Kerala Syllabus Class 9 Physics Chapter 2 Equations of Motion Notes Solutions

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 2 Equations of Motion Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 2 Notes Solutions Equations of Motion

SCERT Class 9 Physics Chapter 2 Notes Solutions Kerala Syllabus Equations of Motion Questions and Answers

Class 9 Physics Chapter 2 Let Us Assess Answers Equations of Motion

Question 1.
A car starts from rest and moves with uniform acceleration. Calculate the acceleration of the car if it covers a distance of 200 m in 20 s.
Answer:
Initial speed u = 0
distance travelled s = 200 m
time taken t = 20 sec
acceleration a = ?
Using the second equation
s = ut + \(\frac{1}{2}\) at²
⇒ 200 = (0)t + \(\frac{1}{2}\) a(20)²
a = 200/200 = 1 m s^-2

Question 2.
If an object starts from rest and moves with an acceleration of 2 m/s², what will be the velocity of the object after 10 s?
Answer:
Given, a = 2 m/s², time = 10 s
Initial velocity u=0
According to the equation of motion:
v = u + at
=0 + 2 × 10
= 20 m/s

Question 3.
Three different graphs related to the motion of a vehicle are given below. Analyse the graphs and find the characteristics of the motion.

Answer:

• Graph 1 shows a vehicle moving with uniform acceleration.
• Graph 2 shows a vehicle moving with uniform deceleration.
• Graph 3 shows a vehicle moving with non-uniform acceleration.

Question 4.
Graph related to the motion of Car A and Car B is given.
a) Which car has more acceleration? Why?
b) Redraw the graph by changing the scale and compare the graphs.

Answer:
a) Car A

When the scale of a graph changes, it affects the appearance and readability of the graph, but the underlying data and relationships remain the same.

Question 5.
Observe the figure.

A child runs along a circular path of circumference 440 m at a constant speed. The radius of the circular path is 70 m. The time taken to run from A and reach A in clockwise direction through B is 80 s. Find the distance, displacement, speed and velocity in each case in the table.
Answer:

Question 6.
A train starts from rest and attains a speed of 72 km/h in 5 minutes. Find the acceleration and displacement of the train.
Answer:
Given,
v = 72 km/h = 20 m/s
(I km/hr = 1000/3600 (m/s))
Time (t) = 5 minutes = 300 s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{20-0}{300}\) = \(\frac{1}{15}\) = m/s
Displacement s= ut + \(\frac{1}{2}\) at²
= 0 × 300 + \(\frac{1}{2}\) × \(\frac{1}{15}\) 300²
= 3000 m

Question 7.
Analyse Table and draw the velocity-time graph.

X Time (s)	0	2	4	6	8	10
Y Velocity(m/s)	10	15	20	20	20	15

a) From the graph, find the time interval during which there is no acceleration.
b) Find the time interval during which deceleration occurs.
c) Find the displacement between the fourth second and the eighth second.
Answer:

a) 4 s to 8 s
b) 8 s to 10 s
c) Time t = 4s
Velocity = 20 m/s
Displacement = Velocity × Time = 20 × 4 = 80 m

Question 8.
Observe the graphs.

In which graph,
a) does the object have uniform acceleration?
b) does the object have uniform velocity?
c) does the object have acceleration and deceleration?
Answer:
a) Graph 2
b) Graph 1
c) Graph 3

Question 9.
The velocity-time graph of an object in straight line motion is given.

Identify the regions of the graph where the object is moving with:
a) acceleration.
b) uniform velocity.
c) deceleration.
Answer:
a) Region OA
b) Region AB
c) Region BC

Question 10.
It is not safe for pedestrians to wear dark coloured clothes at night and in conditions with dim light. The school authorities decided to choose a dark coloured uniform for your school. Record your response to the decision. Justify the answer with respect to road safety.
Answer:
Choosing a dark coloured uniform for our school raises significant safety concerns, particularly regarding pedestrian safety at night and in low- light conditions. This decision is problematic because: visibility issues, road safety. For the safety of all students, it would be better for the school to choose a uniform that includes lighter or reflective elements. This change would help ensure that students are more visible to drivers, particularly in low-light conditions, thereby reducing the risk of accidents and enhancing overall road safety.

