Motion in a Straight Line Notes for NEET — Class 11 NCERT

Introduction

Kinematics is the part of mechanics that describes how things move without asking why they move. Motion in a Straight Line is the simplest case: a particle moves along a single straight line, so we only need one number to describe its position at any instant. Once you can handle this case, you can extend the same ideas to motion in a plane and motion in three dimensions.

For NEET, this chapter is tested heavily. You can expect 2 to 4 questions per year, almost always involving the three kinematic equations, free fall, position-time and velocity-time graphs, or relative velocity. The questions are short and formula-driven, so a careful student can lock in full marks here.

Frame of reference

Motion is always relative. To describe where an object is and how it moves, you first attach a coordinate system, called a frame of reference, to some chosen origin. The same motion looks different from different frames.

A passenger walking inside a moving train sees the seats as stationary. A person on the platform sees both the seats and the passenger moving. For NEET we work mostly in inertial frames, frames that move at constant velocity (or are at rest). A non-rotating ground-fixed frame is a good enough inertial frame for every problem in this chapter.

We also treat the moving object as a point particle: an object whose size is negligible compared to the distances it travels. A car driving from Delhi to Mumbai can be modelled as a point. A car parking in a garage cannot.

Position, distance and displacement

For motion in a straight line, choose the line as the x-axis. The position of the particle at time $t$ is just its x-coordinate, $x (t)$ . Position has a sign: positive on one side of the origin, negative on the other.

Distance vs displacement

Distance (also called path length) is the total length of path traversed. It is always positive and depends on the route taken.

Displacement is the change in position from start to finish:

Δ x = x_{final} - x_{initial} .

Displacement has a sign and depends only on the start and end points, not the path. If you walk $5 m$ east and then $3 m$ west, your distance is $8 m$ but your displacement is $+ 2 m$ (east taken positive).

Distance $\geq$ magnitude of displacement, always. Equality holds only when the motion is in a single direction with no reversal.

Speed and velocity

Average speed

Average speed is the total distance covered divided by the total time taken. It is always non-negative.

Average speed = \frac{Total distance}{Total time} .

Average velocity

Average velocity is the displacement divided by the time interval. It has a sign.

\overset{v}{ˉ} = \frac{Δ x}{Δ t} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}} .

These two are different in general. If you drive $60 km$ north in 1 h and $60 km$ south in the next 1 h, your average speed is $120/2 = 60 km/h$ but your average velocity is $0$ , because the displacement is zero.

For a round trip, average velocity is always zero, but average speed is not. NEET loves this trap.

A car drives d km away at v₁, then drives back the same distance at v₂. Watch the difference between average speed and average velocity.

One-way distance d: 60 km

Outbound speed v₁: 60 km/h

Return speed v₂: 40 km/h

Average speed

48.00 km/h

Average velocity

0 km/h

net displacement = 0

Working

\overset{v}{ˉ}_{speed} = \frac{2 d}{d / v _{1} + d / v _{2}} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}} = \frac{2 \times 60 \times 40}{60 + 40} = 48.00 km/h

Note this is the harmonic mean (48.00), not the arithmetic mean (50.00). NEET tests this distinction every other year.

Instantaneous velocity and speed

Instantaneous velocity is the velocity at a single instant. We get it by shrinking the time interval to zero:

v = Δ t \to 0 lim \frac{Δ x}{Δ t} = \frac{d x}{d t} .

Geometrically, instantaneous velocity is the slope of the position-time graph at that instant.

Instantaneous speed is the magnitude of instantaneous velocity. They are always equal in size, unlike their averages, which can differ.

Acceleration

Acceleration is the rate of change of velocity.

Average acceleration

\overset{a}{ˉ} = \frac{Δ v}{Δ t} = \frac{v _{2} - v _{1}}{t _{2} - t _{1}} .

Instantaneous acceleration

a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}} .

Acceleration has the same sign as the change in velocity. A particle slowing down has a deceleration, which is acceleration opposite to the direction of motion.

