A particle starts from the origin at $t = 0\ \text{s}$ with a velocity of $10.0\,\hat{j}\ \text{m/s}$ and moves in the $x$-$y$ plane with a constant acceleration of $(8.0\,\hat{i} + 2.0\,\hat{j})\ \text{m s}^{-2}$. (a) At what time is the $x$-coordinate of the particle $16\ \text{m}$? What is the $y$-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

A particle starts from the origin at t = 0s with a velocity of 10.0hatjm/s and moves in the x-y plane with a c

Exercises · 3.18

t = 0\ \text{s}

(a) At what time is the

x

-coordinate of the particle

16\ \text{m}

? What is the

y

-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

Kerala DhseTextbookSubjective· 3mImportance★★★★★est

40% · 27/68 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

With constant acceleration $(8.0\hat{i}+2.0\hat{j})\ \text{m/s}^2$ and initial velocity $10.0\hat{j}\ \text{m/s}$ from the origin, the $x$ -coordinate reaches $16\ \text{m}$ at $t = 2.0\ \text{s}$ , when $y = 24\ \text{m}$ and the speed is $2\sqrt{113} \approx 21.3\ \text{m/s}$ .

Setting up

Since acceleration is constant, motion along $x$ and $y$ can be treated independently, each obeying the ordinary constant-acceleration equations:

\vec{r}_0 = 0, \qquad \vec{v}_0 = 10.0\hat{j}\ \text{m/s}, \qquad \vec{a} = 8.0\hat{i}+2.0\hat{j}\ \text{m/s}^2

Step 1 — Position as a function of time

x(t) = x_0 + v_{0x}t + \tfrac{1}{2}a_x t^2 = 0 + 0 + \tfrac{1}{2}(8.0)t^2 = 4.0t^2

y(t) = y_0 + v_{0y}t + \tfrac{1}{2}a_y t^2 = 0 + 10.0t + \tfrac{1}{2}(2.0)t^2 = 10.0t + t^2

Step 2 — Time when $x = 16\ \text{m}$

4.0t^2 = 16 \implies t^2 = 4 \implies t = 2.0\ \text{s} \quad (\text{taking the positive root})

Step 3 — $y$ -coordinate at that time

$y(2.0) = 10.0(2.0) + (2.0)^2 = 20 + 4 = 24\ \text{m}$ …

Start your 14-day free trial to unlock the full solution →

Concept: Get $v_x$ Before $t$ , Using the Time-Free Equation Along $x$

Method: $v_x^2 = v_{0x}^2 + 2a_x x$ First (No Quadratic-in- $t$ Solve), Then $t$ from a Linear Equation

The existing solutions write $x(t)=4.0t^2$ and solve $4.0t^2=16$ directly for $t$ (easy here since there's no linear term, but still a quadratic in form). This method instead finds the $x$ -velocity component first, straight from the time-free kinematic relation along $x$ alone — which never mentions $t$ — and only afterwards gets $t$ from a one-step linear equation.

Step 1 — Identify the $x$ -motion's knowns

x_0 = 0, \qquad v_{0x} = 0\ (\text{initial velocity is purely } \hat j), \qquad a_x = 8.0\ \text{m/s}^2, \qquad \text{target: } x = 16\ \text{m}

Step 2 — Time-free relation along $x$ : find $v_x$ without ever solving for $t$

v_x^2 = v_{0x}^2 + 2a_x x = 0 + 2(8.0)(16) = 256 \quad\Longrightarrow\quad v_x = \sqrt{256} = 16.0\ \text{m/s}

(Positive root: the particle starts at rest in $x$ and $a_x>0$ throughout, so $v_x$ only ever increases from zero.)

Step 3 — Now get $t$ , from the linear velocity equation

v_x = v_{0x}+a_xt \quad\Longrightarrow\quad 16.0 = 0+8.0t \quad\Longrightarrow\quad t = 2.0\ \text{s}

Step 4 — $y$ -coordinate at this time

y(t) = v_{0y}t+\tfrac12a_yt^2 = 10.0(2.0)+\tfrac12(2.0)(2.0)^2 = 20+4 = 24\ \text{m}

Step 5 — $y$ -velocity component at this time

v_y = v_{0y}+a_yt = 10.0+2.0(2.0) = 14.0\ \text{m/s}

Step 6 — Speed

v = \sqrt{v_x^2+v_y^2} = \sqrt{16.0^2+14.0^2} = \sqrt{256+196} = \sqrt{452} = 2\sqrt{113} \approx 21.3\ \text{m/s}

Why finding $v_x$ before $t$ is worth doing …

Start your 14-day free trial to unlock the full solution →

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

KEAM 2026Set eng-2026-04184 marksMCQ
$x$
›Reveal solutionSolution

With displacement $x=1+t^2$ and velocity $v=2t$ , equality gives $(t-1)^2=0$ , i.e. $t=1\,$ s.

