PhysicsICSE Class 9 · Chapter 2

Motion in One Dimension ICSE Class 9: Graphs and Equations of Motion

Updated 15 October 2026 · Fact-checked

Motion in one dimension is motion along a straight line. ICSE Class 9 covers scalars and vectors, how displacement differs from distance and velocity from speed, acceleration, motion graphs, and the equations v = u + at, S = ut + ½at², S = ½(u + v)t and v² = u² + 2aS for uniform acceleration.

What Motion in One Dimension is about

A body is at rest if its position does not change with time relative to its surroundings, and in motion if it does. In one-dimensional motion the body moves along a straight line, for example a car on a straight road or a ball falling vertically.

The CISCE 2028 Class IX syllabus covers: scalar and vector quantities (examples only); rest and motion, including motion under gravity; the difference between distance and displacement, and between speed and velocity; acceleration and its negative, retardation; and graphs of distance against time and velocity against time, with what they tell you; and the four equations of uniformly accelerated motion, derived graphically, with simple numericals. Non-uniform acceleration is excluded.

In the school exam expect definitions and graph-reading MCQs, 3-mark numericals with the equations of motion, 3 to 4-mark graph questions (find acceleration from a slope, distance from an area), and a 3 to 4-mark graphical derivation.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Motion in One Dimension is in the 2. Motion in One Dimension part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Scalars and vectors; distance and displacement

A scalar has only magnitude (size): mass, time, distance, speed, temperature, energy. A vector has magnitude and direction: displacement, velocity, acceleration, force.

Distance is the total length of the path travelled (scalar, never negative).
Displacement is the shortest straight-line distance from the initial to the final position, in a stated direction (vector). It can be zero even when distance is not, for example after a complete round trip.

SI unit of both: metre (m). Magnitude of displacement $\leq$ distance.

Worked example

An athlete runs half a lap of a circular track of radius 7 m. Find the distance covered and the magnitude of the displacement ( $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Distance $= π r = \frac{22}{7} \times 7 = 22$ m.
Displacement $=$ diameter $= 14$ m (from start to the opposite point).

Answer: 22 m; 14 m

Speed and velocity

Speed $= \frac{distance}{time}$ (scalar). Velocity $= \frac{displacement}{time}$ (vector, speed in a stated direction). SI unit: m s⁻¹. Convert: $1$ km h⁻¹ $= \frac{5}{18}$ m s⁻¹.

Uniform speed (or velocity): equal distances (displacements) in equal intervals of time.
Average speed $= \frac{total distance}{total time}$ . It is not the average of the speeds unless the times are equal.

A body moving in a circle at constant speed has changing velocity, because its direction keeps changing.

Worked example

A car travels 60 km at 30 km h⁻¹ and the next 60 km at 60 km h⁻¹. Find its average speed.

Try it first, then show the solutionSolution

Times: $\frac{60}{30} = 2$ h and $\frac{60}{60} = 1$ h.
Average speed $= \frac{120}{3} = 40$ km h⁻¹ (not 45).

Answer: 40 km h⁻¹

Acceleration and retardation

Acceleration is the rate of change of velocity: $a = \frac{v - u}{t}$ , where $u$ is the initial and $v$ the final velocity. SI unit: m s⁻².

Uniform acceleration: velocity changes by equal amounts in equal intervals of time (the only case in this syllabus).
Retardation (deceleration) is negative acceleration: the velocity decreases. If a car slows from 20 m s⁻¹ to 0 in 10 s, $a = - 2$ m s⁻², a retardation of 2 m s⁻².

Motion under gravity: near the Earth, a freely falling body has a uniform acceleration $g \approx 9.8$ m s⁻² downwards (often taken as 10 m s⁻² for simple numbers), whatever its mass, if air resistance is ignored.

Worked example

A train moving at 72 km h⁻¹ is brought to rest in 10 s by applying brakes uniformly. Find the retardation and the distance travelled while braking.

Try it first, then show the solutionSolution

$u = 72 \times \frac{5}{18} = 20$ m s⁻¹, $v = 0$ .
$a = \frac{0 - 20}{10} = - 2$ m s⁻²: retardation 2 m s⁻².
$S = \frac{1}{2} (u + v) t = \frac{1}{2} (20) (10) = 100$ m.

Answer: 2 m s⁻²; 100 m

Distance–time graphs

Time on the x-axis, distance on the y-axis.

