Motion in One Dimension: theory

CONCEPT

Distance, Displacement & Average Speed

In one line
Distance is the path actually covered; displacement is the straight change in position. Average speed uses distance, average velocity uses displacement.

POSITION

2.1.1 Position needs a reference

Observers can describe the same point differently, and all be right.
Position needs two things: distance from the observer and direction.
So position is a vector, called the position vector.

Point O seen from A, B and C: three correct descriptions

r = x î + y ĵ + z k̂

Position vector of P in the xy plane

FRAME

2.1.2 Rest, motion and frame of reference

Frame of reference: axes attached to an observer, used to describe an event.
Rest: position does not change with time, relative to the frame.
Motion: position changes with time, relative to the frame.
A tree is at rest for a person on the platform and moving for a person in a train. Rest and motion are relative.

Motion	Coordinates changing	Example
1-D	one	car on a straight road, free fall
2-D	two	car on a circular turn, billiards ball
3-D	three	flying kite, flying insect

PARTICLE

2.1.3 Particle or point object

A body is a particle if its size is negligible compared with its range of motion.
Planets around the sun can be treated as particles.
All parts of the body then have the same displacement, velocity and acceleration.

DISTANCE

2.1.4 Distance

Actual length of the path covered.
Scalar. Dimension [M⁰L¹T⁰]. SI unit: metre.
Going A to B to C below: distance = 3 + 4 = 7 m.

DISPLACEMENT

2.1.5 Displacement

Change in position vector: from initial to final position.
Vector. Dimension [M⁰L¹T⁰]. SI unit: metre.

Distance 7 m, displacement 5 m

AC = √(3² + 4²) = 5 m

Δr = r_B − r_A

Displacement depends only on the end points; s depends on the path

Net displacement is the vector sum: S = S₁ + S₂ + … + Sₙ.

COMPARE

2.1.6 Distance vs displacement

Distance	Displacement
never negative or zero while moving	can be +, − or zero
depends on the path	single value between two points
never decreases with time	can decrease (moving back)
same as \|displacement\| only on a straight path without turning	minimum possible distance

distance ≥ |displacement|

SPEED

2.1.7 Speed

Rate of covering distance. Scalar. Dimension [M⁰L¹T⁻¹]. SI unit: m/s.
Uniform speed: equal distances in equal intervals, however small.
Variable speed: unequal distances in equal intervals.

Uniform and variable speed

AVERAGE SPEED

2.1.8 Average speed

v_av = total distancetotal time = ΔsΔt

Different times t₁, t₂, t₃ at speeds v₁, v₂, v₃:

v_av = v₁t₁ + v₂t₂ + v₃t₃t₁ + t₂ + t₃

Equal times at v₁ and v₂:

v_av = v₁ + v₂2

Different distances d₁, d₂, d₃ at speeds v₁, v₂, v₃:

v_av = (d₁ + d₂ + d₃) ÷ (d₁v₁ + d₂v₂ + d₃v₃)

Two equal distances at v₁ and v₂:

v_av = 2v₁v₂v₁ + v₂

Three equal distances at v₁, v₂ and v₃:

v_av = 3v₁v₂v₃v₁v₂ + v₂v₃ + v₃v₁

AVERAGE VELOCITY

2.1.9 Average velocity vs average speed

v_av = displacementtime taken = ΔrΔt

Average speed is a scalar; average velocity is a vector.
Back at the start: average velocity = 0, but average speed > 0.
Average velocity can be +, − or zero; average speed cannot be negative.
Equal only when moving straight without turning.

average speed ≥ |average velocity|

Practise PYQs on this topicDistance, Displacement & Average Speed · 23 questions with solutions

CONCEPT

Velocity, Acceleration & Variable Motion

In one line
If position is given as a function of time, differentiate once for velocity and again for acceleration.

VELOCITY

2.2.1 Velocity

Rate of change of position. Vector. Dimension [M⁰L¹T⁻¹]. SI unit: m/s.
Uniform velocity: magnitude and direction both constant. Possible only on a straight line without reversing.
Non-uniform velocity: magnitude or direction or both change.

