Animated Solution for Physics - Kinematics: A particle is projected with velocity $v_0$ along X-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin, i.e. $ma=-\alpha x^2$. The distance at which the particle stops is

Imagine a particle zooming along the X-axis. It starts its journey from the origin with an initial velocity $v_0$. But the universe isn't letting it go that easily! A damping force is actively trying to slow it down.

The interesting part is that this force is not constant. It gets stronger as the square of the distance from the origin. Mathematically, we can write Newton's second law for this particle as $ma=-\alpha x^2$. The negative sign perfectly captures the essence of a damping force—it always acts opposite to the direction of motion.

To find the stopping distance, we need to connect the particle's velocity with its position. Time is not our primary concern here. Do you remember the golden rule of kinematics for variable acceleration?

We can express acceleration $a$ as $v\frac{dv}{dx}$. This is a powerful substitution! Let's plug this into our force equation. We get $m\left(v\frac{dv}{dx}\right)=-\alpha x^2$.

Now, we have a differential equation involving only velocity $v$ and position $x$. The next logical step is to separate the variables. We keep all the velocity terms on the left side and move the position terms to the right side. This yields $vdv=-\frac{\alpha }{m}{x}^{2}dx$.

It's time to integrate both sides. But we must be very careful with our limits. The particle starts at the origin ($x=0$) with an initial velocity $v_0$. It finally stops at some unknown distance $x$, which means its final velocity becomes $0$.

Setting up the definite integrals, we have:

Integrating $v$ gives $\frac{v^2}{2}$, and integrating $x^2$ gives $\frac{x^3}{3}$. Applying the upper and lower limits, we get:

Notice how the negative signs on both sides cancel out beautifully. Now, it is just basic algebra. We rearrange the terms to isolate $x^3$:

Finally, taking the cube root on both sides, we arrive at our stopping distance:

This matches option (d). Always pay close attention to the functional dependence of the force. If the damping force was proportional to velocity instead, the integration would have involved a natural logarithm!