Animated Solution for Physics - Kinematics: The position of a particle as a function of time $t$, is given by $x(t)=at+b{t}^2-c{t}^3$ where $a,b$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be

The study of kinematics is essentially the study of motion's geometry. It is about understanding how position, velocity, and acceleration dance together over time. In this problem, we are given the position of a particle as a function of time, and we are asked to find its velocity at a very specific, special moment: the moment its acceleration becomes zero.

Imagine a particle moving along a straight line. Its position $x$ at any given time $t$ is dictated by a cubic equation:

This equation is the master blueprint of the particle's journey. It tells us exactly where the particle is at any tick of the clock. But to understand how fast it's going, or how its speed is changing, we need to dig deeper. We need to use the tools of calculus.

Velocity is the rate at which position changes. Mathematically, it is the first derivative of the position function with respect to time. Let's differentiate our position equation:

Applying the power rule, we get:

This new equation tells us the particle's velocity at any time $t$.

The problem asks for the velocity when the acceleration is zero. So, our next logical step is to find the acceleration. Acceleration is the rate of change of velocity, which means it is the derivative of the velocity function:

Differentiating again, we find:

Now, we are told that the particle attains zero acceleration. This is our crucial constraint. We set the acceleration equation to zero to find the exact time this event occurs:

Solving for $t$, we get:

This is the golden moment! At $t=\frac{b}{3c}$, the particle's acceleration vanishes.

We know when the acceleration is zero, but the question asks for the velocity at that specific time. To find this, we must substitute our golden time back into the velocity equation:

Let's carefully expand and simplify this expression:

Subtracting the terms, we arrive at our final, elegant answer:

Maximum Velocity

There is a beautiful physical interpretation hidden in this math. In calculus, when the derivative of a function is zero, the function is at a local maximum or minimum. Since acceleration is the derivative of velocity ($a=\frac{dv}{dt}$), setting acceleration to zero means we are finding the extremum of velocity.

If we take the second derivative of velocity (which is the derivative of acceleration, known as jerk), we get $-6c$. Assuming $c$ is positive, this second derivative is negative. A negative second derivative confirms that this point is a local maximum. Therefore, the velocity we just calculated isn't just any velocity; it is the maximum velocity the particle will ever achieve on its journey!