Uniform vs non-uniform motion

Uniform motion: velocity is constant, acceleration is zero.
Uniformly accelerated motion: acceleration is constant, velocity changes linearly with time.
Non-uniform motion: acceleration itself changes with time.

Most NEET problems in this chapter are uniformly accelerated. The kinematic equations below apply only to that case.

Position-time graphs

A position-time graph plots $x$ on the y-axis against $t$ on the x-axis. The slope of the graph at any point is the instantaneous velocity at that instant.

Shape	Motion	Slope (velocity)
Horizontal line	At rest	Zero
Straight slanted line	Uniform velocity	Constant
Curve, slope increasing	Accelerating	Increasing
Curve, slope decreasing	Decelerating	Decreasing

The position-time graph for uniformly accelerated motion is a parabola, since $x = x_{0} + u t + \frac{1}{2} a t^{2}$ is quadratic in $t$ .

A position-time graph can never be vertical. A vertical line would mean infinite velocity, which is impossible.

Graphs

Position-time graph reader

Pick a motion type and slide the cursor to read the position and the instantaneous velocity (the slope of the tangent) at any instant.

Cursor at t = 4.0 s

Position x(t)

12.00 m

Velocity = slope

6.00 m/s

v = \frac{d x}{d t}_{t = 4.0} = 6.00 m/s

The orange dashed line is the tangent at the cursor. Its slope equals the instantaneous velocity.

Try this

On the "At rest" graph, the slope is zero everywhere — velocity is zero at every t.
On the "Uniform v" graph, the slope is constant — velocity does not change with time.
On the "Accelerating" parabola, slide the cursor and watch the slope grow linearly with t.
On the "Decelerating" curve, the slope decreases until it crosses zero (the moment the body stops).

Velocity-time graphs

A velocity-time graph plots $v$ on the y-axis against $t$ . Two facts are crucial.

The slope of a velocity-time graph at any instant equals the instantaneous acceleration.
The area under a velocity-time graph between two times equals the displacement during that interval.

Shape	Motion	Acceleration
Horizontal line above x-axis	Uniform velocity, positive direction	Zero
Straight line, positive slope	Uniform acceleration, speeding up	Constant positive
Straight line, negative slope	Uniform deceleration	Constant negative
Curve	Non-uniform acceleration	Changing

For uniformly accelerated motion, the velocity-time graph is a straight line. The area under it (a trapezium between $u$ , $v$ , $0$ and $t$ ) gives the displacement:

s = \frac{1}{2} (u + v) t .

Graphs

V-T graph: area = displacement

Move the sliders to change u, v and t. The trapezium fills the area under the velocity-time graph — that area equals the displacement.

Initial velocity u: 10 m/s

Final velocity v: 30 m/s

Duration t: 8 s

Displacement = area under v-t graph

s = 160.00 m

s = \frac{1}{2} (u + v) t = \frac{1}{2} (10 + 30) (8) = 160.00 m

Slope of the line = acceleration = 2.50 m/s²

Try this

Set u = 10, v = 30, t = 8 → trapezium with two parallel sides (10 and 30), height 8 → area 160 m.
Set u = v → rectangle (uniform velocity). Slope is zero, no acceleration.
Set v < u → object decelerating. Trapezium tilts down; area still gives the displacement.
Set u = 0, v > 0 → triangle. Area = ½ × base × height = ½ × t × v.

Practice these on the timed test

Try a free 10-question NEET mock test on Motion in a Straight Line — instant results, no sign-up needed.

Kinematic equations of motion

For straight-line motion with constant acceleration $a$ , initial velocity $u$ , final velocity $v$ , displacement $s$ and time $t$ , three equations relate these five quantities.

The three equations

v = u + a t

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

v^{2} = u^{2} + 2 a s

Memorise these three. Every NEET problem on uniformly accelerated motion plugs into one of them. Choose the equation based on which quantity is missing.

