The potential energy (in joule) of a body of mass $2 kg$ moving in the $x - y$plane is given by

                                        $U = 6x + 8y$

where the position coordinates x and y are measured in meter. If the body is at rest at point $(6m, 4m)$ at time t = 0, it will cross the y-axis at time t equal to 

Given $U = 6x + 8y$ Joule and mass $m=2kg.$

Force along x-axis is 

$F_x=-\frac{dU}{dx}=-\frac{d}{dx}(6x+8y)=-6$ newton

Force along y-axis is

$F_y=-\frac{dU}{dy}=-\frac{d}{dy}(6x+8y)=-8$ newton

Therefore, the x and y components of acceleration are

$$\begin{gathered} a_x=\frac{F_x}{m}=\frac{-6}{2}=-3m{s}^{-2} \\ and a_y=\frac{F_y}{m}=\frac{-8}{2}=-4m{s}^{-2} \end{gathered}$$

$\therefore$Resultant acceleration

$a=\sqrt{{a_x}^2+{a_y}^2}=\sqrt{{(-3)}^2+{(-4)}^2}=5m{s}^{-2}$

The x and y coordinates of the body at time t are 

$$\begin{gathered} x=x_0+\frac{1}{2}{a}_x{t}^2=6-\frac{1}{2}\times 3\times t^2 \\ =\left(6-\frac{3}{2}{t}^2\right) metre \end{gathered}$$

and

 $$\begin{gathered} y=y_0+\frac{1}{2}{a}_y{t}^2=4-\frac{1}{2}\times 4\times t^2 \\ =(4-2t^2)metre \end{gathered}$$

The body will cross the y-axis when $x = 0$, i.e., at time t given by $\left(6-\frac{3}{2}{t}^2\right)=0 or t=2s.$