Question 11.
In the figure given below, a child travels from P to S through Q and R and comes back straight to P.

Analyse the figure and complete the table given below.
Answer:

Starting from P	Speed	Velocity
On reaching Q	\(\frac{60}{10}\) = 6 m/s	\(\frac{60}{10}\) = 6 m/s
On reaching R	\(\frac{150}{25}\) = 6 m/s	\(\frac{125}{25}\) = 5 m/s
On reaching S	\(\frac{204}{34}\) = 6 m/s	\(\frac{136}{16}\) = 8.5 m/s
While returning to P	\(\frac{340}{50}\) = 6.8 m/s	\(\frac{0}{50}\) = 0 m/s

Question 12.
A stone is thrown vertically upwards with a velocity of 20 m/s. (a = -10 m/s²)
a) What is the maximum height that the stone has reached?
b) How far will this stone travel in 3 s after it is thrown?
c) How high will the stone be from the ground, 3 s after it is released?
Answer:
Given,
v = 0m/s; u 20 m/s; a = -10 m/s²
a) Using equation of motion, v² = u² + 2as
0 = 20² + 2(-10)h
0 = 400 – 20 h
20 h = 400
h = 400/20 = 20 m

b) s = ut + 1⁄2 at²
20 = (20×t) + 1⁄2 (-10 t²)
20 = 20 t – 5 t²
t² – 4t + 4 = 0
(t-2)² = 0
t = 2s
During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Total distance travelled by the stone in 3 s = 20 + 5 = 25 m.
c) During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Therefore, the height from the ground after 3 seconds: 20 m – 5m = 15 m.

Question 13.
An object is moving with a speed of 40 m/s.
If it is given a deceleration of 8 m/s²,
a) how long will it take to come to rest?
b) what is the displacement of the object in this time?
Answer:
a) Here, v = 0 m/s
u = 40 m/s
a = -8 m/s
By using the equation, v = u + at
0 = 40 + (-8) t
= 40 – 8t
8t = 40
t = 40/8 = 5 s
So, it will take 5 seconds for the object to come to rest.

b)Displacement s = ut + \(\frac{1}{2}\) at²
= (40 × 5 ) + \(\frac{1}{2}\) (-8)5²
= 200 + \(\frac{1}{2}\) (-200)
=200 – 100
= 100 m

Question 14.
An object is moving with a velocity of 20 m/s. This object is given an acceleration of 5 m/s². What is the velocity when the displacement is 120 m?
Answer:
Given,
u = 20 m/s
a = 5 m/s²
s = 120 m
By using equation, v² = u² + 2as
v² = 20² + 2 × 5 × 120
v² = 400 + 1200
v² = 1600
v = √1600 = 40 m/s

Question 15.
A bullet travelling with a velocity of 60 m/s comes to rest after penetrating 2 cm into a wooden block. What is the acceleration of this bullet? How much is its deceleration?
Answer: Given,
Initial velocity (u) = 60 m/s
Final velocity (v) = 0m/s
Distance (s) = 2 cm = 0.02 m.
We know, v² = u² + 2as
0² = 60² + 2 × a × 0.02
0 = 3600 + 0.04 a
a = \(\frac{-3600}{0.04}\) = 90000 m/s²
The deceleration (since it is negative acceleration) is 90000 m/s.

Question 16.
Observe the table illustrating the motion of an object, select an appropriate scale and draw the graph. Interpret the graph and write down the answers to the questions given below.

Time (s)	0	5	10	15	20	25	30
Velocity (m/s)	20	25	30	30	30	25	20

a) Which is the time interval with no acceleration?
b) Which is the time interval with deceleration?
c) Calculate the displacement of this object in 30 s.
Answer:

a) 10s to 20s
b) 20s to 30s
c) Time t = 30 s
Velocity v = 20 m/s
Displacement-Area under velocity-time graph.
= \(\frac{1}{2}\) × (20+30) × 10 + (10×30) + \(\frac{1}{2}\) ×(20+30) × 10
= \(\frac{500}{2}\) + 300 + \(\frac{500}{2}\)
= 250 + 300 + 250
= 800 m

Question 17.
The velocity-time graph of the motion of Car A and Car B is given below.

