Following are four differrent relations about displacement, velocity and acceleration for the motion of a part

NCERT Exemplar · Q13

Q. Following are four differrent relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s) : (Note: more than one of the given options may be correct.)

(a)

\vec{v}_{av} = \dfrac{1}{2}\left[\vec{v}(t_1) + \vec{v}(t_2)\right]

(b)

\vec{v}_{av} = \dfrac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}

(c)

\vec{r} = \dfrac{1}{2}\left(\vec{v}(t_2) - \vec{v}(t_1)\right)(t_2 - t_1)

(d)

\vec{a}_{av} = \dfrac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}

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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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The average velocity is defined as total displacement divided by total time, and average acceleration is total change in velocity divided by total time. Relations (A) and (C) are generally incorrect because they do not hold for all types of motion, particularly when acceleration is not constant.

The incorrect relations are (A) and (C).

The motion of a particle is described by its position vector $\vec{r}(t)$ , velocity vector $\vec{v}(t)$ , and acceleration vector $\vec{a}(t)$ . These quantities are related through differentiation and integration. Understanding the precise definitions of average velocity and average acceleration is crucial for evaluating the given relations.

Understanding Average Velocity and Average Acceleration
- Instantaneous Velocity: The rate of change of position with respect to time, $\vec{v}(t) = \frac{d\vec{r}}{dt}$ .
- Instantaneous Acceleration: The rate of change of velocity with respect to time, $\vec{a}(t) = \frac{d\vec{v}}{dt}$ .
- Average Velocity ( $\vec{v}_{av}$ ): Defined as the total displacement ( $\Delta \vec{r}$ ) divided by the total time interval ( $\Delta t$ ). It is a vector quantity. $\vec{v}_{av} = \frac{\text{Total Displacement}}{\text{Total Time Interval}} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}$
- Average Acceleration ( $\vec{a}_{av}$ ): Defined as the total change in velocity ( $\Delta \vec{v}$ ) divided by the total time interval ( $\Delta t$ ). It is a vector quantity. $\vec{a}_{av} = \frac{\text{Total Change in Velocity}}{\text{Total Time Interval}} = \frac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$
These definitions are fundamental and hold true for any type of motion, regardless of whether acceleration is constant or varying.
Analyzing Option (A): $\vec{v}_{av} = \dfrac{1}{2}\left[\vec{v}(t_1) + \vec{v}(t_2)\right]$

This relation states that the average velocity is the arithmetic mean of the initial and final velocities.
- This statement is only true for motion with constant acceleration. When acceleration is constant, velocity changes linearly with time, and thus the average velocity over an interval is indeed the average of the velocities at the start and end of the interval.
- However, the question asks for relations that hold "in general" for the motion of a particle. In general, acceleration is not constant.
- Consider a case where acceleration is not constant. For example, if $\vec{v}(t) = t^2 \hat{i}$ $v (t) = t^{2} \hat{i}$ . Let $t_1 = 0$ $t_{1} = 0$ and $t_2 = 3$ $t_{2} = 3$ s.
  - $\vec{v}(t_1) = \vec{v}(0) = 0 \hat{i}$
  - $\vec{v}(t_2) = \vec{v}(3) = 3^2 \hat{i} = 9 \hat{i}$
  - RHS of (A): $\frac{1}{2}(0 \hat{i} + 9 \hat{i}) = 4.5 \hat{i}$ .
  - To find the actual average velocity, we first need the displacement: $\vec{r}(t) = \int \vec{v}(t) dt = \int t^2 \hat{i} dt = \frac{t^3}{3} \hat{i} + \vec{C}$ . Assuming $\vec{r}(0) = 0$ , then $\vec{C} = 0$ . So $\vec{r}(t) = \frac{t^3}{3} \hat{i}$ . Displacement $\Delta \vec{r} = \vec{r}(t_2) - \vec{r}(t_1) = \vec{r}(3) - \vec{r}(0) = \frac{3^3}{3} \hat{i} - 0 \hat{i} = 9 \hat{i}$ . Time interval $\Delta t = t_2 - t_1 = 3 - 0 = 3$ s. Actual average velocity (LHS of (A) by definition): $\vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{9 \hat{i}}{3} = 3 \hat{i}$ .
  - Since $4.5 \hat{i} \neq 3 \hat{i}$ , relation (A) is incorrect for general motion.
Analyzing Option (B): $\vec{v}_{av} = \dfrac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}$

This is precisely the definition of average velocity. It represents the total displacement ( $\vec{r}(t_2) - \vec{r}(t_1)$ ) divided by the total time taken ( $t_2 - t_1$ ).
- This relation is always correct by definition.
Analyzing Option (C): $\vec{r} = \dfrac{1}{2}\left(\vec{v}(t_2) - \vec{v}(t_1)\right)(t_2 - t_1)$

Let's analyze this relation. …

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Concept: Recognize the Trapezoidal Rule Hiding Inside Two of the Four Relations

Method: Numerical-Integration Lens -- (A) and (C) Are the Trapezoidal Rule, Exact Only for Constant Acceleration

Rather than testing one specific velocity function to find a numerical counterexample, this method recognizes that options (A) and (C) are both disguised instances of the trapezoidal rule for estimating an integral from its two endpoint values -- a rule from numerical calculus that is exact only when the integrand varies linearly. Since velocity is linear in time exactly when acceleration is constant, this single insight explains, in one stroke, why both (A) and (C) fail in general.

The general definitions (always true, by construction)

\vec v_{av} = \frac{\Delta \vec r}{\Delta t} = \frac{1}{t_2-t_1}\int_{t_1}^{t_2}\vec v(t)\,dt, \qquad \vec a_{av} = \frac{\Delta \vec v}{\Delta t}

These are (B) and (D) respectively -- both are just the definitions of "average" as (change) / (time), true for any motion whatsoever. (B) and (D) are correct, with nothing further to check.

Recognizing (A) as the trapezoidal rule

The trapezoidal rule approximates $\displaystyle\int_{t_1}^{t_2} f(t)\,dt$ by the area of a trapezoid with parallel sides $f(t_1)$ and $f(t_2)$ :

\int_{t_1}^{t_2} f(t)\,dt \approx \frac{1}{2}\big[f(t_1)+f(t_2)\big](t_2-t_1)

This approximation is exact if and only if $f(t)$ is a linear (straight-line) function of $t$ -- a trapezoid's straight top edge matches a straight-line graph perfectly, but not a curved one.

Now apply this with $f(t)=\vec v(t)$ : option (A) claims

\vec v_{av} = \frac{1}{2}\big[\vec v(t_1)+\vec v(t_2)\big]

which is exactly the trapezoidal-rule estimate of $\vec v_{av}=\frac{1}{\Delta t}\int\vec v\,dt$ using only the two endpoint velocities. By the exactness condition above, this equals the true average velocity only when $\vec v(t)$ is linear in $t$ -- i.e., only under constant acceleration. For general (nonlinear) $\vec v(t)$ , it is merely an approximation, not an identity. (A) is not generally correct.

Recognizing (C) as the same rule, one level down

Option (C) claims

$\vec r = \frac12\big(\vec v(t_2)-\vec v(t_1)\big)(t_2-t_1)$ …

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Q.Following are four differrent relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s) : (Note: more than one of the given options may be correct.)

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