Horizontal line: the body is at rest.
Straight sloping line: uniform speed; slope = speed. A steeper line means a faster body.
Curve getting steeper: speed increasing (accelerating); curve flattening: speed decreasing.

A distance–time graph can never slope downwards (distance never decreases), and can never be vertical (that would need infinite speed).

Worked example

A distance–time graph is a straight line from $(0 s, 0 m)$ to $(8 s, 40 m)$ , then horizontal until 12 s. Describe the motion.

Try it first, then show the solutionSolution

0 to 8 s: uniform speed $= \frac{40}{8} = 5$ m s⁻¹.
8 to 12 s: horizontal, so at rest 40 m from the start.

Answer: 5 m s⁻¹ for 8 s, then at rest for 4 s

Velocity–time graphs

Time on the x-axis, velocity on the y-axis.

Horizontal line: uniform velocity (zero acceleration).
Straight line sloping up: uniform acceleration; slope = acceleration.
Straight line sloping down: uniform retardation (negative slope).
Area under the graph = displacement (distance, for motion in one direction).

Split the area into rectangles and triangles (or use the trapezium formula).

Worked example

A car accelerates uniformly from rest to 20 m s⁻¹ in 4 s, moves at 20 m s⁻¹ for 6 s, then slows uniformly to rest in 5 s. Find the accelerations and the total distance.

Try it first, then show the solutionSolution

$a_{1} = \frac{20}{4} = 5$ m s⁻²; $a_{3} = \frac{0 - 20}{5} = - 4$ m s⁻².
Areas: $\frac{1}{2} (4) (20) = 40$ m; $6 \times 20 = 120$ m; $\frac{1}{2} (5) (20) = 50$ m.
Total distance $= 210$ m.

Answer: 5 m s⁻², −4 m s⁻²; 210 m

Equations of motion (graphical derivation)

Draw the velocity–time graph of uniform acceleration: a straight line from $A (0, u)$ to $B (t, v)$ . Drop $B C$ to the time axis and draw $A D$ parallel to the time axis, meeting $B C$ at $D$ ; $O$ is the origin.

1. ** $v = u + a t$ **: slope $= \frac{B D}{A D} = \frac{v - u}{t} = a$ , so $v = u + a t$ .
2. ** $S = \frac{1}{2} (u + v) t$ **: $S$ = area of trapezium $O A B C$ $= \frac{1}{2} (O A + B C) \times O C$ .
3. ** $S = u t + \frac{1}{2} a t^{2}$ **: area = rectangle $O A D C$ ( $u t$ ) + triangle $A B D$ ( $\frac{1}{2} \times t \times a t$ ).
4. ** $v^{2} = u^{2} + 2 a S$ **: $S = \frac{1}{2} (u + v) t$ and $t = \frac{v - u}{a}$ , so $S = \frac{( v + u ) ( v - u )}{2 a}$ , giving $v^{2} - u^{2} = 2 a S$ .

For free fall, use $a = + g$ (downwards positive); for a body thrown up, $a = - g$ while it rises.

Worked example

A ball is thrown vertically upwards at 20 m s⁻¹. Find the maximum height and the time to reach it ( $g = 10$ m s⁻²).

Try it first, then show the solutionSolution

At the top $v = 0$ ; $a = - 10$ m s⁻².
$v^{2} = u^{2} + 2 a S$ : $0 = 400 - 20 S$ , so $S = 20$ m.
$v = u + a t$ : $0 = 20 - 10 t$ , so $t = 2$ s.

Answer: 20 m; 2 s

Physics in real life: Motion in One Dimension

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Braking distance

Suppose a car is moving at 30 m s⁻¹ (108 km h⁻¹) on a highway and the driver brakes, producing a uniform retardation of 5 m s⁻².

How far does the car travel before stopping, and how long does it take?

Try it first, then show the solutionSolution

$v^{2} = u^{2} + 2 a S$ : $0 = 900 - 10 S$ , so $S = 90$ m.
$v = u + a t$ : $0 = 30 - 5 t$ , so $t = 6$ s.

Answer: 90 m in 6 s.

Memory hook: Double the speed, four times the braking distance (S ∝ u²).

Free fall

Imagine a stone is dropped from the top of a 45 m high building (ignore air resistance; take $g = 10$ m s⁻²).

How long does it take to reach the ground, and how fast is it going then?

Try it first, then show the solutionSolution

$S = \frac{1}{2} g t^{2}$ : $45 = 5 t^{2}$ , so $t = 3$ s.
$v = g t = 30$ m s⁻¹.