a) Which car started first?
b) How much time did each car take to attain the same speed?
c) Which car possesses more acceleration?
d) Which car has more displacement?
Answer:
a) Car A
b) 14 s
c) Car B
d) Car A

Class 9 Physics Chapter 2 Extended Activities Answers Equations of Motion

Question 1.
The Science Club organises an awareness class on ‘Road accidents due to over speeding. Prepare the necessary slides for the presentation. (Hints: traffic rules, signboards, traffic rules to be followed by pedestrians, etc.)
Answer:
Hint:
Slide 1:
Title: Road Accidents Due to Over speeding Subtitle: Understanding the Risks and Prevention Measures
Introduction: Brief overview of the importance of road safety and the focus on over speeding.

Slide 2: Introduction
Road accidents are a major cause of injury and death.
Over speeding is a leading factor in many accidents.
Today, we’ll learn about how to stay safe on the road.

Slide 3: Importance of Traffic Rules
Traffic rules are designed to keep everyone safe.
Following these rules helps prevent accidents.

Slide 4: Common Traffic Signs
Stop Sign: Stop your vehicle completely. Speed Limit Sign: Do not exceed the speed indicated.
Pedestrian Crossing Sign: Slow down and be ready to stop for people crossing.

Slide 5: Traffic Rules for Pedestrians
Use pedestrian crossings.
Look both ways before crossing the road. Avoid crossing between parked cars. Walk on sidewalks whenever possible.

Slide 6: Tips for Safe Driving
Always stay within speed limits.
Keep a safe distance from the vehicle in front of you.
Stay alert and focused on the road.

Slide 7: Conclusion
Following traffic rules saves lives.
Over speeding is dangerous for everyone.
Stay safe and be responsible on the road.

Question 2.
An object is thrown upwards with a velocity of 30 m/s. It returns to the same position after some time. Draw the velocity-time graph of this object and display it in the class (Consider the deceleration as 10 m/s²).
Hint:
Initial velocity (u): 30 m/s (upwards).
Final velocity (v) when it comes back to the same
position: -30 m/s (downwards).
Deceleration (a): 10 m/s².
Time to reach the highest point (t): Calculate using V = u + at.
After finding the value, then take the correct scale and draw the graph similar to this.

Prepare a project on whether the safety measures implemented in your area are adequate to reduce road accidents.
Project planning can be done with the help of your teacher.
Bring relevant findings to the attention of the Road Safety Authority

(For more information, the services of the National Transportation Planning and Research Conor (NATPAC) and the Motor Vehicles Department can be availed.

The report should include:

• Introduction
• Hypothesis
• Objectives
• Methodology
• Analysis
• Results
• Conclusion
• Suggestions

Hint:
Introduction
Road accidents are a serious issue affecting our community. This project aims to evaluate whether the current safety measures in our area are sufficient to reduce road accidents.

Hypothesis
The existing safety measures in our area are adequate to significantly reduce road accidents.

Objectives

• To identify the current safety measures implemented in our area.
• To assess the effectiveness of these safety measures.
• To gather opinions from local residents and authorities.
• To make suggestions for improvement based on findings.

Methodology
Planning: Work with our teacher to create a project plan.

Data Collection:

• Conduct surveys with local residents.
• Interview local traffic authorities.
• Observe road conditions and safety measures (e.g., traffic signs, speed limits, pedestrian crossings).

Research: Use resources from NATPAC and the Motor Vehicles Department.
Analysis: Compare collected data with road safety standards.

Analysis

• Survey Results: Summarize the responses from local residents about their perception of road safety.
• Interviews: Key points from interviews with traffic authorities.
• Observations: Note the presence and condition of traffic signs, speed bumps, pedestrian crossings, etc.

Results

• Positive Findings: List of effective safety measures currently in place (e.g., well-marked pedestrian crossings, functional traffic lights).
• Areas of Concern: Identify any shortcomings (e.g., lack of speed bumps in critical areas, poor visibility of traffic signs).

Conclusion
Summarize whether the safety measures are adequate. If they are not, highlight the main issues identified.

Equations of Motion Class 9 Notes Questions and Answers Kerala Syllabus

Activity
Observe figure.

A,B,C and Dare four electric poles erected on the side of the road. The distance between any two adjacent poles is 40 m. A child starts walking from pole B, passes C and reaches D. After that, the child returns from D and reaches the pole C.

