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

NCERT Exemplar · Q7

v_0

(a) The average velocity is not zero at any time.

(b) Average acceleration must always vanish.

(d) Equal path lengths are traversed in equal intervals.

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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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Constant instantaneous speed means the particle moves at a fixed rate along its path, but direction can change. Only the statement about equal path lengths in equal times follows necessarily; the others fail when the particle turns or loops back.

When a particle moves in two dimensions with constant instantaneous speed $v_0$ , we know that at every moment $|\vec{v}(t)| = v_0$ . The magnitude is fixed, but the velocity vector $\vec{v}(t)$ can rotate freely—the particle can curve, spiral, or even reverse direction.

The key distinction here is between speed (a scalar, the magnitude of velocity) and velocity (a vector). Constant speed does not mean constant velocity.

Let me examine each statement by asking what it demands and whether constant speed guarantees it.

Analysis of each option

1. Statement (A): The average velocity is not zero at any time.

Average velocity over a time interval $[0, t]$ is $\vec{v}_{\text{avg}} = \frac{\vec{r}(t) - \vec{r}(0)}{t} = \frac{\Delta \vec{r}}{t}$ .

Consider uniform circular motion: a particle moves around a circle at constant speed $v_0$ . After one complete revolution (time $T = 2\pi R/v_0$ ), the particle returns to its starting point. The displacement $\Delta \vec{r} = \vec{0}$ , so $\vec{v}_{\text{avg}} = \vec{0}$ .

This counterexample shows (A) is false.

Watch out

Constant speed does not prevent a particle from returning to its starting point. Any closed path traversed at constant speed will yield zero average velocity over one complete loop.

2. Statement (B): Average acceleration must always vanish.

Average acceleration over $[0, t]$ is $\vec{a}_{\text{avg}} = \frac{\vec{v}(t) - \vec{v}(0)}{t}$ .

Again, consider uniform circular motion. At $t=0$ , suppose $\vec{v}(0) = v_0 \hat{i}$ . After a quarter circle (time $t = \pi R/(2v_0)$ ), $\vec{v}(t) = v_0 \hat{j}$ . Both have magnitude $v_0$ , but they point in different directions.

The change in velocity is $\Delta \vec{v} = v_0 \hat{j} - v_0 \hat{i} \neq \vec{0}$ , so $\vec{a}_{\text{avg}} \neq \vec{0}$ .

Even though the speed is constant, the direction changes, which requires acceleration. Statement (B) is false.

Note

Acceleration arises whenever velocity changes—either in magnitude or direction. Constant speed eliminates only the tangential component of acceleration, but the normal (centripetal) component can be nonzero whenever the path curves.

3. Statement (C): Displacements in equal time intervals are equal.

Displacement in a time interval $\Delta t$ is $\Delta \vec{r} = \vec{r}(t + \Delta t) - \vec{r}(t)$ .

For displacements to be equal in successive intervals, we would need $\vec{r}(t_1 + \Delta t) - \vec{r}(t_1) = \vec{r}(t_2 + \Delta t) - \vec{r}(t_2)$ for any $t_1, t_2$ . …

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Concept: A Simple "There-and-Back" Path Beats Circular Motion as the Test Case

Method: A Piecewise-Linear Counterexample (No Trigonometry Needed) + a One-Line Integral Proof for (D)

Rather than using uniform circular motion (which needs $\sin$ , $\cos$ , and a period $T=2\pi R/v_0$ ) to falsify (A), (B), (C), this method uses the simplest possible non-straight constant-speed motion: a particle that walks out along a line, then walks straight back the same way, at the same constant speed $v_0$ throughout. No trigonometric functions are needed anywhere, and (D) is settled by the Fundamental Theorem of Calculus applied directly to speed.

The counterexample: "there and back"

A particle starts at the origin, moves along the positive $x$ -axis to $x=d$ at constant speed $v_0$ (taking time $t_1=d/v_0$ ), instantaneously reverses direction, and walks straight back to the origin at the same speed $v_0$ (taking another $t_1=d/v_0$ ). Total elapsed time $2t_1$ . This is a valid 2D motion (a degenerate "V"-shaped path) with constant instantaneous speed $|\vec v(t)|=v_0$ throughout -- exactly the hypothesis of the question -- and needs no circle at all.

Testing (A): average velocity is not zero at any time

Over the full round trip, the net displacement is $\vec r(2t_1)-\vec r(0) = \vec 0$ (back at the start), so

\vec v_{\text{avg}} = \frac{\vec 0}{2t_1} = \vec 0.

(A) is false -- falsified by the simplest possible closed path.

Testing (B): average acceleration must always vanish

Just before the turnaround, $\vec v = v_0\hat i$ ; just after, $\vec v = -v_0\hat i$ . Over any interval straddling the turn,

\vec a_{\text{avg}} = \frac{(-v_0\hat i) - (v_0\hat i)}{\Delta t} = \frac{-2v_0\hat i}{\Delta t} \neq \vec 0.

(B) is false -- a single instantaneous reversal is enough, no smooth curving path is even required.

Testing (C): displacements in equal time intervals are equal

Take two equal-length intervals, one entirely on the outbound leg and one entirely on the return leg. The first has displacement $+v_0\Delta t\,\hat i$ ; the second has displacement $-v_0\Delta t\,\hat i$ -- equal in size but opposite as vectors. (C) is false.

Testing (D): equal path lengths in equal time intervals -- proved for any path, not just this one …

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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?

Analysis of each option

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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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