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:

& 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{aligned}$$

Answer 1

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.

& \dfrac{dx}{dt}=t \\\ & \therefore dx=tdt \\\ \end{aligned}$$ Integrate the above function to find the expression for the x component in terms of ‘t’. $$\begin{aligned} & \int{dx}=\int{t\,dt} \\\ & x=\dfrac{{{t}^{2}}}{2}+C \\\ \end{aligned}$$ When $$x=0,t=0$$ $$\begin{aligned} & 0=\dfrac{{{0}^{2}}}{2}+C \\\ & C=0 \\\ \end{aligned}$$ Therefore, the value of x in terms of ‘t’ is given as follows. $$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{aligned} & \dfrac{dy}{dt}=\dfrac{{{t}^{2}}}{2} \\\ & \therefore dy=\dfrac{{{t}^{2}}}{2}dt \\\ \end{aligned}$$ Integrate the above function to find the expression for the x component in terms of ‘t’. $$\begin{aligned} & \int{dy}=\int{\dfrac{{{t}^{2}}}{2}\,dt} \\\ & y=\dfrac{1}{2}\times \dfrac{{{t}^{3}}}{3}+C \\\ \end{aligned}$$ Therefore, the value of y in terms of ‘t’ is given as follows. $$y=\dfrac{{{t}^{3}}}{6}$$ Now, re-substitute the value of x in the above equation of y. $$\begin{aligned} & y=\dfrac{{{t}^{3}}}{6} \\\ & y=\dfrac{{{\left( \sqrt{2x} \right)}^{3}}}{6} \\\ & 6y=8{{x}^{{}^{3}/{}_{2}}} \\\ \end{aligned}$$ Square on both sides to find the expression for the trajectory of the particle. $$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.