INSTANTANEOUS

2.2.2 Instantaneous velocity and speed

v = lim_Δt→0 ΔrΔt = drdt

speed v = lim_Δt→0 ΔsΔt = dsdt

Instantaneous velocity is always tangential to the path.
Its magnitude equals the instantaneous speed.
Uniform circular motion: speed constant, velocity changing.
Constant velocity: average and instantaneous velocity are equal.

v at each point lies along the tangent to the path

POSITION GIVEN

2.2.3 x(t) given: find v

Differentiate x once.
Example: x = A₀ − A₁t + A₂t² gives:

v = −A₁ + 2A₂t

At t = 0: v = −A₁, speed = A₁.
Moving in x and y: find v_x and v_y separately.

v = √(v_x² + v_y²)

ACCELERATION

2.2.4 Acceleration

Rate of change of velocity. Vector. Dimension [M⁰L¹T⁻²]. SI unit: m/s².
Direction is that of the change in velocity, not of the velocity.
Uniform: magnitude and direction constant. Non-uniform: either changes.

a_av = ΔvΔt = v₂ − v₁Δt

a = dvdt = d²xdt²

What changes	Direction of a	Example
direction only	perpendicular to v	uniform circular motion
magnitude only	along or opposite to v	motion under gravity
both	two components	projectile motion

Angle between a and v

Positive a: velocity increases with time. Zero a: velocity constant.
Negative a (retardation): velocity decreases with time.
A body can have zero velocity and still accelerate (top of a throw).
Constant velocity means speed cannot vary.
Direction of v can change under constant a (projectile).
Force and acceleration: a = Fm.

Angle between a and v: under 90° speeds up, over 90° slows down, exactly 90° keeps speed constant.
In 1-D motion the angle is always 0° or 180°.

USE

2.2.5 Questions from x(t)

Asked	Do this
v at time t	differentiate x once
a at time t	differentiate x twice
time when v = 0	put dxdt = 0
v when a = 0	put d²xdt² = 0, then find v
maximum speed	at the instant a = 0
same velocity of two bodies	equate the two v expressions

Example: x = at² − bt³ gives a = 2a − 6bt.

a = 0 at t = a3b

For a particle that reverses, split the distance at v = 0.
Distance = sum of the parts. Displacement = final x − initial x.

JEE · NEET

Corner

t given as a function of x

t = αx² + βx: invert to find v, then use the chain rule.

a = dvdt = dvdx·dxdt = v dvdx

a = −2αv³

v given as a function of x

v² = u² + kx ⇒ a = k ÷ 2, constant.
v = k√x ⇒ a = k² ÷ 2, and x ∝ t².
Velocity rising at dvdx per metre: a = v dvdx.

Variable acceleration

a = f(t):

v = u + ∫₀ᵗ f(t) dt

s = ut + ∫₀ᵗ (∫ f(t) dt) dt

a = f(x):

v² = u² + 2∫ f(x) dx

a = f(v):

t = ∫ dvf(v) · x = x₀ + ∫ v dvf(v)

v given as a function of t: distance = ∫v dt between the two times.
a = bt from v₀: x = v₀t + bt³ ÷ 6.

Practise PYQs on this topicVelocity, Acceleration & Variable Motion · 50 questions with solutions

CONCEPT

Position-Time Graphs

In one line
The slope of the position-time graph is the velocity. Steeper means faster; a curve means velocity is changing.

SLOPE

2.3.1 Slope gives velocity

Take time t along the x-axis and position along the y-axis.
AB is the x-t graph of a moving particle.

From triangle ABC, tan θ = BC ÷ AC

v = x₂ − x₁t₂ − t₁ = tan θ

Slope of the tangent at a point gives instantaneous velocity.
Two particles with slopes θ₁ and θ₂:

v₁v₂ = tan θ₁tan θ₂

CASES

2.3.2 What each shape means

Position-time graphs and their meaning

Graph	Meaning
parallel to t-axis	θ = 0°: particle at rest
perpendicular to t-axis	θ = 90°: infinite velocity, not possible
straight line, constant slope	uniform velocity, a = 0
bending towards x-axis	θ increasing: v increasing, a > 0
bending towards t-axis	θ decreasing: v decreasing, a < 0
line with θ > 90°	constant negative velocity
straight segments	velocity changes after intervals
two x values at one t	not possible
falls, touches, rises	comes towards origin, then moves away

RULES

2.3.3 Reading any x-t graph

Rising curve: velocity positive. Falling curve: velocity negative.
Top or bottom of a curve: velocity zero.
Same end value as start: average velocity zero.
Later start on the t-axis: that particle started later.
Particle cannot have two positions at one time.
Particle cannot have two velocities at one time, so a circle cannot be a v-t graph.

