Physics
Kinematics
Motion in a Plane

Kinematics · JEE & NEET Physics

Motion in a Plane: notes and previous year questions

Position, velocity and acceleration as vectors, the independence of perpendicular motions, and the split–solve–stitch method.

12 JEE Main questions (2020–2024)
10 NEET questions (2001–2023)

Motion in a Plane in short

A motion in a plane is two independent straight-line motions, along x and along y, happening at the same time.
The position vector r = xî + yĵ points from the origin to the object; its size comes from Pythagoras.
tan⁻¹(y/x) cannot tell opposite quadrants apart: check the signs of x and y.
Displacement is the change in position (a straight chord); distance is the length of the path.

1Why one number is not enough

A plane that climbs as it flies forward cannot be tracked with one number: we need how far forward ( $x$ ) and how high ( $y$ ). Its shadow on the ground moves along a straight line, and so does its shadow on a wall. Each shadow is a one-dimensional motion.

2Position and displacement

The position vector is an arrow from the origin to the object. With $\hat{i}$ and $\hat{j}$ the unit arrows along $x$ and $y$ :

\vec{r} = x\hat{i} + y\hat{j}

|\vec{r}| = \sqrt{x^2 + y^2}

For x = 4 m, y = 3 m: |r| = 5 m.

\theta = \tan^{-1}\!\left(\frac{y}{x}\right)

Measured from the positive x-axis: here about 37°.

The displacement from $\vec{r}_1$ to $\vec{r}_2$ is the change in position; its size is the straight-line distance between the points (Pythagoras):

\vec{s} = \vec{r}_2 - \vec{r}_1 = \Delta x\,\hat{i} + \Delta y\,\hat{j}

|\vec{s}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}

A(2, 3) to B(5, 7): s = 3î + 4ĵ m, |s| = 5 m.

Along a curved path, the distance is the length of the curve, while the displacement is the straight chord. Distance ≥ |displacement|, always.

3Velocity in a plane

\vec{v}_{\text{avg}} = \frac{\Delta\vec{r}}{\Delta t}

It points along the displacement, not along the path.

Example: from A(2, 3) to B(5, 7) in 2 s, $\vec{v}_{\text{avg}} = (3\hat{i} + 4\hat{j})/2 = 1.5\hat{i} + 2\hat{j}$ m/s, of size 2.5 m/s.

The instantaneous velocity comes from differentiating each component separately:

\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}

|\vec{v}| = \sqrt{v_x^2 + v_y^2}

\tan\theta = \frac{v_y}{v_x}

4Acceleration in a plane

\vec{a} = \frac{d\vec{v}}{dt} = a_x\hat{i} + a_y\hat{j}

|\vec{a}| = \sqrt{a_x^2 + a_y^2}

In a plane, the acceleration need not point along the velocity. Split it into two parts: the part along $\vec{v}$ changes the speed; the part perpendicular to $\vec{v}$ changes the direction. That is why paths curve.

Motion	Acceleration
Speeding up in a straight line	along v: only the speed changes
Turning at a steady speed	perpendicular to v: only the direction changes
Speeding up round a bend	both parts

5Independence of perpendicular motions

Drop one ball and push another off a table sideways at the same instant. At every moment they are at the same height, and they land together: the sideways motion has no effect on the fall. Motion along $x$ does not affect motion along $y$ , and vice versa.

Along x	Along y
$v_x = u_x + a_x t$	$v_y = u_y + a_y t$
$x = u_x t + \tfrac{1}{2}a_x t^2$	$y = u_y t + \tfrac{1}{2}a_y t^2$
$v_x^2 = u_x^2 + 2a_x x$	$v_y^2 = u_y^2 + 2a_y y$

6Vector equations and circular motion

With a constant $\vec{a}$ , the equations of motion hold for whole vectors (each vector equation is two ordinary ones):

\vec{v} = \vec{u} + \vec{a}t

\vec{s} = \vec{u}t + \tfrac{1}{2}\vec{a}t^2

v^2 = u^2 + 2\,\vec{a}\cdot\vec{s}

Circular motion from calculus. For $\vec{r} = R\cos(\omega t)\,\hat{i} + R\sin(\omega t)\,\hat{j}$ :

$\vec{v} = -R\omega\sin(\omega t)\,\hat{i} + R\omega\cos(\omega t)\,\hat{j}$ , so the speed $|\vec{v}| = R\omega$ is constant.
$\vec{a} = -\omega^2\vec{r}$ : toward the centre, of size $\omega^2 R$ .
$\vec{v}\cdot\vec{a} = 0$ : velocity and acceleration are perpendicular at every instant, so the acceleration only turns the object.

Summary

Key ideas

A motion in a plane is two independent straight-line motions, along x and along y, happening at the same time.
The position vector r = xî + yĵ points from the origin to the object; its size comes from Pythagoras.
tan⁻¹(y/x) cannot tell opposite quadrants apart: check the signs of x and y.
Displacement is the change in position (a straight chord); distance is the length of the path.
Displacements add as vectors: 3 m east and 4 m north make 5 m, not 7 m.
Average velocity points along the displacement; instantaneous velocity is tangent to the path.
The part of the acceleration along v changes the speed; the part across v changes the direction.
Perpendicular motions are independent: a ball pushed off a table lands at the same time as one dropped.
Split, solve each axis, stitch with Pythagoras; time is the only link between the axes.
With constant acceleration, v = u + at and s = ut + ½at² hold as vector equations.
In uniform circular motion the speed Rω is constant and the acceleration ω²R points to the centre, perpendicular to v.

