In a two dimensional motion, instantaneous speed v₀ is a positive constant. Then which of the following are necessarily true?

Correct Answer: The acceleration of the particle is necessarily in the plane of motion.

Let us analyze the motion of the particle in two dimensions step-by-step:

Step 1: Understand the plane of motion
In a two-dimensional motion, the trajectory of the particle is entirely confined to a single plane. Let us define this plane as the $xy$-plane. Therefore, the position vector of the particle at any instant of time $t$ can be represented as:
$\mathbf{r}\left(t\right)=x\left(t\right)\hat{\mathbf{i}}+y\left(t\right)\hat{\mathbf{j}}$
where $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along the $x$ and $y$ axes, which define the plane of motion.

Step 2: Relate velocity and acceleration to the plane
The velocity vector $\mathbf{v}\left(t\right)$ is the time derivative of the position vector:
$\mathbf{v}\left(t\right)=\frac{d\mathbf{r}}{dt}=\frac{dx}{dt}\hat{\mathbf{i}}+\frac{dy}{dt}\hat{\mathbf{j}}$
Since both unit vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ lie in the $xy$-plane, the velocity vector $\mathbf{v}$ always lies in the plane of motion.

The acceleration vector $\mathbf{a}\left(t\right)$ is the time derivative of the velocity vector:
$\mathbf{a}\left(t\right)=\frac{d\mathbf{v}}{dt}=\frac{d^{2}x}{d{t}^2}\hat{\mathbf{i}}+\frac{d^{2}y}{d{t}^2}\hat{\mathbf{j}}$
Because the components of acceleration are entirely along $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, the acceleration vector $\mathbf{a}$ must also lie completely in the same $xy$-plane. Thus, the acceleration of the particle is necessarily in the plane of motion.

Step 3: Analyze why the other options are not necessarily true
1. The acceleration of the particle is zero: This is incorrect because even though the speed $v_0$ is constant, the direction of the velocity vector can change over time (e.g., in circular motion), resulting in a non-zero centripetal acceleration.
2. The acceleration of the particle is bounded: This is incorrect because the trajectory can have points of extremely high curvature (where the radius of curvature $R$ approaches zero), meaning the centripetal acceleration $a_c=\frac{{v_0}^2}{R}$ can grow without bound.
3. The particle must be undergoing a uniform circular motion: This is incorrect because the particle can follow any arbitrary two-dimensional curved path with constant speed, not just a circle.