CBSE Class 9 Science Chapter 8 Motion Revision Notes

Core Glossary & Definitions

Motion

An object is said to be in motion if it changes its position with respect to a stationary reference point (origin) over time.

Reference Point

A fixed point or origin used to describe the precise position, distance, or direction of an object.

Distance

The actual path length traversed by a moving body from its initial point to its final point, irrespective of direction.

Displacement

The shortest straight-line distance measured from the initial position to the final position of an object, along with its direction.

Acceleration

The rate of change of velocity of an object per unit of time. Mathematically: a = (v - u) / t.

Uniform Circular Motion

The motion of an object moving at a constant speed along a circular path. Its direction changes continuously, making it an accelerated motion.

Key Contrasts: Distance vs. Displacement

Must-Remember Formulas & Core Kinematics Equations

• Average Speed Formula: $\text{Average Speed} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$
• Average Velocity (under uniform acceleration): $v_{\text{avg}} = \frac{u + v}{2}$, where $u$ is initial velocity and $v$ is final velocity.
• Acceleration Equation: $a = \frac{v - u}{t}$, measured in $\text{m/s}^2$.
• First Equation of Motion (Velocity-Time Relation): $v = u + at$
• Second Equation of Motion (Position-Time Relation): $s = ut + \frac{1}{2}at^2$
• Third Equation of Motion (Position-Velocity Relation): $v^2 - u^2 = 2as$
• Uniform Circular Motion Speed Formula: $v = \frac{2\pi r}{t}$ where $r$ is the radius of the circular path.
• Retardation or Deceleration: Negative acceleration that occurs when a body slows down (its final velocity is less than its initial velocity).

Understanding Motion Graphs & Derivations

Graphical analysis is the backbone of Class 9 Kinematics. There are two primary graphs you must master for your CBSE exams:

1. Distance-Time Graphs (s-t): The slope of a distance-time graph represents the Speed of the body. A straight inclined line indicates uniform speed, while a curved line indicates non-uniform speed.

1. Velocity-Time Graphs (v-t): The slope of a velocity-time graph represents Acceleration. A straight horizontal line means zero acceleration (uniform velocity). More importantly, the area under the v-t graph line gives the total Displacement of the body over the specified time interval.

Note on Derivations: When deriving equations of motion graphically, remember that the first equation ($v = u + at$) is derived from the slope of the v-t graph, while the second ($s = ut + \frac{1}{2}at^2$) and third ($v^2 - u^2 = 2as$) equations are derived by calculating the area of the trapezium or the combined area of the rectangle and triangle beneath the velocity-time line.

Step-by-Step Guide to Solving Motion Numericals

1. Identify and List Givens — Read the problem carefully. Note down the values for initial velocity ($u$), final velocity ($v$), acceleration ($a$), time ($t$), and displacement ($s$). Pay attention to key phrases: 'starts from rest' means $u = 0$; 'comes to a stop' means $v = 0$.
2. Convert Units to SI Standard — Ensure all variables are in SI units. If speed is given in $\text{km/h}$, convert it to $\text{m/s}$ by multiplying the value by $\frac{5}{18}$.
3. Select the Appropriate Equation — Choose one of the three equations of motion that matches your given and unknown values. For instance, if you are not given displacement ($s$), use $v = u + at$ to find acceleration or time.
4. Substitute and Calculate — Substitute the values into the equation, manage signs carefully (negative acceleration for brakes/deceleration), and calculate the final answer with proper SI units.

Worked Mini-Examples

• {"title":"Example 1: Displacement on a circular track","markdown":"Question: An athlete completes one round of a circular track of diameter $200\\text{ m}$ in $40\\text{ s}$. What will be the distance covered and the displacement at the end of $2\\text{ minutes } 20\\text{ seconds}$?\n\nSolution:\n Total time = $2\\text{ min } 20\\text{ s} = 140\\text{ s}$.\n Number of rounds completed = $\\frac{140}{40} = 3.5\\text{ rounds}$.\n Distance covered in $1\\text{ round} = 2\\pi r = \\pi d = \\frac{22}{7} \\times 200 = 628.57\\text{ m}$.\n Total Distance covered in $3.5\\text{ rounds} = 3.5 \\times 2\\pi r = 3.5 \\times 628.57 \\approx 2200\\text{ m}$.\n* Since the athlete completes $3.5$ rounds, they end up at the diametrically opposite point of the track. Thus, Displacement = Diameter of the track = $200\\text{ m}$."}
• {"title":"Example 2: Applying the Second Equation of Motion","markdown":"Question: A bus starting from rest moves with a uniform acceleration of $0.1\\text{ m/s}^2$ for $2\\text{ minutes}$. Find the distance travelled.\n\nSolution:\n Initial velocity, $u = 0\\text{ m/s}$ (since it starts from rest)\n Acceleration, $a = 0.1\\text{ m/s}^2$\n Time, $t = 2\\text{ minutes} = 120\\text{ seconds}$\n Using second equation of motion: $s = ut + \\frac{1}{2}at^2$\n $s = 0 \\times 120 + \\frac{1}{2} \\times 0.1 \\times (120)^2$\n $s = 0 + 0.05 \\times 14400 = 720\\text{ m}$.\n* Distance Travelled = $720\\text{ m}$."}

CBSE Board Exam Traps & Cues

• The km/h to m/s Trap: Always check units! If a car traveling at $72\text{ km/h}$ decelerates to rest in $5\text{ seconds}$, you must convert $72\text{ km/h}$ to $20\text{ m/s}$ before using $a = (v-u)/t$.
• Sign Conventions: If brakes are applied, the acceleration is negative (retardation). Substituting a positive $a$ value during deceleration is the most common error that leads to wrong numerical answers.
• Circular Motion Direction: Remember that even if speed is constant in uniform circular motion, velocity is changing continuously because the direction of motion is constantly changing. Hence, it is always an accelerated motion.

Quick Revision Check

• Can an object have zero displacement but non-zero distance? Give an example. Yes. An object has zero displacement and non-zero distance when it returns to its starting point. For example, a runner completing one full lap on a circular track covers a distance equal to the track's circumference, but their displacement is zero.
• What does the slope of a Velocity-Time (v-t) graph represent? The slope of a Velocity-Time graph represents the acceleration of the moving body.
• State the three equations of motion and specify what each symbol stands for. The three equations are: 1) $v = u + at$, 2) $s = ut + \frac{1}{2}at^2$, and 3) $v^2 - u^2 = 2as$. Here, $u$ is initial velocity, $v$ is final velocity, $a$ is constant acceleration, $t$ is time taken, and $s$ is distance/displacement covered.
• Why is uniform circular motion called an accelerated motion? Even though the speed remains constant in uniform circular motion, the direction of travel changes continuously at every point. Since velocity depends on direction, a change in direction means a change in velocity, making it an accelerated motion.

Frequently Asked Questions