DISTANCE-TIME

2.3.4 Distance-time graph

Distance never decreases, so this graph never comes back down.
Such a curve is valid only up to A; after A it is not valid.

A distance-time graph can only keep rising

MATCHING

2.3.5 Matching x-t, v-t and a-t

a-t	v-t	x-t
a = 0	flat line	straight line
a constant +	rising line	curve bending up
a constant −	falling line	curve bending down

Start from the a-t graph, then read off v-t, then x-t.
From rest: v-t passes through the origin.
x₁ − x₂ for accelerating and uniform bodies: an upward parabola that starts negative.

Practise PYQs on this topicGraphs of Motion · 36 questions with solutions

CONCEPT

Velocity-Time Graphs

In one line
Slope of the velocity-time graph is acceleration. Area under it is displacement; total area without sign is distance.

AREA

2.4.1 Area gives distance and displacement

Area above the axis is +, area below is −

displacement = A₁ + A₂ + A₃ = ∫v dt

distance = |A₁| + |A₂| + |A₃| = ∫|v| dt

Add each area with its sign for displacement.
Add every area as positive for distance.
Equal areas above and below: displacement zero, average velocity zero.
Use triangles and rectangles; split the graph at every corner.
Cut the graph where it crosses the t-axis.

SLOPE

2.4.2 Slope gives acceleration

From triangle ABC, tan θ = BC ÷ AC

a = v₂ − v₁t₂ − t₁ = tan θ

Slope of the v-t graph gives acceleration, not the area.
Area under a distance-time graph has no use.

CASES

2.4.3 What each shape means

Velocity-time graphs and their meaning

Graph	Meaning
parallel to t-axis	a = 0, constant velocity
perpendicular to t-axis	a = ∞, not possible
straight line	constant acceleration
bending towards v-axis	acceleration increasing
bending towards t-axis	acceleration decreasing
rising, starts below axis	a > 0, u negative
rising, starts above axis	a > 0, u positive
falling, starts above axis	a < 0, u positive
falling, starts at origin	a < 0, u = 0
falling, starts below axis	a < 0, u negative

A-T

2.4.4 Acceleration-time graph

Δv = area under a-t graph = ∫a dt

From rest: v at any time = area up to that time.
Maximum speed is where a falls to zero (a stays positive up to there).
To draw a-t from v-t: rising part gives +a, flat part gives 0, falling part gives −a.
Straight-line segments give flat a-t pieces.

TWO BODIES

2.4.5 Two v-t graphs on one figure

Gap between two bodies = difference of their areas.
The crossing point of two v-t lines is where they have the same velocity; it is not where they meet.
They meet when their areas become equal (same start).

JEE · NEET

Corner

Velocity-displacement graph

Straight line v = v₀(1 − x ÷ x₀): a is not constant.
a = v dvdx gives a straight a-x graph.
Find a at a point: multiply v by the slope there.

a = v dvdx

v² against x

v² = u² + 2ax: straight line.
Slope = 2a; intercept = u².

Practise PYQs on this topicGraphs of Motion · 36 questions with solutions

CONCEPT

Equations of Motion

In one line
For constant acceleration in a straight line, three equations connect u, v, a, s and t. Pick the one without the quantity you do not need.

SYMBOLS

2.5.1 Symbols

u: velocity at t = 0. v: velocity at time t.
a: acceleration. s: displacement in time t.
Zero acceleration: v = u, and s = ut.

EQUATIONS

2.5.2 Constant acceleration

Acceleration constant means magnitude and direction both constant.
The motion is 1-D if u and a are parallel or antiparallel.

v = u + at

s = ut + 12at²

v² = u² + 2as

s = (u + v2) t

No use of	Equation
s	v = u + at
v	s = ut + ½at²
t	v² = u² + 2as
a	s = ½(u + v)t

VECTOR

2.5.3 Vector form

v = u + at

s = ut + 12at²

v·v − u·u = 2a·s

FROM REST

2.5.4 Starting from rest

s = 12at² ⇒ s ∝ t²

Time doubles: distance becomes 4 times.
Distances in t, 2t, 3t are in the ratio 1² : 2² : 3².
If s = kt², then v = 2kt and a = 2k: acceleration is constant.

