Animated Solution for Physics - Kinematics: The relation between time $t$ and distance $x$ is $t=a{x}^2+bx$, where $a$ and $b$ are constants. The acceleration is

In standard kinematics problems, we are usually given the position of a particle as a function of time, $x(t)$. Finding velocity and acceleration is a straightforward process of differentiating with respect to time. However, this problem flips the script. We are given time as a function of position:

This might seem intimidating at first glance. One might be tempted to solve this quadratic equation for $x$ in terms of $t$ using the quadratic formula. While mathematically valid, differentiating a complex square root expression twice is a recipe for algebraic nightmares and silly mistakes. Instead, we can use the elegant power of implicit differentiation.

Uncovering Velocity

Our goal is to find the acceleration $f$, which is the rate of change of velocity $v$. Therefore, our first milestone is to find an expression for $v$. We know that velocity is defined as $v=\frac{dx}{dt}$.

Let's differentiate our given equation implicitly with respect to time $t$:

The left side is simply $1$. On the right side, we must apply the chain rule because we are differentiating a function of $x$ with respect to $t$.

Now, we substitute our definition of velocity, $v=\frac{dx}{dt}$:

Rearranging this to solve for $v$, we get:

We now have a clean expression for velocity!

The Path to Acceleration

Acceleration $f$ is the derivative of velocity with respect to time, $f=\frac{dv}{dt}$. We will differentiate our velocity expression, again keeping a close eye on the chain rule.

Applying the power rule and the chain rule:

The derivative of the inner term $(2ax+b)$ with respect to $t$ is $2a\cdot \frac{dx}{dt}$. Substituting this back:

Bringing it all Together

We are almost there. The problem asks for the acceleration in terms of velocity $v$. We need to eliminate the $x$ and $\frac{dx}{dt}$ terms from our acceleration expression.

Recall our earlier findings: 1. $\frac{dx}{dt}=v$ 2. $v=(2ax+b{)}^{-1}$, which means squaring both sides gives $v^2=(2ax+b{)}^{-2}$.

Let's substitute these directly into our acceleration equation:

Multiplying the terms together yields our final, elegant result:

The negative sign is physically significant. Assuming $a$ and $v$ are positive, the negative acceleration indicates that the particle is experiencing retardation. Furthermore, this retardation is not constant; it is proportional to the cube of the particle's velocity. As the particle speeds up, the resistive force acting on it increases dramatically!