Motion in One Dimension

Concise Class Notes — For Classroom Use & Printing | Reference: S. Chand Concise Physics (R.K. Bansal)

1. Rest and Motion

Rest Body does not change position with respect to its surroundings over time.

Motion Body changes its position with respect to its surroundings as time passes.

Key Principle Rest and motion are relative — a passenger in a moving train is at rest relative to co-passengers, but in motion relative to someone on the platform.

Types of Motion

Type	Description	Example
Rectilinear	Straight line path	Falling stone, train on track
Circular	Circular path	Earth around Sun, fan blades
Oscillatory	Back and forth about a fixed point	Pendulum, guitar string
Uniform	Equal distances in equal time intervals	Car at constant speed on highway
Non-Uniform	Unequal distances in equal time intervals	Car in city traffic

2. Scalar and Vector Quantities

Scalar — Magnitude Only Distance, Speed, Time, Mass, Energy
No direction needed

Vector — Magnitude + Direction Displacement, Velocity, Acceleration, Force
Represented by arrows

3. Distance and Displacement

Fig. 1 — Distance is the total path length (orange); Displacement is the straight-line shortest path (blue arrow).

Point	Distance	Displacement
Type	Scalar	Vector
Definition	Total path length	Shortest straight-line path from start to end
Value	Always ≥ 0	Can be +, −, or 0
SI Unit	Metre (m)
When equal?	When body moves in a straight line in one direction

Solved Example ICSE A boy walks 4 m East, then 3 m North. Find distance and displacement.

Distance = 4 + 3 = 7 m | Displacement = $\sqrt{4^2+3^2} = \sqrt{25}$ = 5 m (North-East)

4. Speed and Velocity

Fig. 2 — Speed (scalar), Velocity (vector), and Acceleration — formulas and key distinctions.

Speed (Scalar) $$v = \frac{d}{t}$$ Unit: m/s | Always ≥ 0
Avg Speed = Total distance / Total time

Velocity (Vector) $$v = \frac{s}{t}$$ Unit: m/s + direction | Can be −
Avg Velocity = Total displacement / Total time

Unit Conversion — Must Know $1 \text{ km/h} = \dfrac{5}{18} \text{ m/s}$ $1 \text{ m/s} = \dfrac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$

Solved Example Numerical A car travels 150 km in 3 h. Find speed in (a) km/h, (b) m/s.

(a) $v = 150/3 = $ 50 km/h | (b) $50 \times 5/18 = $ 13.9 m/s

5. Acceleration and Retardation

Acceleration Rate of change of velocity: $$a = \frac{v - u}{t}$$ Unit: m/s² | Type: Vector
Positive $a$: speeding up | Negative $a$ (Retardation): slowing down | Zero $a$: constant velocity

Solved Example Numerical A train at 90 km/h comes to rest in 30 s. Find retardation.

$u = 25$ m/s, $v = 0$, $t = 30$ s $a = (0-25)/30 = -0.83$ m/s² Retardation = 0.83 m/s²

6. Equations of Motion (Uniform Acceleration)

Fig. 3 — Three equations of motion derived from the v-t graph. Slope = acceleration; area = displacement.

Notation $u$ = initial velocity | $v$ = final velocity | $a$ = acceleration | $s$ = displacement | $t$ = time

1st Equation: $v = u + at$     (derived from $a = (v-u)/t$)
2nd Equation: $s = ut + \dfrac{1}{2}at^2$     (derived from area under v-t graph)
3rd Equation: $v^2 = u^2 + 2as$     (derived by eliminating $t$)

🧠 Which Equation to Use?

Time given, need $v$ → use $v = u + at$
Time given, need $s$ → use $s = ut + \frac{1}{2}at^2$
No time given/needed → use $v^2 = u^2 + 2as$
Body starts from rest: put $u = 0$ Body stops: put $v = 0$

Solved Examples Numerical ICSE Q1: Bus starts from rest with $a = 2$ m/s². Find velocity and distance after 5 s.

$v = 0 + 2(5) = $ 10 m/s | $s = 0 + \frac{1}{2}(2)(25) = $ 25 m

Q2: Car at 20 m/s brakes with $a = -4$ m/s². Distance to stop?

$0 = 400 + 2(-4)s \implies s = $ 50 m

7. Distance-Time (d-t) Graphs

Fig. 4 — Five types of d-t graphs. Slope = speed. Steeper slope = greater speed. Horizontal = rest.

Graph Shape	Motion Type	Speed
Horizontal line	Rest	Zero
Straight diagonal line	Uniform motion	Constant
Upward curve (concave up)	Acceleration	Increasing
Downward curve (concave down)	Deceleration	Decreasing

Key Rule: Slope of d-t graph = Speed $\left(\text{slope} = \Delta d / \Delta t\right)$

8. Velocity-Time (v-t) Graphs

Fig. 5 — Six types of v-t graphs. Slope = acceleration; area under graph = displacement. Most tested in ICSE!