Answer: 3 s; 30 m s⁻¹.

Memory hook: Dropped means u = 0.

Average speed of a trip

Suppose you cycle 6 km to a friend's house at 12 km h⁻¹ and come back the same way at 18 km h⁻¹.

What is your average speed and your average velocity for the whole trip?

Try it first, then show the solutionSolution

Times: $0.5$ h and $\frac{1}{3}$ h; total $\frac{5}{6}$ h; distance 12 km.
Average speed $= 12 \div \frac{5}{6} = 14.4$ km h⁻¹; displacement is 0, so average velocity is 0.

Answer: 14.4 km h⁻¹; 0.

Memory hook: Speed counts the path; velocity counts only start to finish.

Topics in this chapter

Motion in One Dimension formula sheet

Speed and velocity: $speed = \frac{distance}{time}; velocity = \frac{displacement}{time}$; m s⁻¹; 1 km h⁻¹ = 5/18 m s⁻¹.
Acceleration: $a = \frac{v - u}{t}$; Negative a = retardation.
First equation: $v = u + a t$; No S.
Second equation: $S = u t + \frac{1}{2} a t^{2}$; No v.
Third equation: $v^{2} = u^{2} + 2 a S$; No t.
Average velocity form: $S = \frac{1}{2} (u + v) t$; No a.
Graph facts: $slope_{s - t} = v; slope_{v - t} = a$; Area under v–t = displacement.

Motion in One Dimension: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Velocity–time graph for deriving the equations of motion

Derive v = u + at, S = ut + ½at², S = ½(u + v)t or v² = u² + 2aS graphically; 3 to 4 marks.

Draw

Axes: time t on the x-axis, velocity v on the y-axis, origin O.
Mark A at (0, u) on the velocity axis and B at (t, v).
Join AB (a straight line: uniform acceleration).
Drop BC perpendicular to the time axis; draw AD parallel to the time axis meeting BC at D.

Label

O
A (velocity u)
B (velocity v at time t)
C (time t)
D
OA = u, BC = v, OC = AD = t, BD = v − u

Watch out: Name each area (rectangle OADC, triangle ABD, trapezium OABC) and write its formula before substituting.

How the school exam asks Motion in One Dimension

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Definitions and differences	Distinguish distance and displacement, speed and velocity; classify quantities as scalar or vector.	Vector = magnitude + direction; give one example of each.
Graph interpretation	Describe motion from a distance–time or velocity–time graph; find speed, acceleration or distance.	Slope of s–t = speed; slope of v–t = acceleration; area under v–t = displacement.
Equations of motion	Find v, S, a or t for uniform acceleration, braking or free fall.	List u, v, a, t, S; pick the equation without the unknown you don't need.
Graphical derivation	Derive one of the four equations from a velocity–time graph.	Labelled graph; slope or area; algebra.
Motion under gravity	Dropped or thrown-up objects: maximum height, time of flight, velocity on reaching the ground.	Use a = ±g consistently; at the top v = 0.
Average speed	Two parts of a journey at different speeds.	Total distance ÷ total time.

Tricks to remember Motion in One Dimension

Short memory aids. Say each one aloud once, then use it on a question the same day.

Choose by the missing letter
v = u + at has no S; S = ut + ½at² has no v; v² = u² + 2aS has no t; S = ½(u + v)t has no a.
Use it when: Picking the right equation.
Slope speeds, area travels
Slope of a graph gives a rate (speed or acceleration); area under v–t gives the distance travelled.
Use it when: Graph questions.
Round trip, zero trip
Back where you started? Displacement zero, distance not.
Use it when: Distance vs displacement.
Top of the throw
At the highest point the ball stops for an instant: v = 0.
Use it when: Vertical-throw problems.

Motion in One Dimension: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Slope meaning · 1 mark · MCQExam level

The slope of a velocity–time graph gives:
(a) distance
(b) displacement
(c) acceleration
(d) speed

Try it first, then show the solutionSolution

Slope $= \frac{Δ v}{Δ t}$ , the acceleration.

Answer: (c)

Example 2: Starting from rest · 2 marks · Short answerExam level

A car starts from rest and accelerates uniformly at 2 m s⁻² for 10 s. Find its final velocity and the distance covered.

Try it first, then show the solutionSolution

$v = 0 + 2 \times 10 = 20$ m s⁻¹.
$S = 0 + \frac{1}{2} \times 2 \times 100 = 100$ m.

