A particle moves in the x-y plane with velocity ${{v}_{x}}=8t-2$ and ${{v}_{y}}=2$. If it passes through the point $x=14$and $y=4$ at $t=2\sec $, then the equation of the path is
$\begin{align}
  & \text{A}\text{. }x={{y}^{3}}-{{y}^{2}}+2 \\
 & \text{B}\text{. }x={{y}^{2}}-y+2 \\
 & \text{C}\text{. }x={{y}^{2}}-3y+2 \\
 & \text{D}\text{. }x={{y}^{3}}-2{{y}^{2}}+2 \\
\end{align}$

Hint: The particle moves in x-y plane, means it is exhibiting two dimensional motion. The velocities in individual directions are given. With the help of velocity expressions, we can find the position vector of a particle in individual directions. The equation of path can be obtained by comparing the individual equations of path in $x$ and $y$ directions.

Complete step by step answer:
If a particle is moving in a single direction throughout its journey, then it is said to be moving in one dimension. For example: an ant moving along x-axis is an example of one dimensional motion. Motion is described in terms of displacement$\left( x \right)$, time$\left( t \right)$, velocity$\left( v \right)$, and acceleration$\left( a \right)$.
Motion in two dimensions:
Motion in a plane is described as two dimensional motion. For example: An ant moving on the top surface of a desk is an example of two dimensional motion. Projectile and circular motion are examples for two dimensional motion.
We are given velocity of particle is,
${{v}_{x}}=8t-2$
$\dfrac{dx}{dt}=8t-2$
Integrating both sides,
\[\int{dx}=\int{\left( 8t-2 \right)dt}\]
We get,
$x=4{{t}^{2}}-2t+c$
Where,
$c$ is the constant of integration
As given, $x=14$ at $t=2\sec $
Therefore,
\[\begin{align}
  & 14=4\times {{\left( 2 \right)}^{2}}-2\times 2+c \\
 & 14=16-4+c \\
 & c=2 \\
\end{align}\]
Thus,
$x=4{{t}^{2}}-2t+2$
Also,
${{v}_{y}}=2$
$\dfrac{dy}{dt}=2$
Integrating both sides,
$\int{dy}=\int{2dt}$
$y=2t+c$
Where,
$c$ is the constant of integration
As given, $y=4$ at $t=2\sec $
Therefore,
$\begin{align}
  & 4=2\times 2+c \\
 & c=0 \\
\end{align}$
Thus,
\[y=2t\]
Or,
$t=\dfrac{y}{2}$
We will put value of time in terms of $y$ in the equation obtained for $x$
We have,
$x=4{{t}^{2}}-2t+2$
Put $t=\dfrac{y}{2}$
$\begin{align}
  & x=4{{\left( \dfrac{y}{2} \right)}^{2}}-2\left( \dfrac{y}{2} \right)+2 \\
 & x={{y}^{2}}-y+2 \\
\end{align}$
Equation of motion of particles is given by,
$x={{y}^{2}}-y+2$
Hence, the correct option is B.

Additional information:
Different types of motion exhibited by a particle:
Uniform motion is defined as the motion of particle in which the particle travels in a straight line with uniform speed and its velocity remains constant along that line because it covers equal distances in equal intervals of time, being irrespective of the time duration.
Non-uniform is defined as the motion of a particle in which the object travels with varied speed or velocity and it does not cover the same distance in equal time intervals, irrespective of the time duration.
Uniformly accelerated motion is the motion of a particle in which the acceleration of the particle is uniform throughout the motion. It can move in one dimension, two dimensions, and three dimensions.
Non-uniformly accelerated motion is the one in which the acceleration of the particle is not uniform throughout the motion. The velocity of the particle keeps on changing and hence the acceleration. Acceleration can be time dependent as well.

Note:
When a particle moves in two dimensions, it has different values or expressions for velocity in individual direction. While integrating the expression above to find the position vector of particle in and direction, the constant of integration must be added to get the correct expression.
In above question, when we differentiate velocity vectors to find the expression for acceleration of the particle,
\[\begin{align}
  & {{a}_{x}}=\dfrac{d\left( {{v}_{x}} \right)}{dt} \\
 & {{a}_{x}}=\dfrac{d\left( 8t-2 \right)}{dt} \\
 & {{a}_{x}}=8 \\
\end{align}\]
\[\begin{align}
  & {{a}_{y}}=\dfrac{d\left( {{v}_{y}} \right)}{dt} \\
 & {{a}_{y}}=\dfrac{d\left( 2 \right)}{dt} \\
 & {{a}_{y}}=0 \\
\end{align}\]
The acceleration of particle in $x$ direction is constant and that in $y$ direction is zero. It means that the object is having uniformly accelerated motion in $x$ direction and uniform motion in $y$ direction.