Question 1.
What is the total distance travelled by the child?
Answer:
120 m

Question 2.
What is the distance between the current position C and the initial position B of the child?
Answer:
40 m

Question 3.
If the child travels 40 m from B, which are the possible positions of the child?
Answer:
Child can reach at A or C.

Question 4.
In which direction should the child travel 40 m to reach C from B? (towards the east /towards the west)
Answer:
Towards the east
Here 40 m eastward from B to C is the displacement of the child. So, Displacement is the shortest distance.

Question 5.
Is displacement a vector or a scalar?
Answer:
Vector

Question 6.
If the child travels 40 m in the westward direction from B, the current position of the child is …………
Answer:
A
Note: If the displacement from B in the forward direction (towards the east) is considered as positive, the displacement in the backward direction (towards the west) should be considered as negative. (These can also be considered in the reverse order). Once the direction is determined, the positive and negative directions should not be changed thereafter. Here, B is the initial position and A is the final position. Hence the displacement is negative.

Question 7.
Complete table based on the child’s travel given above.
Answer:

Stages of path covered by the child	Distance covered	Displacement
Directly from B to C	40 m	40 m
Starts from B, reaches D and returns to C	120 m	40 m (from B to C)
Starts from B and reaches D	80 m	80 m (from B to D)
From B to A	40 m	40 m (from B to A)
Starts from B, reaches A and returns to B	80 m	Zero

Question 8.
The child moved from A to D and returned to A. What is the distance covered? What is the displacement? Aren’t the initial position and final position the same?
Answer:
The distance covered is 240 m and the displacement is zero because of initial position and final position are same.

Question 9.
Write down the situations in which the distance covered and displacement are equal.
Answer:

• When you travel directly from the start point to the end point.
• When there are no turns or curves in your path.
• When you move in a straight line without changing direction.

Question 10.
Two different paths taken by a child to move from position P through Q are depicted.

a) What is the distance covered in figure (a)? What is the displacement?
b) What is the distance covered as per the motion in figure (b)? What is the displacement?
c) In which situation is the magnitude of the distance and displacement equal?
Answer:
a) Distance is 100 m. Displacement is 80 m. (From P to R)
b) From P to Q, the distance is 100 m and displacement is also 100 m. (From P to S)
c) In the second case; i.e., Displacement and distance are equal only when an object moves in a straight line without changing direction.

Question 11.
Tabulate the differences between distance and displacement related to the path traversed by a person in Table.
Answer:

Distance	Displacement
Length of the path covered	Straight-lone distance
Can’t be zero	Can be zero
Scalar	Vector

Question 12.
The classrooms and some other locations of a school are depicted.

During the interval, a student from Class 9B went to the staff room. The child then proceeded to the statue of the Father of the Nation in the school garden and returned to the class via the school office. Complete Table based on the path followed by the child.
Answer:

Child’s path	Distance (m)	Displacement (m)
When the child reaches the corridor in front of the staff room from Class 9B	15 m	15 m
When the child reaches the garden near the statue of the Father of the Nation from 9B via the staff room.	85 m	60 m
When the child returns to Class 9B	180 m	0 m

A child travels from P to R via Q in 18 seconds, as shown in the figure.

Question 13.
What is the total distance covered by the child to reach R from P through Q?
Answer:
90 m

Question 14.
What is the speed of the child when the child travels from P to R through Q?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{90}{18}\)
= 5 m/s

Question 15.
What is the displacement of the child?
Answer:
72 m

Question 16.
Isn’t the child’s displacement of 72 m taking place in 18 s?
Answer:
Yes
Let’s find out the displacement in one second.
Displacement in one second = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{72 m}{18 s}\) = 4 m/s
The displacement in unit time is the velocity.

Question 17.
What is its direction?
(P → R/R → P/P → Q → R)
Answer:
P → R
So the direction of displacement and velocity are same.

