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?

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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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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

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

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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 …

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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 question3.17 — 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

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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

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