CBSE Class 9 Science — Motion

Chapter Overview & Weightage

Motion is Chapter 8 in CBSE Class 9 Science (NCERT). This chapter introduces the foundational concepts of kinematics — describing how objects move. It is the most important chapter in Class 9 Physics and the direct foundation for Class 10 and Class 11 mechanics.

Motion typically carries 10–15 marks in CBSE Class 9 annual exams. Equations of motion and numerical problems are the highest-weightage topics. Graphical problems (interpreting velocity-time and distance-time graphs) appear in almost every exam.

What this chapter covers:

Rest and motion, scalar vs vector quantities
Distance vs displacement
Speed vs velocity; uniform vs non-uniform motion
Acceleration
Equations of uniformly accelerated motion
Distance-time graphs and velocity-time graphs
Uniform circular motion

Key Concepts You Must Know

Distance vs Displacement

Distance is the total path length covered — a scalar quantity (magnitude only).

Displacement is the shortest straight-line distance from initial to final position — a vector quantity (magnitude and direction).

A person walking 100 m north and then 100 m south has:

Distance = 200 m
Displacement = 0 m (back to start)

Speed vs Velocity

Speed = Distance / Time (scalar)

Velocity = Displacement / Time (vector)

Average speed = Total distance / Total time

Average velocity = Total displacement / Total time

Uniform and Non-Uniform Motion

Uniform motion: Equal distances in equal time intervals — speed is constant.

Non-uniform motion: Unequal distances in equal intervals — speed changes.

Acceleration

Acceleration = Change in velocity / Time = $(v - u) / t$

Positive acceleration: velocity increasing
Negative acceleration (deceleration/retardation): velocity decreasing
Zero acceleration: constant velocity (uniform motion)

Important Formulas

v = u + at

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

v^2 = u^2 + 2as

s_n = u + \frac{a}{2}(2n-1)

where: $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time, $s$ = displacement

\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}

\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}

\text{Average velocity (uniform acceleration)} = \frac{u + v}{2}

Solved Previous Year Questions

PYQ 1 — Equations of Motion

Q: A car starts from rest and attains a velocity of 18 km/h in 5 s. Find (a) the acceleration, and (b) the distance covered in this time. (CBSE pattern)

Solution:

Convert: $u = 0$ , $v = 18$ km/h $= 18 \times \frac{5}{18} = 5$ m/s, $t = 5$ s.

(a) Using $v = u + at$ : $5 = 0 + a \times 5 \implies a = 1$ m/s²

(b) Using $s = ut + \frac{1}{2}at^2$ : $s = 0 + \frac{1}{2}(1)(5)^2 = 12.5$ m

PYQ 2 — Velocity-Time Graph Interpretation

Q: The velocity-time graph of a moving object is a straight line with positive slope passing through origin. What can you conclude?

Solution:

A straight line through origin: $v = at$ , meaning velocity increases uniformly from zero
Positive slope: positive acceleration
The object starts from rest and accelerates uniformly
Area under the v-t graph = displacement = $\frac{1}{2}at^2$ (triangle area)

PYQ 3 — Braking Distance

Q: A car is moving at 90 km/h. Brakes are applied and the car stops in 10 s. Find the deceleration and distance covered.

Solution:

$u = 90$ km/h $= 25$ m/s, $v = 0$ , $t = 10$ s.

$a = \frac{v - u}{t} = \frac{0 - 25}{10} = -2.5$ m/s² (deceleration)

$s = ut + \frac{1}{2}at^2 = 25 \times 10 + \frac{1}{2}(-2.5)(100) = 250 - 125 = 125$ m

Or using $v^2 = u^2 + 2as$ : $0 = 625 + 2(-2.5)s \implies s = 125$ m ✓

Difficulty Distribution

Difficulty	Question Type	Marks
Easy (30%)	Definitions; unit conversion; identify graph type	1–2 marks
Medium (40%)	One-equation numericals; graph interpretation; distance vs displacement	2–3 marks
Hard (30%)	Multi-step numericals; combined equations; graphical area calculations	4–5 marks

Expert Strategy

Always convert velocities to m/s before using equations of motion. The standard equations assume SI units (m, s, m/s, m/s²). Forgetting to convert km/h to m/s is the #1 error in this chapter.

Conversion: $x$ km/h $= x \times \frac{5}{18}$ m/s. Common conversions: 18 km/h = 5 m/s, 36 km/h = 10 m/s, 72 km/h = 20 m/s.

For velocity-time graphs: the slope = acceleration. The area under the graph = displacement (not distance — unless the velocity doesn’t change sign). If the line goes below the x-axis, the object is moving in the reverse direction.

When to use which equation:

Given $u$ , $a$ , $t$ → find $v$ : use $v = u + at$
Given $u$ , $a$ , $t$ → find $s$ : use $s = ut + \frac{1}{2}at^2$
Given $u$ , $v$ , $s$ → find $a$ : use $v^2 = u^2 + 2as$ (time not needed)
Given $u$ , $v$ , $a$ → find $t$ : use $v = u + at$

Common Traps

Trap 1 — Using $\text{distance} = \text{speed} \times \text{time}$ for non-uniform motion: This formula only applies for uniform (constant speed) motion. For accelerated motion, use the equations of motion.

Trap 2 — Confusing average speed with average velocity: A stone thrown up and caught at the same height has average velocity = 0 (displacement = 0) but average speed ≠ 0 (it travelled a non-zero distance).

Trap 3 — Forgetting to account for sign of acceleration in equations: If a car decelerates, $a$ is negative. Substituting $a$ as positive gives a distance larger than it should be. Write down the sign explicitly: $a = -2.5$ m/s².

Trap 4 — Circular motion has acceleration even at constant speed: In uniform circular motion, the speed is constant but velocity changes direction — so there IS acceleration (centripetal acceleration). Students say “speed is constant, so acceleration = 0” — wrong for circular motion.