Physics
Kinematics
Frame of Reference

Kinematics · JEE & NEET Physics

Frame of Reference: notes and previous year questions

Why rest and motion depend on who is watching, which frames obey Newton's laws as they are, and how pseudo forces handle the rest.

15 JEE Main questions (2017–2026)
7 NEET questions (2006–2025)

Frame of Reference in short

A frame of reference is an origin, a set of axes and a clock: the observer's point of view.
There is no absolute rest; rest and motion mean something only in a chosen frame.
Position, velocity and the path depend on the frame; acceleration and time intervals do not, for frames that are not accelerating.
A ball dropped in a steady train falls straight down in the train and follows a parabola seen from the ground.

1Who is moving?

Sit still in a train moving at 20 m/s. To a friend on the platform you zoom past at 20 m/s. To you, the train is still and the platform slides backward at 20 m/s. Both of you are right: whether something moves depends on who is watching.

A frame of reference is the point of view we describe motion from. It has three parts:

An origin: where the observer measures from.
Coordinate axes ( $x$ , $y$ , $z$ ): the directions to measure along.
A clock: to record when the object is at each place.

2Rest and motion are relative

There is no absolute rest. In your chair you are at rest in the room, but the Earth's spin carries you at about 465 m/s at the equator, the Earth carries you round the Sun at about 30 km/s, and the Sun itself circles the centre of the galaxy. Rest or motion means something only once the frame is chosen.

Depend on the frame: position, velocity and the path (trajectory).
The same in every frame (in classical physics, as long as the frames do not accelerate relative to each other): acceleration and time intervals.

3Inertial and non-inertial frames

An inertial frame is one in which Newton's laws of motion hold as they are. It is at rest or moves at constant velocity: the frame itself does not accelerate, and an object at rest stays at rest.

A room at rest on the Earth (very nearly).
A train moving at a steady speed on a straight track, however fast.
A spaceship drifting far from any star or planet.

A non-inertial frame accelerates: it speeds up, slows down or turns. In it, Newton's laws need an extra pseudo force. Examples: a car taking a turn, a lift speeding up or slowing down, a merry-go-round.

4Pseudo forces

To use Newton's laws inside an accelerating frame, add a pseudo force (also called a fictitious force) to every object:

\vec{F}_{\text{pseudo}} = -m\,\vec{a}_{\text{frame}}

Size: the object's mass times the frame's acceleration. Direction: opposite to the frame's acceleration.

When a bus speeds up, you are thrown backward; when it brakes, you lurch forward. The pseudo force always points opposite to the frame's acceleration. Memory aid: Pseudo forces Appear In Non-inertial frames (PAIN), and "pseudo pushes backward".

5The lift: apparent weight

The weight you feel is the normal force $N$ from the floor (what a scale reads). In the ground frame, for a lift accelerating upward at $a$ :

N - mg = ma \;\Rightarrow\; N = m(g + a)

Lift	Scale reads N	You feel
At rest	$mg$	normal
Moving up or down at constant velocity	$mg$	normal
Accelerating up at $a$	$m(g + a)$	heavier
Accelerating down at $a$	$m(g - a)$	lighter
Free fall (cable snaps, $a = g$ )	$0$	weightless

For a 50 kg person with $g = 10\ \text{m/s}^2$ : 500 N at rest, 600 N accelerating up at $2\ \text{m/s}^2$ , 400 N accelerating down at $2\ \text{m/s}^2$ , 0 N in free fall. Chant: "Up adds, down subtracts, free fall leaves nothing."

6Turning frames, wedges and choosing a frame

A rotating frame accelerates even at constant speed, because the direction of motion keeps changing. From the ground, a rider on a merry-go-round accelerates toward the centre and a real inward force (the grip) holds them. Riding along, they feel flung outward: this is the centrifugal force, a pseudo force of the rotating frame.

F_{\text{centrifugal}} = m\omega^2 r

Directed away from the centre, for an observer turning at angular speed ω at radius r. Example: m = 40 kg, ω = 1 rad/s, r = 3 m gives 120 N.

ExampleA block at rest on an accelerating smooth wedge

A block sits on the smooth slope of a wedge inclined at $\theta$ . In the wedge's frame the forces on the block are $mg$ down, the normal force $N$ and the horizontal pseudo force $ma$ . For the block to stay still, the components along the slope balance:

mg\sin\theta = ma\cos\theta

a = g\tan\theta

For $\theta = 30°$ : $a = 10\tan 30° = 10/\sqrt{3} \approx 5.8\ \text{m/s}^2$ . For $\theta = 45°$ : $a = 10\ \text{m/s}^2$ . Check: a flat surface ( $\theta \to 0$ ) needs $a \to 0$ , and a vertical wall ( $\theta \to 90°$ ) would need an endless acceleration.

Situation	Best frame to use
Ground-based problems	The Earth's surface (inertial)
Problems inside vehicles	The vehicle's frame, with pseudo forces
Relative motion problems	One of the objects as the reference
Rotating systems	The rotating frame, with the centrifugal force

Summary

Key ideas

A frame of reference is an origin, a set of axes and a clock: the observer's point of view.
There is no absolute rest; rest and motion mean something only in a chosen frame.
Position, velocity and the path depend on the frame; acceleration and time intervals do not, for frames that are not accelerating.
A ball dropped in a steady train falls straight down in the train and follows a parabola seen from the ground.
Inertial frames are at rest or move at constant velocity; Newton's laws hold in them as they are.
A frame moving fast but steadily is still inertial: only acceleration makes a frame non-inertial.
The ground is treated as inertial because the Earth's acceleration is tiny (about 0.034 m/s² from its spin).
In an accelerating frame, add a pseudo force $-m\vec{a}_{\text{frame}}$ , opposite to the frame's acceleration, to every object.
Pseudo forces have no agent and no reaction pair; never use them in the ground frame.
Apparent weight follows the acceleration, not the direction of motion: accelerating up feels heavier, down feels lighter, free fall feels weightless.
In a rotating frame the centrifugal force $m\omega^2 r$ points outward; it is a pseudo force.

