Animated Solution for Physics - Kinematics: A particle is moving with a velocity $\mathbf{v}=k(y\hat{\mathbf{i}}+x\hat{\mathbf{j}})$, where $k$ is a constant. The general equation for its path is

The problem asks us to find the general equation of the path for a particle moving with a given velocity vector. This is a classic kinematics problem that beautifully connects vector calculus with coordinate geometry.

We are given the velocity of the particle as:

Here, $k$ is a constant. Notice something interesting? The velocity in the $x$-direction depends on the $y$-coordinate, and the velocity in the $y$-direction depends on the $x$-coordinate. This means the motion is coupled. As the particle moves, its changing position continuously alters its velocity components.

Relating Velocity to Position

To find the path, we need to establish a relationship between the $x$ and $y$ coordinates, completely eliminating time $t$. We know that velocity is the time derivative of the position vector $\mathbf{r}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}$.

Therefore, we can write the velocity vector in terms of its components as:

By comparing this standard definition with our given velocity equation, we can extract the individual rates of change for each coordinate:

The equation of the path is essentially a function $y=f(x)$ or some relation between $x$ and $y$. The slope of this path at any point is given by $\frac{dy}{dx}$.

Using the chain rule (or simply dividing the two rate equations), we can find this slope:

Substituting the expressions we found earlier:

Notice how the constant $k$ neatly cancels out! This tells us that the shape of the path doesn't depend on $k$; $k$ only determines how fast the particle traverses this path.

We now have a simple, variable-separable differential equation:

Let's separate the variables by cross-multiplying:

To find the general equation, we integrate both sides:

Performing the integration yields: where $C_1$ is the constant of integration.

To clean up the equation, let's multiply the entire expression by 2:

Since $2C_1$ is just another constant, we can replace it with a new constant, let's call it $C$:

This is our final answer! If we rearrange it slightly to $y^2-x^2=C$, we can recognize this as the standard equation for a family of hyperbolas. Depending on the initial conditions (which determine the value of $C$), the particle will trace out a specific hyperbola from this family.