Animated Solution for Physics - Kinematics: Three points A, B, and C are on a straight horizontal line with equal distances between adjacent points. At an instant all the three points start moving, the point A begins to move vertically upwards with a constant velocity $u$ and the point C vertically downwards with a constant acceleration $a$ without any initial velocity. How should the point B move vertically so that the three points always remain collinear?

Imagine you are a choreographer directing three dancers: A, B, and C. They start on a straight stage, perfectly spaced out. B is exactly in the middle. Suddenly, the music starts, and they begin to move vertically. A floats upwards with a constant speed, while C drops downwards, accelerating as if in free fall. Your job as the choreographer is to tell B exactly how to move so that the three dancers always form a perfect, straight diagonal line in the air. How do you do it?

This beautiful kinematics problem is all about geometric constraints dictating physical motion. Let's break it down step by step.

Since the points start on a horizontal line with equal spacing, B is the horizontal midpoint of A and C. Because they only move vertically, their horizontal positions never change. Therefore, B will always be the horizontal midpoint.

For three points to remain collinear (on a straight line) when one is horizontally exactly in the middle, its vertical position must also be exactly in the middle. Mathematically, the y-coordinate of B must be the average of the y-coordinates of A and C at any given instant.

This simple equation is the master key to our problem. It bridges the gap between geometry and kinematics.

Now, let's look at the individual motions of A and C.

Point A moves upwards with a constant velocity $u$. Since there is no acceleration, its displacement equation is straightforward:

Point C, on the other hand, starts from rest and moves downwards with a constant acceleration $a$. Using the second equation of motion ($s=ut+\frac{1}{2}a{t}^2$), and taking upwards as positive, its displacement is:

We have the positions of A and C. Let's substitute them into our master geometric constraint to find out what B must do.

Let's split this fraction to see the individual components of B's motion:

To understand what this equation means physically, we need to compare it to the standard kinematic equation for constant acceleration:

Let's rewrite our equation for B to match this structure perfectly:

By comparing the terms, the story becomes crystal clear!

The term multiplied by $t$ represents the initial velocity. For point B, this is $\frac{u}{2}$. Since it's positive, B must start moving upwards with half the initial speed of A.

The term multiplied by $\frac{1}{2}{t}^2$ represents the constant acceleration. For point B, this is $-\frac{a}{2}$. The negative sign indicates it's directed downwards. So, B must have a constant downward acceleration equal to half of C's acceleration.

In physics, a constraint is a geometric or kinematic condition that restricts the possible motions of a system. Think of a bead sliding on a wire, or two blocks connected by a taut string. The wire and the string are constraints. They force the objects to move in very specific ways.

In our problem, the constraint is the invisible straight line connecting A, B, and C. This line isn't a physical rod, but a mathematical requirement. Learning to translate these English sentences ("the three points always remain collinear") into hard mathematical equations is one of the most crucial skills you can develop for exams like JEE and NEET.

Let's dive deeper into the math. We established that:

What if we differentiate this equation with respect to time? The derivative of position is velocity.

This tells us that B's velocity is always the average of A's and C's velocities. Let's find the initial velocities (at $t=0$):

$v_A(0)=u$ $v_C(0)=0$

Therefore, $v_B(0)=\frac{u+0}{2}=\frac{u}{2}$. This confirms our earlier finding!

Now, let's differentiate the velocity equation to find acceleration:

Let's plug in the known accelerations:

$a_A(t)=0$ (since A moves with constant velocity) $a_C(t)=-a$ (constant downward acceleration)

Therefore, $a_B(t)=\frac{0+(-a)}{2}=-\frac{a}{2}$.

This elegant calculus approach yields the exact same result and proves that B's acceleration is constant and directed downwards.

Whether you use the algebraic substitution method or the calculus approach, the physics remains beautifully consistent. Point B acts as a perfect mediator between A and C. It takes half of A's initial enthusiasm (velocity) and half of C's gravitational pull (acceleration) to maintain the perfect balance of the straight line. Next time you see a problem with objects moving together, don't just look at the forces. Look for the hidden geometry. Look for the constraints. They are the invisible strings pulling the puppets!