direction => DELTAy1=1/2 (qE/m) (2d/v0)^2 . When this charge arrives to this point, we call R(xR,yR)=R(2d,3d-DELTAy1) would accumulate a speed in y direction, that we call vR=a1 t1=(qE/m) (2d/v0), this speed would drive the charge a distance of 3d-DELTAy1 in y direction, and it's initial v0 speed in x direction would drive it more 4d-2d=2d in x direction. This means the time required to arrive to point Q(4d,0) is t2=2d/v0. The drive force of the charge is the electric field, so we can express electric field in function of q, m, v0 and d using kinematics equation 3d-DELTAy1=vR t2, because after the charge exits the electric field zone, there would be no more acceleration in y direction. So 3d-DELTAy1=3d-1/2 (qE/m) (2d/v0)^2=(qE/m) (2d/v0) (2d/v0) => E=m (v0)^2/(2qd) this electric field would also move the 2q charge. The 2d charge has v0 speed in x direction and would acquire a speed in y direction at a particular point that we call T(2d,3d-DELTAy2) a vT=a2 (t3) speed. F=m a2=2qE => a2=2qE/m; t3=2d/v0 => vT=(2qE/m) (2d/v0). In y direction the charge would travel to this point a DELTAy2=1/2 (a2) (t3)^2=1/2 (2qE/m) (2d/v0)^2 distance. This means that to arrive to a point S(x,0), the charge would travel more DELTAx=v0 (t4) distance in x direction and 3d-DELTAy2 distance in y direction. After the charge exits the electric field zone would acquire a speed vT which we calculated, and would have no more acceleration in y direction. So 3d-DELTAy2=vT t4 => t4= (3d-DELTAy2)/vT=(3d-1/2 (2qE/m) (2d/v0)^2)/((2qE/m) (2d/v0)), and substituting E in the equation we obtain t4=1/2 (d/v0). DELTAx=v0 (t4)=v0 (1/2) (d/v0)=d/2. Hence x=2d+DELTAx=2d+1/2 d=5d/2