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

NCERT Exemplar · Q8

v_0

(a) The acceleration of the particle is zero.

(b) The acceleration of the particle is bounded.

(d) The particle must be undergoing a uniform circular motion

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Concept understanding — Kinematics Vector Differentiation

Kinematics Vector Differentiation

Imagine you're tracking a drone flying in the sky. At any instant, it has a position — say, 30 metres east and 40 metres north of you. That's a vector: $\vec{r} = 30\hat{i} + 40\hat{j}$ . A second later, it's moved. The question kinematics asks is: how fast is that position changing? That rate of change is velocity, and to get it, you differentiate the position vector.

But here's the key difference from school calculus: in school, you differentiated a scalar function like $y = x^2$ . Here, you're differentiating a vector function — something that has both magnitude and direction, and both can change with time.

The Intuition First

Think of a vector as an arrow. When time passes, that arrow can do two things:

It can get longer or shorter (magnitude changes).
It can rotate (direction changes).

Velocity is the total rate of change of that arrow. If the drone flies straight away from you, only the length changes. If it flies in a circle around you, only the direction changes. Most real motion does both.

So vector differentiation is just: take the derivative of each component separately, because components are independent scalars.

The Precise Statement

If a position vector is written in Cartesian coordinates as:

\vec{r}(t) = x(t)\,\hat{i} + y(t)\,\hat{j} + z(t)\,\hat{k}

where $\hat{i}, \hat{j}, \hat{k}$ are fixed unit vectors (they don't change direction with time), then:

\frac{d\vec{r}}{dt} = \frac{dx}{dt}\,\hat{i} + \frac{dy}{dt}\,\hat{j} + \frac{dz}{dt}\,\hat{k}

That's it. You differentiate each component function $x(t), y(t), z(t)$ exactly as you would in single-variable calculus, and the unit vectors stay put.

\frac{d}{dt}\big(f(t)\,\hat{u}\big) = \frac{df}{dt}\,\hat{u} \quad\text{(if $\hat{u}$ is constant)}

Why This Works

The derivative of a vector is defined the same way as for a scalar — as a limit:

\frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}

The numerator is a vector difference. When you write $\vec{r}$ in components, the difference splits into component differences. The limit then acts on each component separately because the unit vectors are constant. So the definition forces component-wise differentiation.

A Concrete Example

A particle moves such that:

\vec{r}(t) = (3t^2)\,\hat{i} + (5\sin t)\,\hat{j} + (2e^{-t})\,\hat{k}

Its velocity is:

\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t)\,\hat{i} + (5\cos t)\,\hat{j} + (-2e^{-t})\,\hat{k}

Notice: the $x$ -component grows linearly, the $y$ -component oscillates, the $z$ -component decays. Each derivative is just the ordinary derivative of that component's function.

The One Trap: Non-Constant Unit Vectors

The rule above assumes $\hat{i}, \hat{j}, \hat{k}$ are fixed. That's true in Cartesian coordinates. But in polar coordinates, the unit vectors $\hat{r}$ and $\hat{\theta}$ rotate as the particle moves. Differentiating a vector in polar coordinates requires the product rule because the unit vectors themselves depend on time. …

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For a particle moving in two dimensions with constant instantaneous speed, the acceleration is always perpendicular to velocity and lies in the plane of motion, but it need not be constant in magnitude or direction — so only option (C) is necessarily true.

The key idea is that instantaneous speed is the magnitude of the velocity vector. When that magnitude stays constant, the velocity vector can only change direction, not length. That change in direction is what acceleration does — and it must happen perpendicular to the velocity, within the plane where the motion actually occurs.

Let’s work through each option carefully.

Option (A): The acceleration of the particle is zero.

If speed is constant but the direction of motion changes, the velocity vector changes. Any change in velocity means acceleration is present. For example, a particle moving on a circular path at constant speed has nonzero centripetal acceleration. So (A) is false.
Option (B): The acceleration of the particle is bounded.

“Bounded” means there is some finite upper limit on the magnitude of acceleration. Is that guaranteed? Not at all. Consider a particle moving along a path with a very sharp turn — the radius of curvature can be arbitrarily small. Since centripetal acceleration is $v_0^2 / R$ , as $R \to 0$ the acceleration blows up without bound. So (B) is not necessarily true.
Option (C): The acceleration of the particle is necessarily in the plane of motion.

The motion is two-dimensional, so the entire trajectory lies in some plane. The velocity vector always lies in that plane. Acceleration is the time derivative of velocity — if velocity stays in a fixed plane, its derivative also lies in that same plane. (A vector changing within a plane cannot develop a component perpendicular to that plane.) So (C) is necessarily true.
Option (D): The particle must be undergoing uniform circular motion. …

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Concept: The Frenet Frame -- Splitting Acceleration into a Tangential and a Normal Part

Method: Frenet-Serret Decomposition ( $\vec a = \dot v\,\hat T + \dfrac{v^2}{\rho}\,\hat N$ ), Not a Case-by-Case Argument

Rather than reasoning about each option with a different ad-hoc argument, this method invokes one general result from the differential geometry of curves -- valid for any smooth path, not just a circle -- and reads all four verdicts off it at once.

The general result

For a particle moving along any smooth curve, its acceleration always decomposes uniquely into a component along the unit tangent $\hat T$ (which changes the speed) and a component along the unit normal $\hat N$ (which changes the direction, and always lies in the plane the curve is locally curving in):

\vec a = \frac{dv}{dt}\,\hat T + \frac{v^2}{\rho}\,\hat N

where $v=|\vec v|$ is the instantaneous speed and $\rho$ is the local radius of curvature of the path at that instant (small $\rho$ = sharp turn, large $\rho$ = gentle turn, $\rho\to\infty$ = straight line).

Applying it to constant-speed 2D motion

The tangential term vanishes identically. Since $v=v_0$ is constant for all $t$ , $\dfrac{dv}{dt}=0$ at every instant, for any shape of path. Only the normal (curving) term survives:

\vec a = \frac{v_0^2}{\rho(t)}\,\hat N(t)

Option (A): is $\vec a=0$ ? Only if $\rho\to\infty$ everywhere, i.e. the path is a straight line. The problem says nothing about the path being straight -- for any genuinely curved path, $\vec a\neq0$ . (A) is false, since it is not necessarily true.
Option (B): is $|\vec a|$ bounded? $|\vec a| = v_0^2/\rho(t)$ . Nothing in "constant speed" restricts how sharply the particle can turn -- $\rho(t)$ can be made arbitrarily small (an arbitrarily sharp kink in the path) while $v_0$ stays fixed, sending $|\vec a|\to\infty$ . (B) is false.
Option (C): does $\vec a$ stay in the plane of motion? For a curve confined to a single 2D plane, both $\hat T(t)$ and $\hat N(t)$ are defined to lie in that same plane at every point (the normal to a plane curve, by construction, points within the plane, toward the centre of curvature) -- the Frenet formula never introduces a third, out-of-plane direction for genuinely planar motion. So $\vec a = \frac{v_0^2}{\rho}\hat N$ is a vector lying entirely in the plane of motion. (C) is necessarily true. …

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Q.In a two dimensional motion, instantaneous speed v0v_0v0​ is a positive constant. Then which of the following are necessarily true?

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Q.In a two dimensional motion, instantaneous speed $v_0$ is a positive constant. Then which of the following are necessarily true?

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