Animated Solution for Physics - Kinematics: The relation between time $t$ and distance $x$ for a moving body is given as $t=m{x}^2+nx$, where $m$ and $n$ are constants. The retardation of the motion is (when $v$ stands for velocity)

The problem presents us with a fascinating implicit relationship between time $t$ and position $x$: $t=m{x}^2+nx$. Our ultimate goal is to find the retardation of the particle, which is simply the negative of its acceleration.

To embark on this journey, we must remember the fundamental definitions of kinematics. Velocity $v$ is the rate of change of position with respect to time, mathematically expressed as $v=\frac{dx}{dt}$. Acceleration $a$, in turn, is the rate of change of velocity, given by $a=\frac{dv}{dt}$.

Instead of trying to solve the quadratic equation for $x$ in terms of $t$—which would involve messy square roots—we can use the power of implicit differentiation. By differentiating both sides of our equation with respect to time $t$, we can elegantly extract the velocity term.

Let's differentiate $t=m{x}^2+nx$ with respect to $t$. The derivative of $t$ is simply $1$. On the right side, we apply the chain rule. The derivative of $m{x}^2$ is $2mx\cdot \frac{dx}{dt}$, and the derivative of $nx$ is $n\cdot \frac{dx}{dt}$.

Finding Velocity

Putting it all together, we get:

Now, we recognize that $\frac{dx}{dt}$ is exactly our velocity $v$. Substituting $v$ into the equation and factoring it out yields:

Isolating $v$, we arrive at a beautiful expression for velocity in terms of position:

This is our master equation. It tells us exactly how fast the particle is moving at any given position $x$.

Deriving Retardation

To find the acceleration, we must differentiate our velocity expression with respect to time $t$.

Applying the power rule, the $-1$ exponent comes down, and the new exponent becomes $-2$. But we must not forget the chain rule! We multiply by the derivative of the inner function $(2mx+n)$ with respect to $t$.

The derivative of the inner function is $2m\frac{dx}{dt}$, which is simply $2mv$. Substituting this back, we get:

Now, look closely at the term $(2mx+n{)}^{-2}$. From our master equation, we know that $v=(2mx+n{)}^{-1}$. Therefore, squaring both sides gives $v^2=(2mx+n{)}^{-2}$.

Substituting $v^2$ into our acceleration equation, the magic happens:

The question asks for retardation, which is the magnitude of negative acceleration. Thus, we drop the minus sign.

For those who love kinematic shortcuts, there is another way! We know that acceleration can also be written as $a=v\frac{dv}{dx}$.

Starting from $v=(2mx+n{)}^{-1}$, we can differentiate directly with respect to $x$:

Multiplying this by $v$ gives the acceleration:

Both paths lead to the same elegant destination. The beauty of calculus lies in its consistency!