THREE POINTS

2.5.5 Three points on a line

Passing A, B, C at v₁, v₂, v₃ with uniform a:

BCAB = v₃² − v₂²v₂² − v₁²

MIDPOINT

2.5.6 Velocity at the midpoint

M is midway in distance between A and B

v_mid = √(u² + v²2)

Midpoint of the time interval: the velocity is the mean.

v at half time = u + v2

The engine and last compartment of a train passing a post is the same idea.

STAGES

2.5.7 Accelerate, then decelerate

From rest to rest in total time t

v_max = αβtα + β

s = αβt²2(α + β)

t₁t₂ = βα

TWO BODIES

2.5.8 Two bodies on a line

Meeting: same start, equal distance at the same t.
Uniform a meets uniform v:

t = 2va

Equal velocity: gap is largest when they are starting together.
A (velocity u), B (acceleration a):

gap = u²2a

Chasing from rest: use relative acceleration.

gap = 12(a₁ − a₂)t²

Braking towards each other: gap = initial gap − both stopping distances.

Practise PYQs on this topicEquations of Kinematics · 46 questions with solutions

CONCEPT

Stopping Distance & nth Second

In one line
Same brakes: stopping distance grows as the square of the speed. In the nth second, the distance follows the odd numbers from rest.

STOPPING

2.6.1 Stopping distance

v = 0 in v² = u² − 2as:

s = u²2a ⇒ s ∝ u²

Same car, same brakes: a is the same.
Speed becomes n times: distance becomes n² times.
To stop from nu in the same distance: a becomes n² times.

stopping time t = ua

Speed	Stopping distance
u	s
2u	4s
3u	9s
u ÷ 3	s ÷ 9

TRAP

2.6.2 Vehicle stops before the time is over

First find the stopping time.
If the given time is more than t = u ÷ a, the vehicle has already stopped.
Then distance = u² ÷ 2a. Do not put the longer time in s = ut − ½at².
Example: 15 m/s and 0.3 m/s² stops in 50 s, so at 60 s it has still gone only 375 m.

PENETRATION

2.6.3 Bullet penetrating a block

Constant resistance means constant retardation.
Velocity falls from u to u ÷ n in a distance s.

further distance = sn² − 1

n = 2 ⇒ further = s ÷ 3. n = 3 ⇒ further = s ÷ 8.
Comes from v² = u² − 2as applied twice.

NTH SECOND

2.6.4 Distance in the nth second

s_n = u + a2(2n − 1)

This is the distance in that one second, not up to that second.
Find a from two such values:

s_m − s_n = (m − n)a

Consecutive seconds differ by exactly a.

ODD NUMBERS

2.6.5 From rest: the odd-number rule

Distance in successive seconds from rest

s_n = a2(2n − 1) ⇒ s_n ∝ (2n − 1)

s₁ : s₂ : s₃ : s₄ = 1 : 3 : 5 : 7

Same ratio for any equal time steps, not only 1 s.
Total distance up to n seconds ∝ n².

s_ntotal up to n = 2n − 1n² = 2n − 1n²

Second	1st	2nd	3rd	4th
in that second	1	3	5	7
up to that second	1	4	9	16

Practise PYQs on this topicEquations of Kinematics · 46 questions with solutions

CONCEPT

Free Fall

In one line
A dropped body falls with constant acceleration g whatever its mass. Use the equations of motion with u = 0 and a = +g.

GRAVITY

2.7.1 Acceleration due to gravity

Earth pulls bodies with the force of gravity. The resulting acceleration is g.
Without air resistance, all bodies fall with the same acceleration, whatever their size, weight or composition.
Free fall: falling from a small height (h ≪ R) under gravity alone.

g = 9.8 m/s² = 980 cm/s² = 32 feet/s²

DROPPED

2.7.2 Body dropped from a height

Take the start as origin and downward as positive.
u = 0, a = +g.

