Short Intro

“Describing Motion Around Us” chapter explains the concepts of motion, distance, displacement, speed, velocity, acceleration, graphs of motion, kinematic equations, and circular motion. This chapter is very important for exams because it contains conceptual questions, graph-based numericals, and formula applications. Below are complete Unicode solutions prepared for uploading on www.mymockmate.com.

Quick Information Box

Topic	Details
Chapter Name	Describing Motion Around Us
Subject	Science
Grade	Class 9
Main Topics	Motion, Velocity, Acceleration
Important Concepts	Graphs & Kinematic Equations
Exam Importance	Very High
Numerical Problems	Included

Concepts Used (Topics Covered)

• Distance and displacement
• Speed and velocity
• Average velocity
• Average acceleration
• Position-time graph
• Velocity-time graph
• Uniform motion
• Non-uniform motion
• Kinematic equations
• Uniform circular motion
• Motion in a plane

Important Formulas

Average Speed Formula

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Average Velocity Formula

$v_{av}=\frac{s}{t}$

Acceleration Formula

$a=\frac{v-u}{t}$

First Equation of Motion

$v=u+at$

Second Equation of Motion

$s=ut+\frac{1}{2}at^2$

Third Equation of Motion

$v^2=u^2+2as$

Questions & Step-by-step Solutions with Explanation

Question 1

My father went to a shop from home which is located at a distance
of 250 m on a straight road. On reaching there, he discovered that
he forgot to carry a cloth bag. He came home to take it, went to the
shop again, bought provisions and came back home. How much was
the total distance travelled by him? What was his displacement from
home?

Total distance and displacement of father

Given

Distance between home and shop = 250 m

Solution

Father travelled:

• Home → Shop = 250 m
• Shop → Home = 250 m
• Home → Shop = 250 m
• Shop → Home = 250 m

Total Distance

= 250 + 250 + 250 + 250
= 1000 m

Displacement

Initial and final positions are same.

Answer

• Total distance travelled = 1000 m
• Displacement = 0 m

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Question 2

A student runs from the ground floor to the fourth floor of a school
building to collect a book and then comes down to their classroom on
the second floor. If the height of each floor is 3 m, find:
(i) the total vertical distance travelled, and
(ii) their displacement from the starting point

Vertical distance and displacement

Given

Height of each floor = 3 m

Solution

Ground floor to fourth floor:
= 4 × 3 = 12 m

Fourth floor to second floor:
= 2 × 3 = 6 m

Total Distance

= 12 + 6
= 18 m

Displacement

From ground floor to second floor:
= 2 × 3
= 6 m upward

Answer

• Total vertical distance = 18 m
• Displacement = 6 m upward

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Question 3

A girl is riding her scooter and finds that its speedometer reading is
constant. Is it possible for her scooter to be accelerating and if so, how?

Can scooter accelerate with constant speedometer reading?

Answer

Yes.

Explanation

If the scooter changes direction while moving at constant speed, velocity changes and acceleration occurs.

Example:

• Moving on a circular road
• Taking a turn

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Question 4

A car starts from rest and its velocity reaches 24 m s–1 in 6 s. Find the
average acceleration and the distance travelled in these 6 s.

Find acceleration and distance travelled

Given

Initial velocity:
$u=0$

Final velocity:
$v=24\,m/s$

Time:
$t=6\,s$

Step 1: Acceleration

Using formula:
$a=\frac{v-u}{t}$

Substitute values:
$a=\frac{24-0}{6}=4\,m/s^2$

Step 2: Distance

Using:
$s=ut+\frac{1}{2}at^2$

Substitute values:
$s=0+\frac{1}{2}(4)(6^2)=72\,m$

Answer

• Acceleration = 4 m/s²
• Distance travelled = 72 m

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Question 5

A motorbike moving with initial velocity 28 m s^–1 and constant
acceleration stops after travelling 98 m. Find the acceleration of the
motorbike and the time taken to come to a stop.

Motorbike stopping problem

Given

Initial velocity:
$u=28\,m/s$

Final velocity:
$v=0$

Distance:
$s=98\,m$

Step 1: Find acceleration

Using:
$v^2=u^2+2as$

Substitute values:
$0=(28)^2+2(a)(98)$

$a=-4\,m/s^2$

Step 2: Find time

Using:
$v=u+at$

Substitute:
$0=28-4t$

$t=7\,s$

Answer

• Acceleration = –4 m/s²
• Time taken = 7 s

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Question 6

Following Figure shows a position-time graph of two objects A and B that are
moving along the parallel tracks in the same direction. Do objects
A and B ever have equal velocity? Justify your answer

Equal velocity of objects A and B

Answer

No.

Explanation

The slopes of both position-time graphs are different. Different slopes indicate different velocities.

