The potential energy of a particle of mass 1 kg moving in a plane is given by V= 3x + 4y, the position coordinates of the point being x and y. If the particle is at rest at (6,4) ; then find the magnitude of acceleration, speed at which it crosses the y-axis and the coordinate at which it cross the y-axis.

Dear student,
The potential energy of particle is given by V, its x component and y component is given as
$$\begin{gathered} F_x=-\frac{dV}{dx} \\ {F}_y=-\frac{dV}{dy} \\ Hence force actingon particle is \overrightarrow{F}=-\left(3\hat{i}+4\hat{j}\right) N \end{gathered}$$
It means force acting on particle is constant. The particle moves with constant acceleration
$$\begin{gathered} \overrightarrow{a}=\frac{\overrightarrow{F}}{m} \\ \overrightarrow{a } =-\left(3\hat{i}+4\hat{j}\right) m/s^2 \\ The magnitude of accleration is a=\sqrt{3^2+4^2} \\ a=5m/s^2 \end{gathered}$$
When particle is at rest (6,4) , t=0.
The position of particle at time t is given by 
$$\begin{gathered} x=6+\frac{1}{2}{a}_x{t}^2 =\left(6-\frac{3}{2}{t}^2\right) m \to eqn 1 \\ y=4+\frac{1}{2}{a}_y{t}^2=\left(4-2t^2\right) m \to eqn 2 \end{gathered}$$
When the particle cross y axis , x=0
Substituting in above equation we get 
$$\begin{gathered} 6=\frac{3}{2}{t}^2 \\ t=2 sec \\ Speed of particle when it crosses y axis is given as \\ v=at =5\times 2=10m/s \end{gathered}$$

The co ordinate at which it cross x axis at t=1 is 
substituting t=1 in eqn 1 and eqn 2, 
$$\begin{gathered} x=6-\frac{3}{2}{\left(1\right)}^2 =4.5 \\ y=4-2{\left(1\right)}^2=2 \end{gathered}$$
Hence magnitude of acceleration is 5ms^-2,  speed at which it cross y axis is 10m/s, the coordinates x axis is (4.5,2)

Regards.