Dropped from height h

v = gt

h = 12gt²

v² = 2gh

h_n = g2(2n − 1)

t = √(2hg) · v = √(2gh)

Time and speed do not depend on mass.
A heavy and a light body dropped together land together, with the same speed.

GRAPHS

2.7.3 Graphs for a dropped body

Distance, velocity and acceleration against time

RATIOS

2.7.4 Ratios to remember

h ∝ t²: distances in t, 2t, 3t are 1² : 2² : 3².
h_n ∝ (2n − 1): distances in 1st, 2nd, 3rd second are 1 : 3 : 5.
Equal time steps from rest: 1 : 3 : 5.
h ∝ v²: speed 3 times needs height 9 times, so a further 8h.
t ∝ √h: heights 16 m and 25 m give times 4 : 5.
Times for first half and second half of the height:

t₂ = (√2 − 1)t₁

Times for successive equal distances are in the ratio:

√1 : (√2 − √1) : (√3 − √2) : …

FRACTION

2.7.5 Height left at a fraction of the time

Fallen distance after t, out of total time T and height H:

fallen = H (t²T²) · height left = H(1 − t²T²)

At T ÷ 2: fallen H ÷ 4, height left 3H ÷ 4.
At T ÷ 3: fallen H ÷ 9, height left 8H ÷ 9.

DROPS

2.7.6 Drops falling at regular intervals

A drop lands just as drop number (n + 1) begins: n intervals.
A drop that has fallen for k intervals has fallen:

k²n² × h

LAST SECONDS

2.7.7 Distance in the last seconds

Average speed in an interval = speed at the middle of that interval.
Use this to find total time when the last part is given.
Crossing a window of height s in time t: s = ut + ½gt² gives u at its top.

THROWN DOWN

2.7.8 Thrown downward with speed u

v = u + gt

h = ut + 12gt²

v² = u² + 2gh

h_n = u + g2(2n − 1)

Dropped lands at v₀; thrown with u lands at √(u² + v₀²).

Practise PYQs on this topicMotion Under Gravity · 63 questions with solutions

CONCEPT

Bodies Thrown Vertically

In one line
Going up, g slows the body to rest; coming down, g speeds it back. Rise and fall take equal times, and the speed at any height is the same both ways.

THROWN UP

2.8.1 Equations

Start as origin, upward positive: a = −g.

v = u − gt

h = ut − 12gt²

v² = u² − 2gh

h_n = u − g2(2n − 1)

At the top v = 0, but a = −g still

TOP

2.8.2 Maximum height and time

H = u²2g · t_up = ug

T = 2ug

Time up = time down.
Speed on return = speed of projection.
At any point, speed up = speed down.
Reversing the motion: the last t seconds of the rise cover ½gt².
At half the maximum height: v² = gH.

GRAPHS

2.8.3 Graphs for a thrown body

Displacement, velocity and acceleration against time

v-t is one straight line of slope −g, crossing zero at the top.
Displacement, not speed, is what gives the parabola.

CHANGES

2.8.4 Changes during the flight

Quantity	Ground to top	Whole flight
speed	u	0
velocity	u	2u
momentum	mu	2mu
KE	½mu²	0

BOTH WAYS

2.8.5 At a given height h on the way

Reaches h at t₁ going up and takes t₂ more to return to the ground:

h = 12 g t₁t₂ · u = g(t₁ + t₂)2

Passes a height at times t₁ and t₂: their sum is the total flight time.

TOWER

2.8.6 Thrown from a tower

Take the top as origin and up as positive; the ground is at −h.

−h = ut − 12gt²

Thrown up and thrown down with the same speed: both land with the same speed.
Thrown up takes t₁, thrown down t₂, dropped takes t:

t = √(t₁t₂)

TWO BALLS

2.8.7 Two bodies in the air

Both have acceleration g, so relative acceleration = 0.
Relative motion is uniform: one dropped, one thrown up with u from height h:

t_meet = hu

Released at different times: use each body’s own time.
Same distance fallen at the meeting point.

JUGGLER

2.8.8 Balls thrown at regular intervals

Next ball is thrown when the last is at the top: interval = time of rise.
With n balls per second:

h = g2n²

More than two balls in the air at 2 s intervals needs T > 4 s.

