JEE PYQ: Motion in a Plane - Question ID fa93b4574d19 (JEE Main 2019)

In problems involving motion in a plane, the velocity vector $\overrightarrow{v}$ is given as a function of position coordinates $(x, y)$. To find the path (trajectory) of the particle, we relate the components of velocity to the time derivatives of position:

Given $\overrightarrow{v} = K(y\widehat{i} + x\widehat{j})$, we equate components:

To eliminate time $t$, we divide the two equations:

This yields a differential equation in $x$ and $y$ whose solution gives the path equation.

Step 1: Express velocity components

Given: $\overrightarrow{v} = K(y\widehat{i} + x\widehat{j}).$ Equate to $\frac{d\overrightarrow{r}}{dt}$: $\frac{dx}{dt} = Ky, \quad \frac{dy}{dt} = Kx.$

Step 2: Eliminate time $t$

Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{Kx}{Ky} = \frac{x}{y}.$

Step 3: Solve the differential equation

We have: $\frac{dy}{dx} = \frac{x}{y}.$ Separate variables: $y\,dy = x\,dx.$ Integrate both sides: $\int y\,dy = \int x\,dx \quad \Rightarrow \quad \frac{y^2}{2} = \frac{x^2}{2} + C,$ where $C$ is the integration constant.

Step 4: Simplify and match with options

Multiply both sides by 2: $y^2 = x^2 + 2C.$ Let $2C = \text{constant}$, so: $y^2 = x^2 + \text{constant}.$ This matches option C.

Students often confuse the order of $x$ and $y$ when separating variables or forget to introduce the integration constant. Another common mistake is misinterpreting the velocity components, leading to incorrect differential equations. Always double-check the signs and variables when setting up $\frac{dy}{dx}$.

Exam Tip: When given a velocity vector $\overrightarrow{v} = f(x,y)\widehat{i} + g(x,y)\widehat{j}$, always eliminate time by computing $\frac{dy}{dx} = \frac{g(x,y)}{f(x,y)}$ and solve the resulting differential equation to find the path.