Question 18.
Now, let’s consider the doubt raised by the child at the beginning of this lesson. A train of length 200 m travels with a velocity of 20 m/s. What is the time taken by this train to cross a straight bridge of length 800 m?
Answer:

Displacement = Length of the bridge + length of the rain
s = 800 m + 200 m = 1000 m
Velocity (v) = 20 m/s
Time (t) = ?
t = \(\frac{s}{v}\) = \(\frac{1000 m}{20 m/ s}\)
Time = 50 s

Question 19.
Calculate the speed and velocity of the child from P to Q in figure c. What inferences can you draw from this?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
Velocity = \(\frac{\text { Displacement }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
This shows that the speed and velocity is same when it travels through a straight line.

Question 20.
A vehicle travels along a straight line with a velocity of 25 m/s and covers a distance of 400 m. Calculate the time taken for this.
Answer:
Given,
Velocity = 25 m/s
Distance = 400m
Here Speed and Velocity is same (along a straight line)
So, Speed = \(\frac{\text { Distance }}{\text { Time}}\)
Time = \(\frac{\text { Distance }}{\text { Speed}}\)
= \(\frac{400}{25}\) = 16 s.

Question 21.
What is the displacement of an object moving with a velocity of 36 m/s in one minute?
Answer:
Given,
Velocity = 36 m/s
Time = 1 minute = 60 s
Displacement = Velocity × Time
=36 × 60 = 2160 m

Question 22.
Is the velocity of CarA always the same? Why?
Answer:
Yes. Because it travels equal distance in equal interval of time.

Question 23.
What about the velocity of Car B? Why?
Answer:
Not same. Because it travels unequal distance in equal interval of time.

Question 24.
Haven’t you noticed the mud sticking to the tyres of vehicles being thrown off when they rotate? Does the mud splash in the same direction every time?
Answer:
No, the mud does not splash in the same direction every time. It is thrown off tangentially due to the rotation of the tires.

Question 25.
Isn’t the direction of motion of an object moving along a circular path always changing?
Answer:
Yes, the direction of motion of an object moving along a circular path is always changing.

Question 26.
Is the velocity of Car C the same every second? Does the velocity change?
Answer:
Even though the magnitude of the speed does not change, the velocity changes because the direction changes.

Question 27.
Complete the table based on the information in figures (1), (2), (3).

Answer:

Question 28.
Classify the situations given below as uniform velocity and non-uniform velocity. Record it in the science diary.

• Motion of a stone dropped from a height
• When light travels through vacuum
• A bus starts from a bus stop and is moving forward
• A train travelling at a uniform speed in the same direction
• Swinging on a swing

Answer:

Uniform velocity	Non-uniform velocity
When light travels through vacuum A train travelling at a uniform speed in the same direction	Motion of a stone dropped from a height A bus starts from a bus stop and is moving forward Swinging on a swing

Question 29.
The velocity of a bus starting from a bus stop keeps changing. Will the change in velocity be the same in each second?
Answer:
No.

The data related to the straight line motion of the bus is given below. Analyse the information and answer the questions.

Question 30.
Imagine that you are sifting in a bus. When the bus starts and moves forward in a straight line, doesn’t the velocity change?
Answer:
Yes

Question 31.
When the bus travels from A to B, the velocity at A is……………
(initial velocity / final velocity)
Answer:
Initial velocity.

Question 32.
The velocity at B is……………(initial velocity / final velocity)
Answer:
Final velocity.

Question 33.
While considering the motion from B to C, velocity at B is……………………
Answer:
Initial velocity.

Question 34.
Complete the table using the data of the motion of the bus.

Answer:

Question 35.
A car is moving along a straight road with a velocity of 10 m/s. It is given an acceleration of 5 m/s². Calculate the velocity of the car after 2s.
Answer:
Initial velocity u = 10 m/s
Acceleration a = 5 m/s²
time t = 2s
Final velocity v = ?
a = \(\frac{\text { v-u }}{\text { t}}\)
V – u = at
V = u + at = 10 + 5 × 2 = 20 m/s
To calculate the final velocity, we can use the equation v = u + at

Question 36.
The velocity of an object changes from 4 m/s to 28 m/s in 4 s. Calculate the acceleration.
Answer: Given,
Initial velocity (u) = 4 m/s
Final velocity (y) = 28 m/s
Time (t) = 4s
Acceleration = \(\frac{\text { change in velocity }}{\text { t}}\)
a = \(\frac{\text { v-u }}{\text { t}}\)
a = \(\frac{28-4}{4}\) = 6m/s²

Question 37.
Have a look at the scene in an amusement park. List the instances in which acceleration occurs.
Answer:
Motion of a giant wheel
Roller Coasters
Swing Rides: When the swings begin to move, accelerate outward, decelerate back inward.