If you are not given	Use
Displacement $s$	$v = u + a t$
Final velocity $v$	$s = u t + \frac{1}{2} a t^{2}$
Time $t$	$v^{2} = u^{2} + 2 a s$

v = u + a t

Initial velocity u

m/s

Final velocity v

m/s

Acceleration a

m/s²

Time t

Answer

v = 10.000 m/s

v = u + a t = 0 + (2) (5) = 10.00 m/s

Displacement in the n-th second

A useful corollary: the displacement during the $n$ -th second of motion (between $t = n - 1$ and $t = n$ ) is:

s_{n} = u + \frac{a}{2} (2 n - 1) .

This appears in NEET roughly every two years.

Initial velocity u: 0 m/s

Acceleration a: 4 m/s²

Which second n: 5

Formula

s_{n} = u + \frac{a}{2} (2 n - 1) = 0 + \frac{4}{2} (2 \times 5 - 1) = 18.00 m

Total dist at t=4s

32.00 m

Total dist at t=5s

50.00 m

Distance in nth s

18.00 m

Sign conventions

Pick a positive direction at the start of every problem and stick with it. If you take rightward (or upward) as positive, then any velocity, displacement or acceleration in the opposite direction must be entered as a negative number. Mixing signs is the single biggest source of NEET kinematics errors.

Free fall under gravity

Free fall is a special case of uniformly accelerated motion: the acceleration is the local gravitational acceleration $g \approx 9.8 m/s^{2}$ directed downward, ignoring air resistance. NEET usually uses $g = 9.8 m/s^{2}$ or $10 m/s^{2}$ — read the question.

Object dropped from rest

Take downward positive. Then $u = 0$ and $a = + g$ . The kinematic equations become:

v = g t, s = \frac{1}{2} g t^{2}, v^{2} = 2 g s .

Object thrown upward

Take upward positive. Then $u$ is positive and $a = - g$ . At the highest point the velocity is zero, so the time to reach the top is:

t_{up} = \frac{u}{g} .

The maximum height reached is:

H_{max} = \frac{u ^{2}}{2 g} .

The total time of flight (return to the starting level) is $2 u / g$ , and the speed on return equals the speed of projection — same magnitude, opposite direction.

Drop height h: 80 m

Time to ground

4.00 s

Impact speed

40.00 m/s

Working

t = \frac{2 h}{g} = \frac{2 \times 80}{10} = 4.00 s

v = g t = 10 \times 4.00 = 40.00 m/s

Using g = 10 m/s², ignoring air resistance.

Stopping distance

A vehicle moving with velocity $u$ brakes to rest with deceleration $a$ . From $v^{2} = u^{2} - 2 a s$ with $v = 0$ :

s_{stop} = \frac{u ^{2}}{2 a} .

Stopping distance scales as the square of the speed. Doubling the speed quadruples the distance. NEET asks this almost every year.

Initial speed u: 20 m/s (72 km/h)

Deceleration a: 5 m/s²

Stopping distance

s = 40.00 m

s = \frac{u ^{2}}{2 a} = \frac{2 0 ^{2}}{2 \times 5} = 40.00 m

t = \frac{u}{a} = \frac{20}{5} = 4.00 s

Doubling the speed quadruples the distance. At 40 m/s on the same road, you would need 160.00 m to stop — exactly 4× the current value, because s ∝ u².

Relative velocity (one dimension)

The velocity of object A as measured by an observer on object B is:

v_{A B} = v_{A} - v_{B} .

Both velocities must be measured in the same external frame (usually the ground). The sign of each velocity matters.

Same direction

Two cars move in the same direction at $60$ and $40 km/h$ . The relative velocity of the faster car with respect to the slower is $60 - 40 = 20 km/h$ .

Opposite directions

If the same two cars move toward each other, take one direction positive and the other negative:

v_{A B} = 60 - (- 40) = 100 km/h .

