NCERT Solutions for Class 9 Science Chapter 8: Get PDF - Embibe
  • Written By Ravikant_Yadav
  • Last Modified 02-08-2022
  • Written By Ravikant_Yadav
  • Last Modified 02-08-2022

NCERT Solutions for Class 9 Science Chapter 8

NCERT Solutions for Class 9 Science Chapter 8 deals with the topic ‘Motion’. It is an important chapter in the CBSE Class 9 Science syllabus. It explains everything about Motion, Velocity, and important concepts related to them. This article will provide you with detailed Class 9 Science Chapter 8 NCERT Solutions.

Academic experts design Embibe’s Motion Class 9 NCERT Solutions after extensive research and analysis. These Class 9 NCERT Solutions will help students better understand the 9th Class concepts. It will also help them understand various approaches to a question and how to answer them. Students can keep the Motion Class 9 NCERT Solutions handy for their annual exam.

NCERT Class 9 Science Chapter 8 Solutions: Overview

Every day, we see innumerable objects in motion, like the birds flying, the fish swimming, our bodies in motion, cars moving, wind blowing, etc. Sometimes, an object may not seem to be moving to you, but it might be moving in reality. It is called relative motion. 

There are different types of motion like linear motion, where the object moves in one direction; circular motion, where it is rotating along the circular path; and vibration motion, where the object moves forward and backwards constantly. In this chapter, you will learn about the various concepts related to motion.

Let us have a look at the topics and sub-topics covered in NCERT Solutions for Class 9 Science Chapter 8.

SectionTopic Name
8.1 Describing Motion
8.2 Measuring The Rate Of Motion
8.3 Rate of Change of Velocity
8.4 Graphical Representation of Motion
8.5 Equations of Motion By Graphical Method
8.6Uniform Circular Motion

NCERT Solutions for Class 9 Science Chapter 8: Free PDF

We have provided detailed NCERT Solutions for all the in-text questions and exercise problems of Class 9 Science chapter 8. The students can also download the CBSE Class 9 solutions for this chapter in PDF format so that they can easily practice the questions offline.

NCERT Solutions for Class 9 Chapter 8: Solved Questions

Q.1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s?
Solution:
Given that diameter of the circular track, d=200 m
The radius of the track, r=d2=100 m
Time of running, t=2 min 20 s=140 s We know that circumference of a circular path is C=2πr=2π(100)=200π m And given that athlete completes one round in 40 s. In 40 s, athlete covers a distance of 200π m.

In unit time, the athlete will cover a distance =200π40 m.
The total distance covered by athlete in 140 s =200×2240×7×140=2200 m.
When the athlete runs for 140 s, he completes 3 full round and a half-round around the circle. That means after 140 s he will be at a point which is diametrically opposite to the starting point.
Therefore, the displacement will be equal to the diameter of the circular path which is 200 m.

Q.2. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth?
Solution:
As artificial satellite is moving along a circular path, in one revolution it will cover the circumference of a circle which is given by (2πr) where r is the radius of the circular orbit.
Radius of the circular orbit, r=42250 km.
Time taken to revolve around the earth, t=24 h. Speed of a satellite, v=Distancetime =
2πrt.⇒v=2×3.14×4225024=11055.41 km/h = 3.07 km/s.
Hence, the speed of the artificial satellite is 3.07 km/s.

Download CBSE Class 9 Science Chapter 8 Solutions PDF

Important Terms in Motion Class 9 NCERT Solutions

We advise students to go through the Go important concepts of Class 9 Science Chapter 8 before preparing for the annual exams. It is a good practice to maintain notes for all the below points while revising for the exams.

  • Motion is a change of position; it can be described as the distance moved or the displacement.
  • The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. 
  • The speed of an object is the distance covered per unit of time, and velocity is the displacement per unit of time. 
  • The acceleration of an object is the change in velocity per unit of time. 
  • Uniform and non-uniform motions of objects can be shown through graphs. 
  • The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely v = u + at, s = ut + ½ at^2, 2as = v^2 – u^2, where u is the initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity, and s is the distance it travelled in time t. 
  • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

Benefits of NCERT Solutions for Class 9 Science Motion

There are various benefits of using NCERT Solutions for Class 9 Science Chapter 8. Check out a few benefits:

  • NCERT Solutions uses a simple and easy-to-understand method to make students familiar with topics.
  • Offers comprehensive answers to all of the problems in the Motion Class 9 NCERT Book.
  • These solutions would be beneficial for CBSE Term I exams, Science Olympiads, and other competitive examinations.

FAQs on Class 9 Science NCERT Solution for Chapter 8

Q.1: A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms–2 for 8.0 s. How far does the boat travel during this time?

Ans: Initial velocity of the boat = 0 m/s (Given)
Acceleration of the boat = 3 ms-2
Time period = 8s
As per the second motion equation, s = (ut + 1/2 at2)
Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8)2
= 96 meters
Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

Q.2: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h-1 in 10 minutes. Find its acceleration.

Ans: Initial velocity (u) = 80km/hour (Given)
= 80000m/3600s
= 22.22 ms-1
Final velocity (v) = 60km/hour
= 60000m/3600s
= 16.66 ms-1
Time frame, t = 5 seconds.
Therefore, acceleration (a) = (v – u)/t
= (16.66 ms-1 – 22.22 ms-1)/5s
= -1.112 ms-2
Therefore, the total acceleration of the bus is -1.112 ms-2. Please note that the negative sign indicates that the velocity of the bus is decreasing.

Q.3: A trolley, while going down an inclined plane, has an acceleration of 2 cms-2. What will be its velocity 3 s after the start?

Ans: Initial velocity (u) = 0 (the trolley begins from the rest position)
Acceleration (a) = 0.02 ms-2
Time (t) = 3s
As per the first motion equation, v = (u+at)
Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms-2)(3s)
= 0.06 ms-2
Therefore, the velocity of the trolley after 3 seconds will be 6 cms-2

Q.4: Which of the following is true for displacement?
(i) It cannot be zero.
(ii) Its magnitude is greater than the distance travelled by the object.

Ans: Both of these statements are false. Statement (i) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.
Statement (ii) is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Q.5: What does the odometer of an automobile measure?

Ans: The odometer measures the total distance travelled by automobile.

Also, Check:

CBSE Class 9 English SyllabusNCERT Class 9 English Book
CBSE Class 9 Maths SyllabusNCERT Class 9 Maths Book
CBSE Class 9 Science SyllabusNCERT Class 9 Science Book
CBSE Class 9 Social Science SyllabusNCERT Class 9 Social Science Book
CBSE Class 9 Hindi SyllabusNCERT Class 9 Hindi Book

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