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Question 7

A graph in Fig. 4.28 shows the change in position with time for two
objects A and B moving in a straight line from 0 to 10 seconds. Choose
the correct option(s).
(i) The average velocity of both over the 10 s time interval is equal
since they have the same initial and final positions.
(ii) The average speeds of both over the 10 s time interval are equal
since both cover equal distance in equal time.
(iii) The average speed of A over the 10 s time interval is lower than
that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than
that of B since B’s speed is lower than A’s in some segments.

Correct options from graph

Answer

Correct options:

• (i)
• (iii)

Explanation

Average velocity depends on displacement, while average speed depends on total distance travelled.

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Question 8

A truck driver driving at the speed of 54 km h^–1 notices a road sign
with a speed limit of 40 km h^–1 (Fig. 4.29) for trucks. He slows down
to 36 km h^–1 in 36 s. What was the distance travelled by him during this
time? Assume the acceleration to be constant while slowing down.

Truck slowing down

Given

Initial velocity:
54 km/h = 15 m/s

Final velocity:
36 km/h = 10 m/s

Time:
36 s

Distance

Using:
$s=\frac{(u+v)}{2}t$

Substitute:
$s=\frac{(15+10)}{2}\times36=450\,m$

Answer

Distance travelled = 450 m

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Question 9

A car starts from rest and accelerates uniformly to 20 m s–1 in
5 seconds. It then travels at 20 m s^–1 for 10 seconds and finally
applies the brake (with uniform acceleration) to stop in 6 seconds.
Find the total distance travelled.

Total distance travelled by car

Part 1: Acceleration phase

$u=0,\ v=20\,m/s,\ t=5\,s$

Distance:
$s_1=\frac{(0+20)}{2}\times5=50\,m$

Part 2: Constant velocity

$s_2=20\times10=200\,m$

Part 3: Braking

$s_3=\frac{(20+0)}{2}\times6=60\,m$

Total Distance

$s=50+200+60=310\,m$

Answer

Total distance travelled = 310 m

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Question 10

A bus is travelling at 36 km h^–1 when the driver sees an obstacle
30 m ahead. The driver takes 0.5 seconds to react before pressing the
brake. Once the brake is applied, the velocity of the bus reduces with
constant acceleration of 2.5 m s^–2. Will the bus be able to stop before
reaching the obstacle?

Will bus stop before obstacle?

Given

Initial velocity:
36 km/h = 10 m/s

Reaction time:
0.5 s

Distance during reaction time

$s=vt=10\times0.5=5\,m$

Braking distance

Using:
$v^2=u^2+2as$

Substitute:
$0=(10)^2+2(-2.5)s$

$s=20\,m$

Total stopping distance

$5+20=25\,m$

Answer

Yes, the bus stops before obstacle because stopping distance is 25 m which is less than 30 m.

Question 11

A student said, “The Earth moves around the Sun”. In this context,
discuss whether an object kept on the Earth can be considered to be
at rest.

Solution

Yes, an object kept on Earth can be considered at rest with respect to Earth, because it does not change its position relative to Earth.

But with respect to the Sun, the object is in motion, because Earth itself moves around the Sun.

Conclusion: Rest and motion are relative terms. An object may be at rest for one observer and in motion for another.

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Question 12

The velocity-time graph from 0 s to 120 s for a cyclist is shown in Following Figure
Shade the areas (in different colours) representing the displacement of
the cyclist
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s
time interval

Solution

(i) Constant velocity

From 20 s to 100 s, velocity is constant.

(ii) Decreasing velocity

From 100 s to 120 s, velocity is decreasing.

Displacement

Area under velocity-time graph:

From 0 to 20 s:
= 1/2 × 20 × 3 = 30 m

From 20 to 100 s:
= 80 × 3 = 240 m

From 100 to 120 s:
= 1/2 × (3 + 2) × 20 = 50 m

Total displacement:
= 30 + 240 + 50 = 320 m

Average acceleration

Initial velocity = 0 m/s
Final velocity = 2 m/s
Time = 120 s

Average acceleration = (2 – 0) / 120
= 1/60 m/s²
= 0.0167 m/s²

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Question 13

A girl is preparing for her first marathon by running on a straight
road. She uses a smartwatch to calculate her running speed at
different intervals. The graph (Shown in Following Figure) depicts her velocity versus
time. Estimate the distance she ran based on the graph.

Solution

Distance = area under velocity-time graph.

Approximate area from graph:

0 to 2 h:
= 1/2 × (7 + 7.5) × 2 = 14.5 km

2 to 4 h:
= 1/2 × (7.5 + 7) × 2 = 14.5 km

4 to 6 h:
= 1/2 × (7 + 5.5) × 2 = 12.5 km

6 to 7 h:
= 1/2 × (5.5 + 4.5) × 1 = 5 km

Total distance ≈ 46.5 km

So, the girl runs approximately 46 km to 47 km.

