The position of a particle is given by $\vec{r} = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}\ \text{m}$ where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres. (a) Find the $\vec{v}$ and $\vec{a}$ of the particle? (b) What is the magnitude and direction of velocity of the particle at $t = 2.0\ \text{s}$?

The position of a particle is given by vecr = 3.0t – 1st PUC Physics

Exercises · 3.17

\vec{r} = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}\ \text{m}

(a) Find the

\vec{v}

and

\vec{a}

of the particle?

(b) What is the magnitude and direction of velocity of the particle at

t = 2.0\ \text{s}

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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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Velocity is the time derivative of position, and acceleration is the derivative of velocity. For $\vec{r} = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}$ , we get $\vec{v} = 3.0\,\hat{i} - 4.0t\,\hat{j}$ m/s and $\vec{a} = -4.0\,\hat{j}$ m/s². At $t = 2.0$ s, $\vec{v} = 3.0\,\hat{i} - 8.0\,\hat{j}$ m/s, with magnitude $8.54$ m/s and direction $69.4^\circ$ below the $+x$ axis.

This is a straightforward application of kinematics vector differentiation. In physics, when position is given as a vector function of time, velocity and acceleration are simply its first and second time derivatives — component by component. No chain rule tricks, no product rule; each coordinate is independent.

The key insight: differentiate each component separately, treating $\hat{i}$ , $\hat{j}$ , $\hat{k}$ as constant unit vectors. The $t$ in the $x$ -component is linear, the $t^2$ in the $y$ -component is quadratic, and the $z$ -component is constant — so its derivative is zero.

Step-by-step solution

1. Write down the position vector clearly

\vec{r}(t) = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k} \quad \text{(metres)}

All coefficients already have the right units: $3.0$ is m/s, $-2.0$ is m/s², $4.0$ is m.

2. Find velocity $\vec{v}$ by differentiating $\vec{r}$ with respect to $t$

\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.0t)\,\hat{i} + \frac{d}{dt}(-2.0t^{2})\,\hat{j} + \frac{d}{dt}(4.0)\,\hat{k}

$x$ -component: $\frac{d}{dt}(3.0t) = 3.0$ m/s
$y$ -component: $\frac{d}{dt}(-2.0t^{2}) = -4.0t$ m/s
$z$ -component: $\frac{d}{dt}(4.0) = 0$

So:

\vec{v}(t) = 3.0\,\hat{i} - 4.0t\,\hat{j} \quad \text{m/s}

Note

The $z$ -component of velocity is zero at all times — the particle never moves in the $k$ direction.

3. Find acceleration $\vec{a}$ by differentiating $\vec{v}$

\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3.0)\,\hat{i} + \frac{d}{dt}(-4.0t)\,\hat{j}

$x$ -component: derivative of constant $3.0$ is $0$
$y$ -component: $\frac{d}{dt}(-4.0t) = -4.0$ m/s²

Thus:

\vec{a}(t) = -4.0\,\hat{j} \quad \text{m/s}^2

\vec{v}(t) = 3.0\,\hat{i} - 4.0t\,\hat{j} \quad \text{m/s}, \qquad \vec{a}(t) = -4.0\,\hat{j} \quad \text{m/s}^2

4. Evaluate velocity at $t = 2.0$ s

Substitute $t = 2.0$ into $\vec{v}(t)$ :

\vec{v}(2.0) = 3.0\,\hat{i} - 4.0(2.0)\,\hat{j} = 3.0\,\hat{i} - 8.0\,\hat{j} \quad \text{m/s}

5. Find magnitude of this velocity …

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Concept: Confirm the Derivative Numerically, via a Shrinking- $\Delta t$ Table

Method: Finite-Difference Convergence (Compute $\Delta\vec r/\Delta t$ at Several Small $\Delta t$ and Watch It Converge), Not Symbolic Differentiation

Both existing solutions differentiate $\vec r(t)$ symbolically, term by term. This method instead demonstrates the same result numerically: it evaluates $\vec r(t)$ at $t=2.0\ \text{s}$ and at several nearby times, computes the average rate of change $\Delta\vec r/\Delta t$ for each, and shows the sequence of numbers visibly closing in on the values the symbolic derivative would give — making concrete exactly what "the derivative at $t=2$ " means as a limit.

