For any arbitrary motion in space, which of the following relations are true: (a) vecvaverage = (1)/(2)(vecv(t

Exercises · 3.20

Q. For any arbitrary motion in space, which of the following relations are true:

(a)

\vec{v}_{average} = \frac{1}{2}\,(\vec{v}(t_1) + \vec{v}(t_2))

(b)

\vec{v}_{average} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}

(c)

\vec{v}(t) = \vec{v}(0) + \vec{a}\,t

(d)

\vec{r}(t) = \vec{r}(0) + \vec{v}(0)\,t + \frac{1}{2}\,\vec{a}\,t^{2}

(e)

\vec{a}_{average} = \frac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}

. (The ‘average’ stands for average of the quantity over the time interval

t_1

t_2

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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 arbitrary (general) motion, only the definitional relations hold: (b) and (e). The others assume constant acceleration or uniform motion, which is not guaranteed for arbitrary motion.

The key is to separate definitions from formulas that assume a specific type of motion. In kinematics, average velocity and average acceleration are defined in terms of the net change in position or velocity over the total time. The other relations — like the arithmetic mean of velocities, or the equations of motion — are only true under special conditions (like constant acceleration).

Let’s examine each statement one by one.

Statement (a): $\vec{v}_{average} = \frac{1}{2}(\vec{v}(t_1) + \vec{v}(t_2))$

This would be true only if the velocity changes linearly with time (i.e., constant acceleration). For arbitrary motion, velocity can vary in any complicated way — speeding up, slowing down, changing direction — so the average is not simply the arithmetic mean of the endpoints.

Watch out

A common mistake is to assume average velocity is always the mean of initial and final velocities. That is only true for uniform acceleration.

False for arbitrary motion.
Statement (b): $\vec{v}_{average} = \frac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}$

This is the definition of average velocity. It always holds, regardless of the path or how the motion varies.
$\vec{v}_{\text{average}} = \frac{\Delta \vec{r}}{\Delta t}$
True for any motion.
Statement (c): $\vec{v}(t) = \vec{v}(0) + \vec{a}\,t$

This is the first equation of motion for constant acceleration. For arbitrary motion, acceleration $\vec{a}$ itself may change with time, so this linear relation does not hold in general.

Note

If acceleration is not constant, you would need $\vec{v}(t) = \vec{v}(0) + \int_0^t \vec{a}(\tau)\,d\tau$ , not a simple product.

False for arbitrary motion. …

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Concept: Disprove the False Ones with One Concrete Non-Uniform Motion

Method: A Single Explicit Counterexample ( $\vec r(t)=t^3\hat i$ , So Acceleration Genuinely Varies) — Numbers, Not General Argument

The existing solutions argue each statement in words (definitions always hold; constant-acceleration formulas don't hold in general). This method instead picks one explicit, concrete non-uniformly-accelerated motion, computes both sides of every relation with real numbers for that motion, and shows directly — by arithmetic mismatch — exactly which relations break down.

Step 1 — Choose a concrete arbitrary (non-uniformly-accelerated) motion

Let $\vec r(t) = t^3\,\hat i$ (metres, $t$ in seconds) — deliberately chosen so the acceleration is not constant:

\vec v(t) = \frac{d\vec r}{dt} = 3t^2\,\hat i, \qquad \vec a(t) = \frac{d\vec v}{dt} = 6t\,\hat i \quad(\text{genuinely varies with } t)

Step 2 — Evaluate everything at $t_1=0$ and $t_2=2\ \text{s}$

\vec r(0)=\vec 0, \quad \vec r(2)=8\,\hat i; \qquad \vec v(0)=\vec 0, \quad \vec v(2)=12\,\hat i; \qquad \vec a(0)=\vec 0, \quad \vec a(2)=12\,\hat i

Step 3 — Test (b): the definition of average velocity

\vec v_{\text{avg}} = \frac{\vec r(2)-\vec r(0)}{2-0} = \frac{8\,\hat i}{2} = 4\,\hat i\ \text{m/s}

This is definitionally what "average velocity" means — it cannot fail, for any motion. (b) holds.

Step 4 — Test (a): is the average of the endpoint velocities the same number?

\frac{1}{2}\big(\vec v(0)+\vec v(2)\big) = \frac{1}{2}(0+12\,\hat i) = 6\,\hat i\ \text{m/s}

Compare to the true average velocity from Step 3: $6\,\hat i \ne 4\,\hat i$ . Numeric mismatch — (a) is false for this arbitrary motion.

Step 5 — Test (c): does $\vec v(t)=\vec v(0)+\vec a t$ hold, using the actual (non-constant) $\vec a$ ?

Using the initial acceleration $\vec a(0)=\vec 0$ as " $\vec a$ " in the constant-acceleration formula:

\vec v(2) \overset{?}{=} \vec v(0)+\vec a(0)(2) = \vec 0+\vec 0(2) = \vec 0

But the actual $\vec v(2) = 12\,\hat i \ne \vec 0$ . Numeric mismatch — (c) is false.

Step 6 — Test (d): does $\vec r(t)=\vec r(0)+\vec v(0)t+\tfrac12\vec a t^2$ hold?

\vec r(2) \overset{?}{=} \vec 0+\vec 0(2)+\tfrac12(\vec 0)(2)^2 = \vec 0

But the actual $\vec r(2)=8\,\hat i \ne \vec 0$ . Numeric mismatch — (d) is false. …

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Q.For any arbitrary motion in space, which of the following relations are true:

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