Animated Solution for Physics - Kinematics: The coordinates of a particle moving in a plane are given by $x(t)=a\cos (pt)$ and $y(t)=b\sin (pt)$ where $a,b$ ($<a$) and $p$ are positive constants of appropriate dimensions. Then,

Imagine you are tracking a mysterious particle in a two-dimensional plane. You don't know its exact path, but you have a clock, and you know its coordinates at any given time $t$. The coordinates are given by the parametric equations:

To understand the true geometric path of this particle, we must remove time from the equation. We need a direct relationship between $x$ and $y$. By rearranging the equations, we get:

Now, we pull out the most famous weapon in our trigonometric arsenal: the Pythagorean identity. We know that for any angle, the sum of the squares of its sine and cosine is exactly $1$.

Substituting our expressions into this identity, we reveal the hidden path:

This is the unmistakable signature of an ellipse! Since $a>b$, it is an ellipse stretched along the x-axis. Thus, the path of the particle is indeed an ellipse, making our first option perfectly correct.

Now that we know the stage (the ellipse), let's analyze the dance. The position vector of the particle at any time $t$ is simply its coordinates attached to the directional unit vectors:

Velocity is the rate at which this position changes. We differentiate the position vector with respect to time $t$. Remember, the derivative of $\cos (pt)$ is $-p\sin (pt)$, and the derivative of $\sin (pt)$ is $p\cos (pt)$.

To find the acceleration, we differentiate the velocity vector. The rate of change of velocity tells us how the particle is being pushed or pulled.

The problem asks us to investigate a very specific moment: $t=\frac{\pi }{2p}$. Let's plug this time into our equations. At this instant, the angle $pt$ becomes exactly $\frac{\pi }{2}$ radians, or $90^{\circ }$.

We know that $\cos (\frac{\pi }{2})=0$ and $\sin (\frac{\pi }{2})=1$. Substituting these values into our velocity and acceleration vectors, we get:

Look at these vectors! The velocity is pointing purely along the negative x-axis, while the acceleration is pointing purely along the negative y-axis. They are completely independent of each other. To prove they are perpendicular mathematically, we take their dot product:

Since their dot product is zero, the velocity and acceleration vectors are perfectly normal (perpendicular) to each other at this instant. This confirms that the second option is also correct.

Let's take a closer look at the general acceleration vector we derived earlier. What happens if we factor out $-p^2$ from the expression?

Do you recognize the term inside the brackets? It is exactly our original position vector $\mathbf{r}(t)$! This allows us to write a beautiful, elegant relationship:

This equation is profound. It tells us that the acceleration is always directly proportional to the position vector, but points in the exact opposite direction (due to the negative sign). Since the position vector $\mathbf{r}$ originates from the center of the coordinate system $(0,0)$, the acceleration must always point back towards the origin.

Here is the critical trap: The origin is the center of the ellipse, not the focus. The foci of an ellipse are located at $(\pm ae,0)$. Therefore, stating that the acceleration is directed towards a focus is technically incorrect. However, in the historical context of this specific JEE paper, the option was marked as correct, likely due to a phrasing error by the examiners who confused the center with the focus. As elite students, we must understand the true physics while being aware of such historical anomalies.

Finally, let's evaluate the distance traveled by the particle from $t=0$ to $t=\frac{\pi }{2p}$.

At $t=0$, the position is $(a\cos (0),b\sin (0))$, which simplifies to $(a,0)$. At $t=\frac{\pi }{2p}$, the position is $(a\cos (\frac{\pi }{2}),b\sin (\frac{\pi }{2}))$, which simplifies to $(0,b)$.

The particle has traveled from the rightmost tip of the ellipse to the topmost tip. It has traced out exactly one-quarter of the elliptical perimeter. Is this distance equal to $a$?

Absolutely not! The straight-line displacement between these two points would be $\sqrt{a^2+b^2}$. The actual distance traveled is the arc length of the ellipse, which is a curved path and is strictly greater than the straight-line displacement. It requires a complex elliptic integral to calculate exactly, but it is definitively not equal to $a$. Thus, the final option is incorrect.