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CBSE Class 9 Science, Motion: Chapter notes (Part-II)

Here you get the CBSE Class 9 Science chapter 8, Motion: Chapter Notes (Part-II). These notes cover three equations of motion and their derivation. These chapter notes can be really helpful while preparing for the CBSE Class 9 Science Exam 2017-2018.

Oct 23, 2017 17:38 IST
    Class 9 Science Chapter Notes
    Class 9 Science Chapter Notes

    This article brings you the CBSE Class 9 Science notes on chapter 8 ‘Motion’ (Part-II) . In previous part, i.e., CBSE Class 9 Science notes on chapter 8 ‘Motion’ (Part-I), you learned about motion and its various attributes. In this part you will learn about the three equations of motion. These chapter notes are prepared by the subject experts and cover every important topic from the chapter. At the end of the notes you can try the questions asked from the discussed set of topics. These questions will help you to track your preparation level and get a hold on the subject.

    CBSE Class 9 Science Syllabus 2017-2018

    Main topics covered in this part of CBSE Class 9 Science,  Motion: Chapter Notes, are:

    • Three Equations of Motion
    • Derivation of Three Equations of Motion

                By simple methametical method

                By graphical method

    • Circular Motion 

    Key notes for Chapter - The Fundamental Unit of Life, are:

    Equations of Uniformly Accelerated Motion:

    There are three equations of motion for bodies travelling with uniform acceleration. These are explained below:

    1. First Equation of Motion:

    (Velocity-Time Relation)

                v = u +at

    Where,  v = Final velocity of body

                u = Initial velocity of body

                a = Acceleration

    And     t = Time


    Using formula for acceleration:

    2. Second Equation of Motion:

    (Position-Time Relation)

    3.  Third Equation of Motion:

    Equations of Motion by Graphical Methods:

    1. First Equation of Motion:

                v = u +at

    Consider the following velocity time graph:


    2. Second Equation of Motion:

    Let s be the distance covered.

    As we know that distance covered by an object is given as area enclosed by the graph,

    (iii) Third Equation of Motion:

                v2u2 = 2as

    Consider the velocity-time graph:

    Here, Distance, s = Area of trapezium OABC

                s = Area of rectangle ADCO + Area of triangle ABD

    Circular Motion
    1. Uniform circular motion
    : When an object moves on a circular path with constant or uniform speed, it is said to have uniform circular motion. For example: motion of hands of a clock, motion of moon revolving around earth.
    2. Non uniform circular motion: When an object moves on circular path with varying speed, it is said to have non-uniform circular motion.
    When an object is in circular motion, direction of its velocity keeps on changing.

    Speed of an Object Moving on a Circular Path:

    Circular Motion of Object

    Suppose a body moves on a circular path with radius r.

    Try the following questions:

    Q1. How can you find distance of an object from its speed – time graph?

    Q2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110m. Determine the acceleration of the car.

    Q3. Study the speed time graph

    Q4. Find following based on the graph

    (i) Which path has the constant speed?

    (ii) When is the maximum speed reached?

    (iii) Along which path motion gets accelerated?

    (iv) Along which path motion gets deaccelerated?

    Q5. A car a moving at rate of 72km/h and applies brakes which provide a retardation of 5ms-2.

    (i) How much time does the car takes to stop.

    (ii) How much distance does the car cover before coming to rest?

    (iii) What would be the stopping distance needed if speed of the car is doubled?


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