Animated Solution for Physics - Kinematics: Particle A moves with a constant speed $2\text{ m/s}$ on a circular path of radius $4\text{ m}$ whereas particle B moves along a straight line through the centre of the circular path always maintaining a constant distance $4\text{ m}$ from the particle A. Which of the following conclusions can be drawn?

Imagine you are standing at the center of a giant circular track. Particle A is running along this track at a constant speed, while Particle B is restricted to moving back and forth along a straight line passing right through where you are standing. The catch? The distance between A and B is always exactly equal to the radius of the track.

At first glance, finding the exact motion of B seems like a nightmare of algebraic constraints. But physics rewards those who look at the geometry. Let's draw a line from the origin $O$ to particle $A$, and another line from $A$ to particle $B$.

What do we see? The distance $OA$ is the radius $R$. The problem states the distance $AB$ is also $R$. This means $△OAB$ is an isosceles triangle!

Because $△OAB$ is isosceles, if we drop a perpendicular from $A$ down to the straight line (the $x$-axis), it will perfectly bisect the base $OB$. Let's call this midpoint $M$.

If particle A is at an angle $\theta =\omega t$, the $x$-coordinate of this midpoint $M$ is simply $R\cos (\omega t)$. Since $M$ is exactly halfway to $B$, the position of $B$ is just twice that distance:

Suddenly, the complex constraint collapses into a beautiful, simple equation. Particle B is executing Simple Harmonic Motion (SHM)!

Now, we can easily find its velocity and acceleration by differentiating with respect to time.

Given $R=4\text{ m}$ and $v_A=R\omega =2\text{ m/s}$, we find $\omega =0.5\text{ rad/s}$. Substituting these values, the maximum speed of B is $4\text{ m/s}$ and its maximum acceleration is $2{\text{ m/s}}^2$.

During one full revolution of A, particle B completes one full oscillation, traveling from $8\text{ m}$ to $-8\text{ m}$ and back, covering a total distance of $4\times 8=32\text{ m}$.

What about the relative velocity? We could subtract their velocity vectors, but there is a profound physical shortcut.

Think about the motion from the perspective of particle B. In B's frame of reference, particle A is always at a constant distance $R$. If a particle is always at a constant distance from you, it must be moving in a circle around you!

For an object moving in a circle, its velocity is purely tangential, and its speed is constant if the angular rate is constant. The math confirms this intuition perfectly. The magnitude of the relative velocity is strictly $R\omega =2\text{ m/s}$.

Every single option provided in the question is a beautiful consequence of this geometric dance.