Answer: 20 m s⁻¹; 100 m

Example 3: Reading a velocity–time graph · 3 marks · Short answerExam level

A velocity–time graph is a straight line from (0 s, 5 m s⁻¹) to (10 s, 25 m s⁻¹). Find the acceleration and the distance covered in 10 s.

Try it first, then show the solutionSolution

$a = \frac{25 - 5}{10} = 2$ m s⁻².
Area (trapezium) $= \frac{1}{2} (5 + 25) \times 10 = 150$ m.
Check: $S = 5 \times 10 + \frac{1}{2} \times 2 \times 100 = 150$ m ✓.

Answer: 2 m s⁻²; 150 m

Example 4: Thrown up and back · 3 marks · Short answerHarder exam level

A ball is thrown vertically upwards with a velocity of 30 m s⁻¹ ( $g = 10$ m s⁻²). Find (a) the maximum height (b) the total time before it returns to the thrower's hand (c) its velocity on return.

Try it first, then show the solutionSolution

(a) $0 = 900 - 20 S$ , $S = 45$ m.
(b) Time up $= \frac{30}{10} = 3$ s; time down is also 3 s; total 6 s.
(c) 30 m s⁻¹ downwards (same speed, opposite direction).

Answer: 45 m; 6 s; 30 m s⁻¹ downwards

Example 5: Finding u and a · 4 marks · Long answerHarder exam level

A body moving with uniform acceleration covers 16 m in the first 2 s and 70 m in the first 5 s. Find its initial velocity and acceleration.

Try it first, then show the solutionSolution

$S = u t + \frac{1}{2} a t^{2}$ : $2 u + 2 a = 16$ , i.e. $u + a = 8$ .
$5 u + 12.5 a = 70$ .
Substitute $u = 8 - a$ : $40 - 5 a + 12.5 a = 70$ , so $7.5 a = 30$ and $a = 4$ m s⁻².
$u = 4$ m s⁻¹.

Answer: u = 4 m s⁻¹; a = 4 m s⁻²

Example 6: Deriving S = ut + ½at² · 3 marks · Long answerHarder exam level

Derive $S = u t + \frac{1}{2} a t^{2}$ from a velocity–time graph for uniformly accelerated motion.

Try it first, then show the solutionSolution

Draw the v–t graph $A B$ from $(0, u)$ to $(t, v)$ ; $B C ⊥$ time axis; $A D ∥$ time axis.
Displacement $S$ = area $O A B C$ = area of rectangle $O A D C$ + area of triangle $A B D$ .
$= u \times t + \frac{1}{2} \times t \times (v - u)$ , and $v - u = a t$ .
So $S = u t + \frac{1}{2} a t^{2}$ .

Answer: Derived.

Example 7: Two cars on one road · 4 marks · Long answerHOTS / competency

Car A starts from rest at a signal with uniform acceleration 2 m s⁻². At the same moment car B passes the signal at a uniform 20 m s⁻¹ in the same direction. When and where does A catch up with B, and what is A's speed then?

Try it first, then show the solutionSolution

A: $S = \frac{1}{2} \times 2 \times t^{2} = t^{2}$ . B: $S = 20 t$ .
Equal: $t^{2} = 20 t \Rightarrow t = 20$ s (t = 0 is the start).
Distance $= 400$ m from the signal; A's speed $= 2 \times 20 = 40$ m s⁻¹.

Answer: After 20 s, 400 m from the signal; 40 m s⁻¹

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 2 + 3 + 4 marks.

A metro train between stations

A metro train starts from rest at station P and accelerates uniformly to 18 m s⁻¹ in 12 s. It then runs at 18 m s⁻¹ for 40 s and then slows uniformly, stopping at station Q 9 s later.

(i) · 2 marks
Find the acceleration and the retardation.
Answer
Answer: 1.5 m s⁻²; 2 m s⁻²
$a = \frac{18}{12} = 1.5$ m s⁻². Retardation $= \frac{18}{9} = 2$ m s⁻².
(ii) · 3 marks
Sketch the velocity–time graph and find the distance between P and Q.
Answer
Answer: 909 m
Trapezium-shaped graph: $\frac{1}{2} (12) (18) = 108$ m; $40 \times 18 = 720$ m; $\frac{1}{2} (9) (18) = 81$ m. Total $= 909$ m.
(iii) · 4 marks
Find the average speed for the whole journey. Why is it less than 18 m s⁻¹?
Answer
Answer: About 14.9 m s⁻¹
Total time $= 12 + 40 + 9 = 61$ s. Average speed $= \frac{909}{61} \approx 14.9$ m s⁻¹. It is less than 18 m s⁻¹ because the train moves more slowly than 18 m s⁻¹ while speeding up and slowing down.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Distinguish between distance and displacement, giving one example where the displacement is zero but the distance is not. (2 marks)