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Question 14

On entering a state highway, a car continues to move with a constant
velocity of 6 m s^–1 for 2 minutes and then accelerates with a constant
acceleration 1 m s^–2 for 6 seconds. Find the displacement of the car
on the state highway in the 2 min 6 s time interval by drawing a
velocity-time graph for its motion.

Solution

Given:
Initial velocity = 6 m/s
Time at constant velocity = 2 min = 120 s

Distance in first 120 s:
= 6 × 120 = 720 m

Now acceleration = 1 m/s²
Time = 6 s

Distance in next 6 s:
s = ut + 1/2 at²
= 6 × 6 + 1/2 × 1 × 6²
= 36 + 18
= 54 m

Total displacement:
= 720 + 54
= 774 m

Velocity-time graph

• From 0 to 120 s: horizontal line at 6 m/s
• From 120 to 126 s: straight line rising from 6 m/s to 12 m/s

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Question 15

Two cars A and B start moving with a constant acceleration from rest,
in a straight line. Car A attains a velocity of 5 m s^–1 in 5 s. Car B attains
a velocity of 3 m s^–1 in 10 s. Plot the velocity-time graphs for both the
cars in the same graph. Using the graph, calculate the displacement
in the two time intervals mentioned (Hint: Calculate the acceleration
in both cases. Then calculate their velocities at five instants of time
to plot the graph).

Solution

Car A

Initial velocity = 0
Final velocity = 5 m/s
Time = 5 s

Acceleration = 5 / 5 = 1 m/s²

Distance from 0 to 5 s:
= 1/2 × 5 × 5
= 12.5 m

Distance from 5 to 10 s:
Velocity at 10 s = 10 m/s

Area = 1/2 × (5 + 10) × 5
= 37.5 m

Car B

Initial velocity = 0
Final velocity = 3 m/s
Time = 10 s

Acceleration = 3 / 10
= 0.3 m/s²

Distance from 0 to 5 s:
Velocity at 5 s = 1.5 m/s

Area = 1/2 × 5 × 1.5
= 3.75 m

Distance from 5 to 10 s:
= 1/2 × (1.5 + 3) × 5
= 11.25 m

Final Answers

Car A acceleration = 1 m/s²
Car B acceleration = 0.3 m/s²

Car A displacement 0–5 s = 12.5 m
Car A displacement 5–10 s = 37.5 m

Car B displacement 0–5 s = 3.75 m
Car B displacement 5–10 s = 11.25 m

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Question 16

Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip
of the minute’s hand of the wall clock. During the given time interval,
what is its:
(i) distance travelled,
(ii) displacement,
(iii) speed, and
(iv) velocity.
The length of the minute’s hand is 7 cm (Shown in Following Figure).

Solution

Time from 6:00 PM to 7:30 PM = 90 minutes

Minute hand makes 1 full revolution in 60 minutes.
In 90 minutes, it makes 1.5 revolutions.

Radius = 7 cm

(i) Distance travelled

Distance = 1.5 × circumference
= 1.5 × 2πr
= 1.5 × 2 × π × 7
= 21π cm
≈ 66 cm

(ii) Displacement

At 6:00, minute hand is at 12.
At 7:30, minute hand is at 6.

Displacement = diameter
= 2 × 7
= 14 cm

(iii) Speed

Speed = distance / time
= 66 / 90
≈ 0.73 cm/min

(iv) Velocity

Velocity = displacement / time
= 14 / 90
≈ 0.156 cm/min

Direction: from 12 to 6, vertically downward.

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Common Mistakes

• Confusing speed with velocity
• Forgetting SI units
• Using wrong sign for acceleration
• Mixing distance and displacement
• Incorrect graph slope calculations

Exam Tips

• Memorize all three equations of motion.
• Practice graph-based numericals daily.
• Always convert km/h into m/s before solving.
• Draw diagrams for displacement problems.
• Remember: slope gives velocity or acceleration.

Practice MCQs

1. SI unit of acceleration is:

A. m
B. m/s
C. m/s²
D. km/h

Answer

C. m/s²

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2. Which quantity has direction?

A. Speed
B. Distance
C. Velocity
D. Path

Answer

C. Velocity

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3. Area under velocity-time graph gives:

A. Speed
B. Distance
C. Displacement
D. Acceleration

Answer

C. Displacement

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4. Uniform circular motion has:

A. Zero acceleration
B. Constant direction
C. Changing velocity direction
D. Zero speed

Answer

C. Changing velocity direction

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FAQ Section

Q1. What is displacement?

Displacement is the shortest distance between initial and final position with direction.

Q2. What is acceleration?

Acceleration is the rate of change of velocity.

Q3. What does slope of velocity-time graph represent?

It represents acceleration.

Q4. Which graph gives displacement?

Area under velocity-time graph gives displacement.

Q5. What is uniform motion?

Motion with constant velocity is called uniform motion.

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