Every equation

Pseudo force: $\vec{F}_{\text{pseudo}} = -m\,\vec{a}_{\text{frame}}$
Lift at rest or constant velocity: $N = mg$
Lift accelerating up: $N = m(g + a)$
Lift accelerating down: $N = m(g - a)$
Lift in free fall: $N = 0$
Effective gravity in a lift accelerating up: $g_{\text{eff}} = g + a$
Pendulum in an accelerating vehicle: $\tan\theta = \frac{a}{g}$
Effective gravity in that vehicle: $g_{\text{eff}} = \sqrt{g^2 + a^2}$
Block still on a smooth wedge: $a = g\tan\theta$
Centrifugal force: $F = m\omega^2 r$
Fall time from height h: $t = \sqrt{2h/g}$

Previous year questions with solutions

Real JEE and NEET questions on frame of reference. Try each one before you open the solution.

Q1JEE Main 2026One correct option

Two cars $A$ and $B$ are moving in the same direction along a straight line with speeds $100\ \mathrm{km/h}$ and $80\ \mathrm{km/h}$ respectively, such that car $A$ is moving ahead of car $B$ . A person in car $B$ throws a stone with a speed $v$ so that it hits car $A$ with a speed of $5\ \mathrm{m/s}$ . The value of $v$ is __________ $\mathrm{km/h}$ .

A $18$
B $28$
C $38$
D $48$

Show answer and solution

Answer: Option C

First make the units agree: $5\ \mathrm{m/s} = 5 \times 3.6 = 18\ \mathrm{km/h}$ , and that is the stone's speed relative to $A$ . The stone is thrown at $v$ relative to $B$ , so its ground speed is $80+v$ and its speed relative to $A$ is $(80+v)-100$ . Setting that to $18$ gives $v = 38\ \mathrm{km/h}$ .

Q2NEET 2025One correct option

Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ minutes. A girl driving a scooty at $60\ \mathrm{km/h}$ in the direction $X$ to $Y$ notices that a bus goes past her every $30$ minutes in the direction of her motion, and every $10$ minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed of the buses.

A $10\ \mathrm{min}$ , $90\ \mathrm{km/h}$
B $15\ \mathrm{min}$ , $120\ \mathrm{km/h}$
C $9\ \mathrm{min}$ , $40\ \mathrm{km/h}$
D $25\ \mathrm{min}$ , $100\ \mathrm{km/h}$

Show answer and solution

Answer: Option B

Consecutive buses are a fixed distance $vT$ apart. Catching her from behind, they close at $v-60$ and take $30$ minutes: $(v-60)(30) = vT$ . Coming the other way they close at $v+60$ and take $10$ : $(v+60)(10) = vT$ . Equating gives $30v-1800 = 10v+600$ , so $v = 120\ \mathrm{km/h}$ , and then $T = 15$ minutes.

Q3JEE Main 2025Numerical answer

A person travelling on a straight line moves with a uniform velocity $v_1 = 5\ \mathrm{m/s}$ for a distance $x$ , and with a uniform velocity $v_2$ for the next $\dfrac{3}{2}x$ . The average velocity for this motion is $\dfrac{50}{7}\ \mathrm{m/s}$ .

Find $v_2$ , in $\mathrm{m/s}$ .

Show answer and solution

Answer: 10

Everything runs one way, so the displacement is $x + \dfrac{3}{2}x = \dfrac{5}{2}x$ and the time is $\dfrac{x}{5} + \dfrac{3x}{2v_2}$ . Setting $\dfrac{5x/2}{\frac{x}{5}+\frac{3x}{2v_2}} = \dfrac{50}{7}$ and cancelling $x$ gives $\dfrac{1}{5} + \dfrac{3}{2v_2} = \dfrac{7}{20}$ , so $\dfrac{3}{2v_2} = \dfrac{3}{20}$ and $v_2 = 10\ \mathrm{m/s}$ .

Practice questions, easy to hard

Three questions from the frame of reference practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Average speed is the total distance divided by the total time — one number for a whole journey, however the journey actually went.

A cyclist covers $900\ \mathrm{m}$ in $300\ \mathrm{s}$ . What is her average speed, in $\mathrm{m/s}$ ?

Show answer and solution

Answer: 3

$900 / 300 = 3\ \mathrm{m/s}$ . It does not claim she rode at $3\ \mathrm{m/s}$ at any moment — it is the steady speed that would have taken exactly as long.

Q5One correct option

Relative motion works for accelerations too: in the frame of $B$ , body $A$ has acceleration $a_A - a_B$ .

Two cars start together from rest, one with $5\ \mathrm{m/s^2}$ and the other with $3\ \mathrm{m/s^2}$ . As seen from the slower car, the faster one:

AStarts from rest and accelerates at $2\ \mathrm{m/s^2}$
BMoves at a steady $2\ \mathrm{m/s}$
CAccelerates at $8\ \mathrm{m/s^2}$
DStays level with it

Show answer and solution

Answer: Option A

Relative velocity at the start is $0 - 0 = 0$ , and relative acceleration is $5 - 3 = 2\ \mathrm{m/s^2}$ . So from the slow car, the fast one pulls away as if it alone were starting from rest at $2\ \mathrm{m/s^2}$ — and all three equations of motion can be used in that frame.