Step 1 — The position vector and its value at $t=2.0\ \text{s}$

\vec r(t) = 3.0t\,\hat i - 2.0t^2\,\hat j + 4.0\,\hat k \quad\text{m}, \qquad \vec r(2.0) = 6.0\,\hat i - 8.0\,\hat j + 4.0\,\hat k\ \text{m}

Step 2 — Compute $\dfrac{\Delta\vec r}{\Delta t}$ for shrinking $\Delta t$ , centred at $t=2.0\ \text{s}$

$\Delta t = 0.1\ \text{s}$ : $\vec r(2.1) = 6.3\,\hat i - 2.0(2.1)^2\,\hat j + 4.0\,\hat k = 6.3\,\hat i - 8.82\,\hat j + 4.0\,\hat k$ .

\Delta \vec r = (6.3-6.0)\hat i + (-8.82-(-8.0))\hat j = 0.30\,\hat i - 0.82\,\hat j \quad\Longrightarrow\quad \frac{\Delta\vec r}{\Delta t} = 3.0\,\hat i - 8.20\,\hat j

$\Delta t = 0.01\ \text{s}$ : $\vec r(2.01) = 6.03\,\hat i - 2.0(2.01)^2\,\hat j + 4.0\,\hat k = 6.03\,\hat i - 8.0802\,\hat j+4.0\,\hat k$ .

\frac{\Delta\vec r}{\Delta t} = \frac{0.03\,\hat i - 0.0802\,\hat j}{0.01} = 3.0\,\hat i - 8.02\,\hat j

$\Delta t = 0.001\ \text{s}$ : repeating the same computation gives $\dfrac{\Delta \vec r}{\Delta t} = 3.0\,\hat i - 8.002\,\hat j$ .

The $\hat i$ -component is exactly $3.0$ at every step (the $x$ -motion is already linear in $t$ , so no limiting is even needed there); the $\hat j$ -component visibly closes in on $-8.0$ as $\Delta t$ shrinks: $-8.20 \to -8.02 \to -8.002 \to \dots \to -8.0$ .

Step 3 — Read off the velocity as the limiting value

\vec v(2.0) = \lim_{\Delta t \to 0}\frac{\Delta\vec r}{\Delta t} = 3.0\,\hat i - 8.0\,\hat j\ \text{m/s}

This matches the symbolic derivative $\vec v(t) = 3.0\,\hat i - 4.0t\,\hat j$ evaluated at $t=2.0$ , but was found here purely from a sequence of shrinking numerical estimates.

Step 4 — Same check for acceleration …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$y=9 x^2$
›Reveal solutionSolution
The key idea is that constant horizontal velocity means zero horizontal acceleration, so the given acceleration is purely vertical. Using the path constraint $y=9x^2$ and differentiating twice with respect to time gives a relation between $v_x$ and the vertical acceleration, yielding $v_x = \frac{1}{3}\,\text{m/s}$ .

We are told the particle moves on the parabola $y = 9x^2$ , and its $x$ -component of velocity is constant. That means $v_x = \dot{x}$ is constant, so $\ddot{x} = 0$ . The acceleration vector is given as $2\,\hat{j}\,\text{m/s}^2$ , meaning the only acceleration is vertical. This is a classic constrained-motion problem: the path geometry links $x$ and $y$ , so we can differentiate that relation to connect velocities and accelerations.

Write the path equation and differentiate once The path is $y = 9x^2$ . Differentiate with respect to time $t$ :

$\frac{dy}{dt} = 9 \cdot 2x \frac{dx}{dt} = 18x \, v_x.$
So the vertical velocity is $v_y = 18x\,v_x$ .