The position/displacement is $x=1+t^2$ (in metres) and the velocity is
$v=\frac{dx}{dt}=2t\ \text{m/s}.$
Set the magnitudes equal: …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04194 marksMCQ
$y$
›Reveal solutionSolution

The position is quadratic, so acceleration is constant $=2\ \text{m s}^{-2}$ ; the average equals this constant value.

Given $y=t^2+2t+3$ .

Velocity: $v=\dfrac{dy}{dt}=2t+2$ .

Acceleration: $a=\dfrac{dv}{dt}=2\ \text{m s}^{-2}$ , which is constant. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04204 marksMCQ
$\vec{x} = (t^3 - 3t^2 + 2)\,\hat{\imath}$
›Reveal solutionSolution

Differentiate the position to get velocity, then set it to zero.

$\vec{x}=(t^3-3t^2+2)\,\hat{\imath}$ , so $v=\dfrac{dx}{dt}=3t^2-6t=3t(t-2)$ . …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04214 marksMCQ
$\vec{r} = 2t\hat{i} + \sqrt{3}t^{2}\hat{j} + 5\hat{k}$
›Reveal solutionSolution

Differentiate to get $\vec v$ ; at $t=1$ the components give $\tan\theta=\sqrt3$ , so $\theta=60^\circ$ .

$\vec r=2t\hat i+\sqrt3\,t^2\hat j+5\hat k\Rightarrow \vec v=\dfrac{d\vec r}{dt}=2\hat i+2\sqrt3\,t\,\hat j$ .

At $t=1$ s: $\vec v=2\hat i+2\sqrt3\,\hat j$ (the $\hat k$ term is constant, contributes nothing). …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04234 marksMCQ
$ms^{-1}$
›Reveal solutionSolution
a = dv/dt = 3.0 j, so |a| = 3 m/s^2.

Concept and Intuition

Acceleration is the time derivative of velocity. Only the time-dependent component of v contributes.

Step-by-Step Solution

v = 2.0 i + 3.0t j.

a = dv/dt = 0 i + 3.0 j.

Magnitude |a| = 3.0 m/s^2.

Common Mistakes …
Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0429F4 marksMCQ
$t$
›Reveal solutionSolution

Acceleration is the second time-derivative of position; for $\vec r = 8t\hat i + 3t^2\hat j + 3\hat k$ only the $3t^2$ term contributes, giving $\vec a = 6\hat j$ (magnitude 6).

Given $\vec r = 8t\hat i + 3t^2\hat j + 3\hat k$ m, the velocity is
$\vec v = \frac{d\vec r}{dt} = 8\hat i + 6t\hat j,$
and the acceleration is

$\vec a = \frac{d\vec v}{dt} = 6\hat j\;ms^{-2}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06054 marksMCQ
$\mathbf{r} = 3\,\mathbf{i} + 2t^{2}\,\mathbf{j}$
›Reveal solutionSolution

Differentiate $\mathbf{r}$ : $\mathbf{v}=4t\,\mathbf{j}$ , so at $t=5$ s speed $=20\ \mathrm{ms^{-1}}$ .

Step 1: Velocity is the time derivative of position:

$\mathbf{v}=\frac{d}{dt}\left(3\mathbf{i}+2t^2\mathbf{j}\right)=0\,\mathbf{i}+4t\,\mathbf{j}.$ …
Start your 14-day free trial to see the full solution →

←Previous question3.17 — The position of a particle is given by vecr = 3.0thati - 2.0t²hatj + 4…→Next question3.19 — hati and hatj are unit vectors along x- and y-axis respectively. What …

Q.A particle starts from the origin at t=0 st = 0\ \text{s}t=0 s with a velocity of 10.0 j^ m/s10.0\,\hat{j}\ \text{m/s}10.0j^​ m/s and moves in the xxx-yyy plane with a constant acceleration of (8.0 i^+2.0 j^) m s−2(8.0\,\hat{i} + 2.0\,\hat{j})\ \text{m s}^{-2}(8.0i^+2.0j^​) m s−2.

Setting up

Step 1 — Position as a function of time

Step 2 — Time when x=16 mx = 16\ \text{m}x=16 m

Step 3 — yyy-coordinate at that time

🎓Unlock everything free for 14 days

Q.A particle starts from the origin at $t = 0\ \text{s}$ with a velocity of $10.0\,\hat{j}\ \text{m/s}$ and moves in the $x$ - $y$ plane with a constant acceleration of $(8.0\,\hat{i} + 2.0\,\hat{j})\ \text{m s}^{-2}$ .

Step 2 — Time when $x = 16\ \text{m}$

Step 3 — $y$ -coordinate at that time

Unlock everything free for 14 days