Step	Marks
Distance: total path length, a scalar; displacement: shortest distance from start to finish in a given direction, a vector	1
Example: one full lap of a circular track: distance $= 2 π r$ , displacement $= 0$	1

A scooter moving at 54 km h⁻¹ is brought to rest in 5 s. Find the retardation and the distance covered while stopping. (3 marks)

Step	Marks
$u = 54 \times \frac{5}{18} = 15$ m s⁻¹	1
$a = \frac{0 - 15}{5} = - 3$ m s⁻²: retardation 3 m s⁻²	1
$S = \frac{1}{2} (15 + 0) \times 5 = 37.5$ m	1

Derive $v^{2} = u^{2} + 2 a S$ using a velocity–time graph. (4 marks)

Step	Marks
Labelled v–t graph: line from $(0, u)$ to $(t, v)$ , trapezium $O A B C$	1
$S$ = area of trapezium $= \frac{1}{2} (u + v) t$	1
Slope $a = \frac{v - u}{t}$ , so $t = \frac{v - u}{a}$	1
$S = \frac{( v + u ) ( v - u )}{2 a}$ , so $v^{2} = u^{2} + 2 a S$	1

Where students lose marks in Motion in One Dimension

The slips examiners see most often, and what to write instead.

Mistake: Not converting km h⁻¹ to m s⁻¹
Mixing km h⁻¹ with seconds and metres gives nonsense.
Wrong working
$a = \frac{72}{10} = 7.2$ m s⁻²
Right working
$72$ km h⁻¹ $= 20$ m s⁻¹; $a = 2$ m s⁻²
Do this: Multiply km h⁻¹ by 5/18 first.
Usually costs: The answer.
Mistake: Averaging speeds
Average speed is total distance ÷ total time, not (v₁ + v₂) ÷ 2, unless the times are equal.
Do this: Find the time for each part first.
Usually costs: 1 to 2 marks.
Mistake: Wrong sign for retardation or upward throws
Using $a = + g$ for a body moving up makes it speed up for ever.
Do this: Choose a positive direction and keep to it; opposite quantities are negative.
Usually costs: The answer.
Mistake: Unlabelled graph in a derivation
Without labels for $u$ , $v$ , $t$ and the areas, the derivation cannot be followed.
Do this: Label every point and length used.
Usually costs: 1 mark.
Mistake: Using the equations for non-uniform acceleration
The equations hold only when acceleration is uniform.
Do this: Check the question says 'uniform acceleration' or 'uniformly'.
Usually costs: Method marks.