Graph Shape	Motion Type	Acceleration
Horizontal line (above x-axis)	Uniform velocity	Zero
Straight line going up	Uniform acceleration	Constant positive
Straight line going down	Uniform deceleration	Constant negative
Horizontal on x-axis (v=0)	Rest	Zero
Curved line	Non-uniform motion	Varying

Two Golden Rules Slope of v-t graph = Acceleration ($a$) | Area under v-t graph = Displacement ($s$)

Solved Example Important ICSE A v-t graph: 0→30 m/s in 6 s, constant for 4 s, 30→0 in 3 s. Find total displacement.

Phase 1 (▲): $\frac{1}{2}(6)(30)=90$ m | Phase 2 (□): $(4)(30)=120$ m | Phase 3 (▲): $\frac{1}{2}(3)(30)=45$ m
Total = 90 + 120 + 45 = 255 m

9. Free Fall and Gravity

Fig. 6 — A freely falling body accelerates at g = 9.8 m/s² downward. All objects (regardless of mass) fall at the same rate.

Key Facts $g = 9.8$ m/s² $\approx 10$ m/s² (downward)
All bodies fall at same acceleration
No air resistance in free fall

Equations (dropped from rest, u=0) $v = gt$ $h = \dfrac{1}{2}gt^2$ $v^2 = 2gh$
Thrown upward: $H = u^2/(2g)$, $T = 2u/g$

Solved Example Numerical ICSE Stone dropped from 80 m cliff. Find: (a) time, (b) impact velocity. ($g = 10$ m/s²)

(a) $80 = \frac{1}{2}(10)t^2 \implies t^2 = 16 \implies t = $ 4 s
(b) $v = gt = 10 \times 4 = $ 40 m/s

10. Formula Quick Reference

Quantity	Formula	Unit
Speed	$v = d/t$	m/s (Scalar)
Velocity	$v = s/t$	m/s (Vector)
Acceleration	$a = (v-u)/t$	m/s² (Vector)
1st Equation	$v = u + at$	—
2nd Equation	$s = ut + \frac{1}{2}at^2$	—
3rd Equation	$v^2 = u^2 + 2as$	—
Free fall (u=0)	$h = \frac{1}{2}gt^2$, $v = gt$, $v^2=2gh$	—
Max height (up)	$H = u^2/(2g)$, Total time $T = 2u/g$	—
Conversion	$\times 5/18$ → km/h to m/s $\times 18/5$ → m/s to km/h	—

11. Practice Problems

Q1 — Conceptual ICSE A person walks 3 km East then 4 km North. Find distance and displacement.

Distance = 7 km | Displacement = $\sqrt{9+16}$ = 5 km North-East

Q2 Numerical A car starts from rest with $a = 4$ m/s². Find velocity after 6 s and distance covered.

$v = 0+4(6) = $ 24 m/s | $s = 0+\frac{1}{2}(4)(36) = $ 72 m

Q3 Numerical A ball is thrown up at 20 m/s. Find: (a) max height, (b) time of flight. ($g = 10$ m/s²)

(a) $H = 20^2/(2\times10) = $ 20 m | (b) $T = 2(20)/10 = $ 4 s

Q4 — Train Problem Numerical A train 200 m long passes a pole in 10 s. How long to pass a bridge 800 m long?

Speed = $200/10 = 20$ m/s | Time $= (200+800)/20 = $ 50 s

Q5 — v-t Graph Important ICSE Velocity rises 0→20 m/s in 5 s, stays 20 m/s for 3 s, drops to 0 in 2 s. Find: (a) all accelerations, (b) total distance.

(a) $a_1 = 4$ m/s², $a_2 = 0$, $a_3 = -10$ m/s² (retardation = 10 m/s²)
(b) $\frac{1}{2}(5)(20)+(3)(20)+\frac{1}{2}(2)(20) = 50+60+20 = $ 130 m

12. Common Mistakes & Exam Tips

⚠️ Avoid These Mistakes

Using $a = +g$ for upward throw — must use $a = -g$
Not converting km/h → m/s before substituting
Thinking average speed = $(v_1+v_2)/2$ always — only valid for equal time intervals
Confusing slope of d-t (= speed) with slope of v-t (= acceleration)
Setting $u = 0$ instead of $v = 0$ when a body "comes to rest"
Thinking heavier objects fall faster — all objects fall at same $g$

🎯 ICSE Exam Strategy

Every numerical: Write Given → Find → Formula → Substitution → Answer with unit
Definitions: Always include the formula and SI unit (1–2 marks each)
Derivations: Know all 3 equations — 4 to 5 mark questions
Graphs: Slope and area rules appear every year in ICSE
$g$ value: Use 9.8 m/s² unless the question says to use 10 m/s²