For overtaking and crossing problems, the relative velocity sets the time taken. If two trains of length $L_{1}$ and $L_{2}$ cross each other, the time is $(L_{1} + L_{2}) / v_{rel}$ where $v_{rel}$ is the magnitude of the relative velocity.

Velocity of A (v_A): 20 m/s

Speed of B (|v_B|): 10 m/s

Relative velocity of A w.r.t. B

v_AB = 10 m/s

v_{A B} = v_{A} - v_{B} = 20 - (10) = 10 m/s

In the same direction, you subtract magnitudes. Negative means B is faster.

Crossing/overtaking time (treating A and B as trains)

Length of A: 100 m

Length of B: 200 m

Time to overtake = 30.00 s (300 m / 10 m/s)

Worked NEET problems

NEET-style problem · Average velocity

Question

A car travels the first half of a journey at

60 km/h

and the second half at

40 km/h

. What is its average speed for the whole journey?

Solution

Step 1. Let the total distance be $2 d$ , so each half is $d$ .

Step 2. Time for the first half is $t_{1} = d /60$ . Time for the second half is $t_{2} = d /40$ .

Step 3. Average speed is total distance over total time:

\overset{v}{ˉ} = \frac{2 d}{d /60 + d /40} = \frac{2}{1/60 + 1/40} = \frac{2}{( 2 + 3 ) /120} = \frac{2 \times 120}{5} = 48 km/h .

The harmonic mean, not the arithmetic mean, is what you need when the distances are equal but the speeds differ.

NEET-style problem · Free fall

Question

A stone is dropped from a tower of height

80 m

. Take

g = 10 m/s^{2}

. Find the time to reach the ground and the speed on impact.

Solution

Take downward positive. $u = 0$ , $a = g = 10 m/s^{2}$ , $s = 80 m$ .

From $s = \frac{1}{2} g t^{2}$ :

80 = \frac{1}{2} (10) t^{2} \Rightarrow t^{2} = 16 \Rightarrow t = 4 s .

Final speed: $v = g t = 10 \times 4 = 40 m/s$ .

NEET-style problem · Stopping distance

Question

A car moving at

20 m/s

can stop in

40 m

on a certain road. If the same car moves at

30 m/s

on the same road, the stopping distance is:

Solution

From $v^{2} = u^{2} - 2 a s$ with $v = 0$ , we get $s = u^{2} / (2 a)$ . The deceleration $a$ is the same for the same road, so $s \propto u^{2}$ .

\frac{s _{2}}{s _{1}} = (\frac{u _{2}}{u _{1}})^{2} = (\frac{30}{20})^{2} = \frac{9}{4} .

So $s_{2} = 40 \times 9/4 = 90 m$ .

NEET-style problem · n-th second

Question

A particle starts from rest with constant acceleration

a = 4 m/s^{2}

. Find the distance it covers in the 5th second.

Solution

Use $s_{n} = u + \frac{a}{2} (2 n - 1)$ with $u = 0$ , $a = 4 m/s^{2}$ , $n = 5$ :

s_{5} = 0 + \frac{4}{2} (2 \times 5 - 1) = 2 \times 9 = 18 m .

NEET-style problem · Relative velocity

Question

Two trains of length

100 m

and

200 m

move on parallel tracks in the same direction at

20 m/s

and

10 m/s

respectively. How long does the faster train take to overtake the slower one completely?

Solution

The relative velocity of the fast train with respect to the slow one is $20 - 10 = 10 m/s$ . To overtake completely, the fast train must cover its own length plus the slow train's length:

L_{rel} = 100 + 200 = 300 m .

t = \frac{300}{10} = 30 s .