Practice questions on Motion in One Dimension

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
Which of the following is a vector quantity?
(a) speed
(b) distance
(c) displacement
(d) time
Hint
Needs direction.
Answer
(c)
Q2 · 1 mark · MCQExam level
A body moving along a straight line has a distance–time graph that is a curve getting steeper. Its motion is:
(a) uniform
(b) at rest
(c) accelerated
(d) retarded
Hint
Slope = speed.
Answer
(c)
Q3 · 2 marks · Short answerExam level
A cyclist moves at 18 km h⁻¹ for 20 minutes. Find the distance in metres.
Hint
18 km h⁻¹ = 5 m s⁻¹.
Answer
$5 \times 1200 = 6000$ m.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): A body moving in a circle with constant speed is accelerating.
Reason (R): Its velocity changes because its direction changes.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Acceleration = change of velocity.
Answer
(c)
Q5 · 3 marks · Short answerHarder exam level
A stone is dropped from a height of 80 m. Find the time it takes to reach the ground and its velocity just before impact ( $g = 10$ m s⁻²).
Hint
u = 0.
Answer
$80 = 5 t^{2}$ , $t = 4$ s; $v = 40$ m s⁻¹.
Q6 · 3 marks · Short answerHarder exam level
A bus accelerates from 10 m s⁻¹ to 25 m s⁻¹ while covering 175 m. Find its acceleration and the time taken.
Hint
v² = u² + 2aS.
Answer
$625 = 100 + 350 a$ , $a = 1.5$ m s⁻²; $t = \frac{15}{1.5} = 10$ s.
Q7 · 3 marks · Short answerHarder exam level
A velocity–time graph shows a body moving at 10 m s⁻¹ for 5 s, then accelerating uniformly to 30 m s⁻¹ in the next 5 s. Find the total distance and the average velocity over 10 s.
Hint
Rectangle + trapezium.
Answer
$50 + \frac{1}{2} (10 + 30) (5) = 50 + 100 = 150$ m; average $= 15$ m s⁻¹.
Q8 · 3 marks · Short answerHarder exam level
A ball thrown vertically upwards reaches a height of 20 m. With what velocity was it thrown, and how long was it in the air ( $g = 10$ m s⁻²)?
Hint
v = 0 at the top.
Answer
$0 = u^{2} - 400$ , $u = 20$ m s⁻¹; time up 2 s, total 4 s.
Q9 · 2 marks · Short answerExam level
Derive $v = u + a t$ from a velocity–time graph.
Hint
Slope of the graph.
Answer
Slope of the line from $(0, u)$ to $(t, v)$ is $\frac{v - u}{t}$ , which is the acceleration $a$ ; so $v = u + a t$ .
Q10 · 3 marks · Short answerHarder exam level
A train starting from rest reaches 72 km h⁻¹ in 2 minutes. Find its acceleration and the distance covered in this time.
Hint
20 m s⁻¹ in 120 s.
Answer
$a = \frac{20}{120} = \frac{1}{6} \approx 0.17$ m s⁻²; $S = \frac{1}{2} (0 + 20) (120) = 1200$ m.
Q11 · 4 marks · Long answerHOTS / competency
A car moving at 20 m s⁻¹ sees an obstacle 50 m ahead. The driver takes 0.5 s to react and then brakes with a retardation of 5 m s⁻². Does the car stop before the obstacle? Show working.
Hint
Distance during reaction + braking distance.
Answer
Reaction distance $= 20 \times 0.5 = 10$ m. Braking: $0 = 400 - 10 S$ , $S = 40$ m. Total $= 50$ m: it stops just at the obstacle, with no margin.
Q12 · 3 marks · Short answerHOTS / competency
Explain, using the area under a velocity–time graph, why doubling the speed of a car quadruples its braking distance for the same retardation.
Hint
Triangle area with base t ∝ u.
Answer
The v–t graph while braking is a triangle of height $u$ and base $t = \frac{u}{a}$ ; area $= \frac{u ^{2}}{2 a}$ . Doubling $u$ doubles both height and base, so the area (distance) becomes 4 times.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Give two examples each of scalar and vector quantities.Show answer
Scalars: mass, speed. Vectors: displacement, velocity.
2.SI unit of acceleration?Show answer
m s⁻².
3.What does the area under a velocity–time graph give?Show answer
Displacement (distance for one-way motion).
4.What does a horizontal distance–time graph show?Show answer
The body is at rest.
5.Which equation has no time in it?Show answer
v² = u² + 2aS.
6.Value of g near the Earth?Show answer
About 9.8 m s⁻².
7.Convert 36 km h⁻¹ to m s⁻¹.Show answer
10 m s⁻¹.
8.What is retardation?Show answer
Negative acceleration: the velocity decreases with time.

Best order to study it

Learn scalars and vectors, distance and displacement, speed and velocity.
Learn acceleration, retardation and motion under gravity.
Practise reading distance–time and velocity–time graphs.
Learn the four graphical derivations.
Practise numericals: braking, free fall, thrown-up bodies, two-stage motion.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you give the difference between distance and displacement?
Do you convert km h⁻¹ to m s⁻¹?
Can you find acceleration from a v–t slope and distance from its area?
Can you derive all four equations graphically?
Do you choose the equation by the missing quantity?
Do you use v = 0 at the top of a vertical throw?
Do you find average speed as total distance ÷ total time?

Motion in One Dimension: frequently asked questions

Is non-uniform acceleration in the Class IX syllabus?

No. CISCE excludes non-uniform acceleration; all numericals use uniform acceleration.

Do I have to derive the equations of motion?

Yes. The syllabus says the equations are to be derived graphically, using a velocity–time graph.

Can speed be negative?

No. Speed is a scalar magnitude. Velocity can be negative, meaning it points in the opposite direction to the one chosen as positive.

Should I use g = 9.8 or 10 m s⁻²?

Use the value the question gives. If none is given, 9.8 m s⁻² is the standard value.

Why does a heavier stone not fall faster?

Without air resistance, all bodies fall with the same acceleration g, whatever their mass.

Based on: CISCE ICSE 2028 Science (52) Physics syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028