Every equation

Position vector: $\vec{r} = x\hat{i} + y\hat{j}$
Its size: $|\vec{r}| = \sqrt{x^2 + y^2}$
Its direction: $\theta = \tan^{-1}(y/x)$
Displacement: $\vec{s} = \vec{r}_2 - \vec{r}_1$
Its size: $|\vec{s}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
Average velocity: $\vec{v}_{\text{avg}} = \Delta\vec{r}/\Delta t$
Velocity: $\vec{v} = d\vec{r}/dt = v_x\hat{i} + v_y\hat{j}$
Speed: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$
Direction of velocity: $\tan\theta = v_y/v_x$
Acceleration: $\vec{a} = d\vec{v}/dt = a_x\hat{i} + a_y\hat{j}$
Constant a: velocity: $\vec{v} = \vec{u} + \vec{a}t$
Constant a: displacement: $\vec{s} = \vec{u}t + \tfrac{1}{2}\vec{a}t^2$
Constant a: no time: $v^2 = u^2 + 2\,\vec{a}\cdot\vec{s}$
Circle: speed: $|\vec{v}| = R\omega$
Circle: acceleration: $\vec{a} = -\omega^2\vec{r}$

Previous year questions with solutions

Real JEE and NEET questions on motion in a plane. Try each one before you open the solution.

Q1JEE Main 2024Numerical answer

A vector has magnitude same as that of $\vec{A} = 3\hat{i} + 4\hat{j}$ and is parallel to $\vec{B} = 4\hat{i} + 3\hat{j}$ . The $x$ and $y$ components of this vector in first quadrant are $x$ and $3$ respectively where $x =$ __________.

Show answer and solution

Answer: 4

The direction comes from $\vec{B}$ and the size from $\vec{A}$ , so build it out of both.

$|\vec{A}| = \sqrt{9 + 16} = 5$ , and $\hat{B} = \dfrac{4\hat{i} + 3\hat{j}}{5}$ since $|\vec{B}| = 5$ as well.

The vector wanted is $5\hat{B} = 4\hat{i} + 3\hat{j}$ , so $x = 4$ — and the $y$ -component of $3$ quoted in the question confirms it.

Q2NEET 2016One correct option

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

A $45^{o}$
B $180^{o}$
C $0^{o}$
D $90^{o}$

Show answer and solution

Answer: Option D

Square both statements and set them side by side: $A^{2} + B^{2} + 2AB\cos\theta = A^{2} + B^{2} - 2AB\cos\theta$ . Everything cancels except the middle terms, leaving $4AB\cos\theta = 0$ . Neither vector has zero length, so it must be $\cos\theta$ that vanishes, and $\theta = 90^{\circ}$ . In the picture it is the parallelogram whose two diagonals come out the same length, which happens only when it is a rectangle.

Q3JEE Main 2024One correct option

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at angle $\theta$ , then

A $|\vec{A}+\vec{B}|=2R\cos (\frac{\theta }{2})$
B $|\vec{A}-\vec{B}|=2R\cos (\frac{\theta }{2})$
C $|\vec{A}-\vec{B}|=\sqrt{2}R\sin (\frac{\theta }{2})$
D $|\vec{A}+\vec{B}|=2R\sin (\frac{\theta }{2})$

Show answer and solution

Answer: Option A

Here $R$ is the shared magnitude, not the resultant. Put both magnitudes equal to $R$ : $|\vec{A}+\vec{B}|=\sqrt{R^{2}+R^{2}+2R^{2}\cos\theta}=R\sqrt{2(1+\cos\theta)}$ , and $1+\cos\theta=2\cos^{2}(\theta/2)$ , which gives $2R\cos(\theta/2)$ .

If that identity does not come to mind, test the special angles. At $\theta=0^{\circ}$ the vectors coincide, so $|\vec{A}+\vec{B}|=2R$ and $|\vec{A}-\vec{B}|=0$ , which kills B and D. At $\theta=180^{\circ}$ the difference is $2R$ but option C offers only $\sqrt{2}R$ . A is left.

Practice questions, easy to hard

Three questions from the motion in a plane practice ladder: one easy, one medium, one hard.

Q4Numerical answer

A walker goes $6\ \mathrm{m}$ east, then $8\ \mathrm{m}$ north. Head to tail, these two displacements make a right-angled triangle.

How long is the closing side, in $\mathrm{m}$ ?

Show answer and solution

Answer: 10

The closing side is the sum of the two displacements, and in a right-angled triangle its length is $\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\ \mathrm{m}$ . It is not $14\ \mathrm{m}$ : vectors only add like plain numbers when they point the same way.

Q5Numerical answer

Two vectors drawn from the same point do not add end to end unless they already line up. Complete the parallelogram on them: the sum is the diagonal through that point, and its length is $R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$ , where $\theta$ is the angle between them.

Two vectors of magnitudes $5$ and $3$ have $60^{\circ}$ between them. What is the magnitude of their sum?

Show answer and solution

Answer: 7

$\cos 60^{\circ} = \tfrac12$ , so $R = \sqrt{25 + 9 + 2 \times 5 \times 3 \times \tfrac12} = \sqrt{49} = 7$ . The other diagonal of that same parallelogram is the difference $\vec{A} - \vec{B}$ , and it carries the same formula with the sign flipped, $\sqrt{A^{2} + B^{2} - 2AB\cos\theta}$ , because reversing $\vec{B}$ turns $\theta$ into $180^{\circ} - \theta$ and the cosine changes sign with it. Here that would give $\sqrt{19}$ , about $4.36$ .