Question 38.
Find instances of acceleration in your daily life and record them in the science diary.
Answer:

• Motion of a coconut falling from a coconut tree
• A car speeding up
• A bicycle starting from rest
• An airplane taking off
• An elevator starting to ascend or descend

Question 39.
Can you think of some instances ¡n everyday life where the velocity decreases? Expand the list by giving more examples.
Answer:

• Train arrivingata station
• The upward motion of a stone thrown upwards
• A car approaching a red traffic light
• A ball rolling up hill
• A parachutist landing

Question 40.
Observe the figure, and complete the table.


Answer:

Question 41.
Should negative sign be given while writing the value of retardation?
Answer:
Yes, a negative sign should be given while writing the value of retardation, as it indicates a decrease in velocity.

Question 42.
An object starts from rest and attains a velocity of 10 m/s in 5 s.
a) What is its acceleration?
b) What is the acceleration if it comes to rest in 5 s? What is the retardation?
Answer:
Given,
a) Initial velocity (u)=0 m/s
Final velocity (v) = 10 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 10-0 }}{\text { 5}}\) = 2 m/s²

b) Here,
initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 0-10 }}{\text { 5}}\) = -2 m/s²
Retardation is simply the magnitude ofthe negative acceleration.
Thus, retardation = 2 m/s²

Question 43.
A vehicle travelling at a speed of 5 m/s is brought to. rest ¡n 2 s by applying brakes. Calculate the retardation of the vehicle.
Answer:
Initial velocity u = 5 m/s
Final velocity v = 0
Time taken = 2 s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 0-5 }}{\text {2}}\) = -2.5 m/s²
Thus, retardation = 2.5 m/s²

Question 44.
If the velocity of an object in the 2nd second is 40 m/s and 30 m/s in the 4th second, what is its acceleration? What is its retardation? What is its velocity at the 8th second?
Answer:
Velocity at the 2^nd second u = 40 m/s
Velocity at the 4^th second v = 30 m/s
Time (t) = 2s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 30 – 40 }}{\text {2}}\) = -5 m/s²
So retardation is 5 m/s².
To find velocity at 8^th second.
Initiai velocity (u) = 40 m/s
Time (t) = 6 s
So, a = \(\frac{\text { v-u }}{\text { t}}\)
V = u + at
= 40 + (-5 × 6)
=40 – 30 = 10 m/s.

Question 45.
Was the acceleration obtained as per Table 1 the same on each occasion?
Answer:
Yes

Question 46.
What about the acceleration obtained as per Table 2?
Answer:
Not same and also have negative sign.
You have learned about speed, velocity, and acceleration. Over speeding of vehicles can cause accidents. We must follow traffic rules to reduce accidents. However, accidents aren’t caused only by over speeding. Pedestrians should also follow traffic rules.

Question 47.
Which are the traffic rules for pedestrians to follow?
Answer:

• Pedestrians should walk along the right side of the road.
• Cross the road only at the zebra crossing, obeying the traffic signal.
• Use sidewalks
• Avoid crossing between parked cars

Question 48.
In addition to the signs mentioned above, collect more symbols for each category, prepare separate posters and display them on the school bulletin board.
Answer:
Hint: An example shown below.

Question 49.

What information can be gathered from the above graph?
Answer:
From 2020 to 2022 there is a significant spike in accidents. The highest number of accidents happened in 2022. From 2019 to 2020 the number of accidents decreased.

Question 50.
List down your findings.
Answer:
The number of road accidents is generally increasing, this might indicate issues with traffic management, increasing vehicle numbers, or lack of road safety measures. Significant drops in 2019 to 2020 could be attributed to the COVID-19 pandemic, which resulted in lockdowns and reduced road traffic.

