Animated Solution for Physics - Kinematics: The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $x^2=a{t}^2+2bt+c$. If the acceleration of the particle depends on $x$ as $x^{-n}$, where $n$ is an integer, the value of $n$ is ...... .

Imagine a particle moving along a straight line. Usually, in kinematics, we are handed the position of a particle explicitly as a function of time, like $x(t)=5t^2+3t$. But in this problem, we are given an implicit relationship:

This subtle difference changes our entire approach. We are asked to find how the acceleration $A$ depends on the position $x$. To do this, we must embark on a journey of implicit differentiation, peeling back the layers of velocity and acceleration hidden within this quadratic equation.

Uncovering Velocity

To find acceleration, we must first find velocity. Velocity is the rate of change of position with respect to time, $v=\frac{dx}{dt}$. We differentiate our given equation with respect to time $t$.

Here is where many students fall into a trap. The derivative of $x^2$ with respect to $t$ is not simply $2x$. Because $x$ is a function of $t$, we must use the Chain Rule.

Since $\frac{dx}{dt}$ is our velocity $v$, we can substitute it in and divide the entire equation by 2 to simplify our lives:

This is a beautiful, compact relation between position, velocity, and time. But we aren't done yet. We need acceleration.

The Path to Acceleration

Acceleration is the rate of change of velocity, $A=\frac{dv}{dt}$. To reach it, we must differentiate our new equation, $xv=at+b$, with respect to time once more.

On the left side, we have the product of two functions of time: $x$ and $v$. This calls for the Product Rule ($d(uv)=udv+vdu$).

Now, we substitute our kinematic definitions back into the equation. We know $\frac{dv}{dt}=A$ and $\frac{dx}{dt}=v$.

Eliminating Time

We have successfully brought acceleration $A$ into the picture. However, the problem asks for $A$ strictly as a function of $x$. Currently, our equation is polluted with $v$, which implicitly contains time $t$. We need to isolate $A$ and eliminate $v$.

First, let's isolate the term with $A$:

Now, recall our earlier velocity equation: $xv=at+b$, which means $v=\frac{at+b}{x}$. Let's substitute this expression for $v$ into our acceleration equation:

We still have $t$ floating around in the numerator. How do we get rid of it? This is where the magic happens. We look back at the very first equation given to us in the problem: $x^2=a{t}^2+2bt+c$. We can substitute this entire expression for $x^2$ directly into the numerator!

Let's expand the terms in the numerator carefully:

Look at that! The terms containing $t^2$ ($a^2{t}^2-a^2{t}^2$) and the terms containing $t$ ($2abt-2abt$) perfectly cancel each other out. The time variable vanishes completely, leaving behind only constants.

Dividing both sides by $x$, we arrive at our final, pristine expression for acceleration:

Since $a,b,$ and $c$ are constants, the numerator $(ac-b^2)$ is just a constant. Therefore, we can write the proportionality:

The problem states that $A\propto x^{-n}$. By directly comparing our result with the given form, it is crystal clear that the integer value we are looking for is $n=3$.

This problem is a masterclass in implicit differentiation and algebraic substitution. It teaches us that sometimes, the most elegant way forward is not to solve for a variable explicitly, but to manipulate the relationships between them.