Question

The velocity vector of a particle moving in the xy plane is given by \[v=ti+xj\]. If initially, the particle was at origin then the equation of trajectory of the particle is:
\[\begin{align}
  & A.\,4{{x}^{2}}-9y=0 \\
 & B.\,9{{x}^{2}}-2{{y}^{3}}=0 \\
 & C.\,16{{x}^{3}}-9{{y}^{2}}=0 \\
 & D.\,9{{x}^{3}}-2{{y}^{2}}=0 \\
\end{align}\]

Answer 1

Hint: We will separately integrate the x and y components of the given equation of the velocity vector of a particle moving in the xy plane and then combine to find the expression for the trajectory of the particle.

Complete step-by-step answer:
From the given question statement, we have the data as follows.
The velocity vector of a particle moving in the xy plane, \[v=ti+xj\].

The component of x plane is, \[{{v}_{x}}=t\]
The component of y plane is, \[{{v}_{y}}=x\]

Consider the component of x plane.
\[{{v}_{x}}=t\]
Represent the above equation in terms of the differentiation.
\[\begin{align}
  & \dfrac{dx}{dt}=t \\
 & \therefore dx=tdt \\
\end{align}\]
Integrate the above function to find the expression for the x component in terms of ‘t’.
\[\begin{align}
  & \int{dx}=\int{t\,dt} \\
 & x=\dfrac{{{t}^{2}}}{2}+C \\
\end{align}\]
When \[x=0,t=0\]
\[\begin{align}
  & 0=\dfrac{{{0}^{2}}}{2}+C \\
 & C=0 \\
\end{align}\]
Therefore, the value of x in terms of ‘t’ is given as follows.
\[x=\dfrac{{{t}^{2}}}{2}\]


Consider the component of y plane.
\[{{v}_{y}}=x\]
Substitute the value of x in the above equation.
\[{{v}_{y}}=\dfrac{{{t}^{2}}}{2}\]
Represent the above equation in terms of the differentiation.
\[\begin{align}
  & \dfrac{dy}{dt}=\dfrac{{{t}^{2}}}{2} \\
 & \therefore dy=\dfrac{{{t}^{2}}}{2}dt \\
\end{align}\]
Integrate the above function to find the expression for the x component in terms of ‘t’.
\[\begin{align}
  & \int{dy}=\int{\dfrac{{{t}^{2}}}{2}\,dt} \\
 & y=\dfrac{1}{2}\times \dfrac{{{t}^{3}}}{3}+C \\
\end{align}\]
Therefore, the value of y in terms of ‘t’ is given as follows.
\[y=\dfrac{{{t}^{3}}}{6}\]

Now, re-substitute the value of x in the above equation of y.
\[\begin{align}
  & y=\dfrac{{{t}^{3}}}{6} \\
 & y=\dfrac{{{\left( \sqrt{2x} \right)}^{3}}}{6} \\
 & 6y=8{{x}^{{}^{3}/{}_{2}}} \\
\end{align}\]

Square on both sides to find the expression for the trajectory of the particle.
\[9{{y}^{2}}=16{{x}^{3}}\]

\[\therefore \] The equation of trajectory of the particle is, \[9{{y}^{2}}=16{{x}^{3}}\].

So, the correct answer is “Option D”.

Note: The expression for the trajectory of the particle using the velocity vector of a particle moving in the plane is found by integrating the given equation, as, by integrating the velocity equation, we get the expression of the particle.