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Summary cheat sheet

Distance vs displacement: distance is path length (always positive), displacement is $x_{final} - x_{initial}$ (can be negative or zero).
Average velocity: $\overset{v}{ˉ} = Δ x /Δ t$ . Average speed: total distance over total time. They are not the same.
Instantaneous velocity: $v = d x / d t$ , the slope of the position-time graph.
Acceleration: $a = d v / d t$ , the slope of the velocity-time graph.
Three kinematic equations: $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , $v^{2} = u^{2} + 2 a s$ .
Distance in n-th second: $s_{n} = u + \frac{a}{2} (2 n - 1)$ .
Free fall: dropped from rest, $v = g t$ , $s = \frac{1}{2} g t^{2}$ . Thrown up: $H_{max} = u^{2} / (2 g)$ , time of flight $2 u / g$ .
Stopping distance: $s_{stop} = u^{2} / (2 a)$ . Doubling speed quadruples stopping distance.
Relative velocity (1D): $v_{A B} = v_{A} - v_{B}$ . Use sign convention strictly.
VT graph areas = displacement. PT graph slopes = velocity. VT graph slopes = acceleration.

Next: try the 8 interactive widgets for kinematic equations, free fall, graphs and relative velocity, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Motion in a Straight Line in NEET 2027?

You can expect 2 to 4 questions from Motion in a Straight Line in NEET 2027, often combined with Motion in a Plane and Laws of Motion. The chapter has high PYQ frequency, with kinematic equations, free fall and graph-based questions being the most commonly tested.

Is Motion in a Straight Line important for NEET Physics?

Yes. It is the gateway to all of NEET mechanics. Every problem you solve in Motion in a Plane, Laws of Motion, Work-Energy and even Rotational Motion uses the kinematic equations and graph reasoning you learn here. Master this chapter first.

What are the three kinematic equations I must memorise?

For motion with uniform acceleration: v = u + at, s = ut + (1/2)at squared, and v squared = u squared + 2as. Here u is initial velocity, v is final velocity, a is acceleration, s is displacement and t is time. These three equations solve almost every NEET problem in this chapter.

What is the difference between distance and displacement?

Distance is the total path length you travel and is always positive. Displacement is the straight-line vector from your starting position to your ending position and can be positive, negative or zero. If you walk 5 m east and then 3 m west, your distance is 8 m but your displacement is 2 m east.

How do I read a velocity-time graph for NEET?

On a velocity-time graph, the slope at any point equals the instantaneous acceleration, and the area under the curve between two times equals the displacement during that interval. A horizontal line means constant velocity; a straight slanted line means uniform acceleration.

How should I solve free fall problems in NEET?

Pick a sign convention first. Take downward as positive (or negative) and stick with it. For an object dropped from rest, u = 0 and a = +g (downward positive). For an object thrown up, u is positive and a is negative g. Then plug into the kinematic equations like any other constant-acceleration problem.

What is relative velocity in one dimension?

The velocity of object A as seen by an observer on object B is v_AB = v_A minus v_B. Both velocities must be measured in the same reference frame. If two cars move in the same direction at 60 and 40 km per hour, their relative velocity is 20 km per hour. If they move opposite, it is 100 km per hour.

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Class 11 · Ch 1

Units and Measurements

Motion in a Plane

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Motion in a Straight LineNEET Physics · Class 11 · NCERT Chapter 2

Introduction

Frame of reference

Position, distance and displacement

Distance vs displacement

Speed and velocity

Average speed

Average velocity

Instantaneous velocity and speed

Acceleration

Average acceleration

Instantaneous acceleration

Uniform vs non-uniform motion

Position-time graphs

Position-time graph reader

Velocity-time graphs

V-T graph: area = displacement

Kinematic equations of motion

The three equations

Displacement in the n-th second

Sign conventions

Free fall under gravity

Object dropped from rest

Object thrown upward

Stopping distance

Relative velocity (one dimension)

Same direction

Opposite directions

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Motion in a Straight Line in NEET 2027?

Is Motion in a Straight Line important for NEET Physics?

What are the three kinematic equations I must memorise?

What is the difference between distance and displacement?

How do I read a velocity-time graph for NEET?

How should I solve free fall problems in NEET?

What is relative velocity in one dimension?

Continue with the next chapter notes

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