Differentiate again to get acceleration Differentiate $v_y = 18x\,v_x$ with respect to time. Since $v_x$ is constant, we treat it as a constant factor:

$a_y = \frac{d}{dt}(18x\,v_x) = 18 v_x \frac{dx}{dt} = 18 v_x \cdot v_x = 18 v_x^2.$
Notice: we used $\frac{dx}{dt} = v_x$ and $v_x$ constant, so no product rule term from $v_x$ itself.

Equate to given acceleration The problem states the acceleration is $2\,\hat{j}\,\text{m/s}^2$ , so $a_y = 2$ . Thus:
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KCET 2023Set A-31 markMCQ
$v_0$
›Reveal solutionSolution

Average velocity is displacement over time; use $v^2-v_0^2=2as$ to express the displacement.

Step 1 — Definition.
$\bar v = \frac{\text{displacement}}{\text{time}} = \frac{s}{t}.$
Step 2 — Get $s$ from a kinematic equation.

For uniform acceleration the time-free equation is
$v^{2} = v_0^{2} + 2as \;\Longrightarrow\; s = \frac{v^{2}-v_0^{2}}{2a}.$
We use this particular equation because the options are written in terms of $v^2$ and $v_0^2$ .

Step 3 — Divide by $t$ .
$\bar v = \frac{s}{t} = \frac{v^{2}-v_0^{2}}{2at}.$
Step 4 — Cross-check with the standard result.

For constant acceleration the velocity–time graph is a straight line, so

$\bar v = \frac{v+v_0}{2}.$ …
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KCET 2019Set A-11 markMCQ
$\vec{F} = (-3\hat{i} + 4\hat{j})$
›Reveal solutionSolution
The key idea is that the velocity component along the x-axis must become zero for the velocity to be purely along the y-axis. Using Newton’s second law and kinematic equations, the time comes out to be 10 s.

The problem asks for the time when the velocity is only along the y-axis. That means the x-component of velocity must be zero at that instant. The force is constant, so the acceleration is constant — this is a straightforward application of the equations of motion under uniform acceleration.

Let’s break it down.

Find the acceleration vector. Newton’s second law: $\vec{F} = m \vec{a}$ . Given $m = 5$ kg and $\vec{F} = (-3\hat{i} + 4\hat{j})$ N,

$\vec{a} = \frac{\vec{F}}{m} = \left( \frac{-3}{5} \hat{i} + \frac{4}{5} \hat{j} \right) \text{ m/s}^2.$

Write the velocity as a function of time. For constant acceleration,

$\vec{v}(t) = \vec{u} + \vec{a} t,$
where initial velocity $\vec{u} = (6\hat{i} - 12\hat{j})$ m/s.

So
$\vec{v}(t) = \left( 6 - \frac{3}{5}t \right) \hat{i} + \left( -12 + \frac{4}{5}t \right) \hat{j}.$

Set the x-component to zero. For velocity to be purely along the y-axis, the $\hat{i}$ component must vanish:

$6 - \frac{3}{5}t = 0.$
Solve:
$\frac{3}{5}t = 6 \quad \Rightarrow \quad t = 6 \times \frac{5}{3} = 10 \text{ s}.$

Check the y-component at that time. At $t = 10$ s, $v_y = -12 + \frac{4}{5}(10) = -12 + 8 = -4 \text{ m/s}.$ …
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←Previous question3.16 — Read each statement below carefully and state, with reasons, if it is …→Next question3.18 — A particle starts from the origin at t = 0s with a velocity of 10.0hat…

Q.The position of a particle is given by r⃗=3.0t i^−2.0t2 j^+4.0 k^ m\vec{r} = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}\ \text{m}r=3.0ti^−2.0t2j^​+4.0k^ m where ttt is in seconds and the coefficients have the proper units for r⃗\vec{r}r to be in metres.

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Q.The position of a particle is given by $\vec{r} = 3.0t\,\hat{i} - 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}\ \text{m}$ where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres.

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