Question 51.
In which year is the number of accidents the least?
Answer:
2020

Question 52.
How many accidents occurred in 2019?
Answer:
43000

Question 53.
What is the nature of the graph obtained?
Answer:
An inclined straight line

Question 54.
By what name is this graph known?
Answer:
Position-Time graph

Question 55.
From the shape of the graph obtained, what is the nature of the velocity of the object?
Answer:
Uniform velocity

Question 56.
What is the displacement of the object in 5 s?
Answer:
2.5 m

Question 57.
What is the time taken to travel 1.5 m?
Answer:
3 s
Let’s consider another situation.
Position-Time graph of a moving car.

Question 58.
How can we find the velocity of the car from A to B from above graph?
Answer:
Here,
Displacement=30 m, Time=6 s
Velocity = Displacement/Time = \(\frac{30}{6}\) = 5 m/s

Question 59.
What is the displacement of the car from A to B in the graph?
Answer:
30 m

Question 60.
What is the time taken by the car to travel from A to B?
Answer:
6 s

Question 61.
What is the position of the car at the sixth second in the graph?
Answer:
30 m

Question 62.
Which type of velocity does this car have?
Answer:
Uniform velocity

Question 63.
Find out the velocity of the car between 6 s and 8 s from the graph.
Answer:
Here,
Displacement = 10m
Time = 2 m
Velocity = Displacement/ Time
= \(\frac{10}{2}\) = 5 m/s

A velocity-time graph is a graph that plots velocity on the Y-axis and time on the X-axis.

Question 64.
From the graph, find the displacement of the vehicle between the fourth and tenth second.
Answer:
Velocity = Displacement/Time
Then, displacement = velocity × time
On the graph, it will be equal to AB × AD (Equal to the area of the rectangle ABCD)
Displacement = 40 × 6 = 240 m

Question 65.
Isn’t this equal to the area of the portion below BC on the graph?
Answer:
Yes.
Area = 40 × 6 = 240 m

Question 66.
Find the displacement during the first 4 s from graph.
Answer:
displacement Area of triangle AOB
= \(\frac{1}{2}\) × 40 × 4 = \(\frac{160}{2}\) = 80 m

Question 67.
What is the change in velocity in the first 4 s? What is the acceleration?
Answer:
Initial velocity (u) = 0 m/s
Final velocity (v) = 40 m/s
Change in velocity = v – u – 0 – 40 = -40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{-40}{4}\) = -10 m/s²

Question 68.
What is the acceleration of this vehicle between 4 s and 10 s?
Answer:
Initial velocity (u) = 40 m/s
Final velocity (v) = 40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{40-40}{6}\) = 0 m/s²

Question 69.
A body starts from rest and acquires a velocity of 20 m/s in 2 s and 40 m/s in 6 s. What is the displacement of the object during this time interval?
Answer:
Initial velocity, u = 20 m/s
Time t = t₂ – t₁ = 6s – 2s = 4 s
Final velocity, v = 40 m/s
Acceleration, a = \(\frac{v-u}{t}\)
= \(\frac{40 \mathrm{~m} / \mathrm{s}-20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}\)
= 5 m/s²
Displacement s = ut + \(\frac{1}{2}\) at²
= (20 m/s × 4 s) + [\(\frac{1}{2}\) × 5 m/s² (4 s)²]
= 80 m + 40 m
=120 m

Question 70.
If the velocity of a car increases from 6 m/s to 16 m/s in 10 s.
a) calculate the acceleration of the car.
b) what is the displacement of the car during this time?
Answer:
a) u = 6 m/s
v = 16 m/s
t = 10 s
Acceleration a = \(\frac{v-u}{t}\)
= \(\frac{16 \mathrm{~m} / \mathrm{s}-6 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= \(\frac{10 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= 1 m/s²

b) s = ut + \(\frac{1}{2}\) at²
=(6 m/s × 10 s) + [\(\frac{1}{2}\) × 1 m/s² (10 s)²]
=60 m + 50 m
= 110 m

Question 71.
The velocity of a train that started from a railway station becomes 90 km/h in 10 minutes. Calculate the acceleration of the train.
Answer:

Question 72.
An object falls down from rest and moves with an acceleration 10 m/s, hits the ground with a velocity 20 m/s. From what height does the object fall?
Answer:
u = 0
a = 10 m/s²
v = 20 m/s
v² = u² + 2as
(20 m/s)² = 0² + 2 × 10 × s
400 = 20 × s
s = \(\frac{400}{20}\) = 20 m