BOUNCE

2.8.9 Elastic bounce

Ball released from h rises back to h each time.

period = 2√(2hg)

Motion is periodic, but it is not simple harmonic motion.
Rebound: average acceleration = change in velocity ÷ contact time.

JEE · NEET

Corner

Dropped from a balloon or helicopter

The object starts with the velocity of the balloon, not zero.
Rising at v: −h = vt − ½gt².
The balloon goes on moving up: its height = h + vt.

With air resistance

Retardation a from air acts opposite to the motion, always.

t_up = ug + a · h = u²2(g + a)

t_down = u√((g + a)(g − a))

Time up < time down, because (g + a) > (g − a).
Height is less than u² ÷ 2g; return speed is less than u.
Part of the KE is used to overcome the friction.

Elastic rebound graphs

v against t: straight lines of slope g, with a sudden jump from −v to +v at each bounce.
h against t: repeated parabolas.

Practise PYQs on this topicMotion Under Gravity · 63 questions with solutions

CONCEPT

Relative Velocity

In one line
Velocity depends on who measures it. Velocity of A with respect to B is the velocity of A minus the velocity of B.

DEFINITION

2.9.1 Relative velocity on a line

v_AB = v_A − v_B

v_BA = v_B − v_A = −v_AB

Two bodies on parallel lines

Direction	Relative speed
same	v_A − v_B
opposite	v_A + v_B

Pick one direction as positive and give each velocity its sign.
Velocity of ground w.r.t. B = −v_B.
Velocity of A w.r.t. ground = its own velocity.

TRAINS

2.9.2 Crossing and overtaking

Train A of length L₁ crossing train B of length L₂ covers L₁ + L₂ at the relative speed.

t = L₁ + L₂v₁ + v₂ (opposite)

t = L₁ + L₂v₁ − v₂ (same direction)

Passenger or small bird crossing a train: only the train length counts.
Train crossing a tunnel or bridge: speed is its own; distance = train + tunnel.
Convert km/h to m/s: multiply by 5 ÷ 18.

CHASE

2.9.3 Chasing

Relative speed needed = gap ÷ time allowed.
Actual speed = relative speed + speed of the one in front.

THROW

2.9.4 Throwing from a moving body

Stone thrown from B: speed v relative to B.
Ground speed of stone = v + v_B (same direction).
Find relative speed with the target using its ground speed.

JEE · NEET

Corner

Escalator

Walk alone t₁; ride alone t₂: speeds add.

t = t₁t₂t₁ + t₂

Buses every T minutes

Gap between buses = v_bus T.
Same direction: gap ÷ (v_bus − v). Opposite: gap ÷ (v_bus + v).

Practise PYQs on this topicRelative Velocity · 13 questions with solutions

EXAM-DAY CARD

Just see this before the exam

v_av = 2v₁v₂v₁ + v₂

equal distances

v = dxdt, a = dvdt

calculus

slope = v, area = s

graphs

v² = u² + 2as

equations

h = ½gt², T = 2ug

gravity

v_AB = v_A − v_B

relative

FORMULA SHEET

Whole chapter on one page

Speed

distance ≥ |displacement|

equal t: v₁ + v₂2

equal d: 2v₁v₂v₁ + v₂

3 equal d: 3v₁v₂v₃v₁v₂ + v₂v₃ + v₃v₁

Calculus

v = dxdt, a = dvdt

a = v dvdx

maximum v when a = 0

x = ∫v dt, v = ∫a dt

Graphs

x-t slope = v

v-t slope = a

v-t area = displacement

a-t area = change in v

Equations

v = u + at

s = ut + ½at²

v² = u² + 2as

s_n = u + (a2)(2n − 1)

Shortcuts

s_stop = u²2a, s ∝ u²

v_mid = √(u² + v²2)

from rest: 1 : 3 : 5 : 7

bullet: further = sn² − 1

Gravity

t = √(2hg), v = √(2gh)

H = u²2g, T = 2ug

t = √(t₁t₂), h = ½gt₁t₂

juggler: h = g2n²

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Equal distances: taking the mean

Using t past the stopping time

v = 0 at the top, so a = 0

v-t area is always distance

CORRECT ✓

Equal distances: 2v₁v₂v₁ + v₂

Find stopping time first: t = ua

a = g even at the top

Signed area = displacement