How quickly does velocity change? Acceleration answers that question — and its direction tells you whether an object is speeding up or slowing down. From NCERT Chapter 4 (Exploration edition) Class 9 Science. Aligned with CBSE syllabus 2026-27.
Q. Have you ever felt a sudden jolt when a bus starts moving or brakes sharply?
When you are standing in a bus and it suddenly starts from rest, you feel pushed backwards. When it brakes sharply, you lurch forward. That physical sensation — the jolt — is your body's experience of acceleration. It happens whenever velocity changes.
Recall that velocity is a vector: it has both magnitude (speed) and direction. Velocity can change in three ways:
In all three cases, the velocity has changed, so the object has undergone acceleration. Acceleration, therefore, is the physical quantity that measures how rapidly velocity changes.
Velocity told us how quickly position changes. Acceleration tells us how quickly velocity changes. Just as we needed velocity to describe moving objects, we need acceleration to describe objects whose motion is changing — which is most real-world motion. Without acceleration, we cannot describe a car speeding up, a ball falling, or a rocket launching.
Q. How do we define average acceleration mathematically?
The average acceleration of an object is the change in its velocity divided by the time interval over which that change occurs.
If an object has velocity $u$ at time $t_1$ and velocity $v$ at time $t_2$, then:
$$a = \frac{v - u}{t_2 - t_1} \quad \cdots \text{(Eq. 4.3c)}$$Where:
SI unit: metre per second squared (m s⁻² or m/s²)
Type: Vector quantity — acceleration has both magnitude and direction
Q. Why is the unit of acceleration m s⁻² (metres per second squared)?
Acceleration is change in velocity per unit time. Velocity is measured in m s⁻¹. So:
$$\text{Unit of acceleration} = \frac{\text{m s}^{-1}}{\text{s}} = \text{m s}^{-2}$$An acceleration of 2 m s⁻² means the velocity of the object increases (or decreases) by 2 m/s every second. After 1 s, it is 2 m/s faster; after 2 s, it is 4 m/s faster — and so on (provided the acceleration is constant).
The formula is often simplified to $a = \dfrac{v - u}{t}$ when the initial time is taken as zero. Both forms are equivalent. In NCERT Equation 4.3c, the general form with $t_2 - t_1$ is used to emphasise that it is a time interval, not just a single moment.
Q. How do we determine the direction of acceleration?
Because acceleration is a vector, its direction is determined by the direction of the change in velocity (i.e., $v - u$), not the direction of velocity itself. This leads to two important cases for motion along a straight line.
When an object speeds up in the positive direction, its final velocity $v$ is greater than its initial velocity $u$. So $v - u > 0$, which means acceleration is positive — in the same direction as the motion (velocity).
Example: A car starting from rest and moving forward — acceleration is directed forward, same as velocity.
When an object slows down, its final velocity $v$ is less than its initial velocity $u$. So $v - u < 0$, which means acceleration is negative — directed opposite to the direction of velocity.
Example: A car braking to a stop — velocity is forward, but acceleration is backward (opposing the motion).
Even if speed is constant, if the direction of velocity changes, the object is accelerating. For example, a car moving at constant speed around a circular bend — its speed is fixed, but the velocity direction keeps changing. This gives rise to centripetal acceleration, which is covered in the chapter on uniform circular motion.
Yes. Speed is the magnitude of velocity. If the direction of velocity changes (even with constant speed), the velocity vector changes — so the object accelerates. This is exactly what happens in uniform circular motion: the object moves in a circle at constant speed, but is continuously accelerating towards the centre.
Q. What is constant (uniform) acceleration?
An object has constant acceleration (also called uniform acceleration) if its velocity changes by equal amounts in equal intervals of time, no matter how small those intervals are. The value of acceleration remains the same throughout the motion.
Constant acceleration is important because the three kinematic equations of motion (NCERT Eq. 4.4a, 4.4b, 4.4c) are derived specifically for this case. They cannot be used when acceleration is changing.
| Time (s) | Velocity (m/s) (starting from rest, a = 2 m s⁻²) |
Change in Velocity per Second |
|---|---|---|
| 0 | 0 | — |
| 1 | 2 | +2 m/s (constant) |
| 2 | 4 | +2 m/s (constant) |
| 3 | 6 | +2 m/s (constant) |
| 4 | 8 | +2 m/s (constant) |
Each second, the velocity increases by exactly 2 m/s — this is constant acceleration.
A bus is moving along a straight highway. In the first phase, the driver presses the accelerator and the bus speeds up from 10 m s⁻¹ to 15 m s⁻¹ in 10 seconds. In the second phase, the driver applies the brakes and the bus slows down from 15 m s⁻¹ to rest (0 m s⁻¹) in 5 seconds. Find the acceleration in each phase.
Given:
Using Eq. 4.3c:
$$a = \frac{v - u}{t} = \frac{15 - 10}{10} = \frac{5}{10} = +0.5 \text{ m s}^{-2}$$Result: Acceleration = +0.5 m s⁻² (positive — in the direction of motion; bus is speeding up)
Given:
Using Eq. 4.3c:
$$a = \frac{v - u}{t} = \frac{0 - 15}{5} = \frac{-15}{5} = -3 \text{ m s}^{-2}$$Result: Acceleration = −3 m s⁻² (negative — opposite to direction of motion; this is retardation). The magnitude, 3 m s⁻², tells us the bus loses 3 m/s of speed every second during braking.
Q. What is free fall, and how does it introduce us to the acceleration due to gravity?
When an object is dropped from a height and only gravity acts on it (no air resistance), it undergoes free fall. The object accelerates downward at a constant rate because Earth's gravitational pull is essentially uniform near the surface. This constant acceleration is called the acceleration due to gravity, symbolised by $g$.
Symbol: $g$ | Value (near Earth's surface): $g \approx 9.8$ m s⁻² | Direction: downward (towards Earth's centre)
Every freely falling object gains approximately 9.8 m/s of downward velocity for every second it falls.
Consider an object dropped from rest (initial velocity = 0). Let us take downward as positive. The velocity at each second is:
| Time (s) | Velocity (m s⁻¹) (downward positive) |
Change in Velocity in that Second | Acceleration (m s⁻²) |
|---|---|---|---|
| 0 | 0 | — | — |
| 1 | 9.8 | 9.8 − 0 = 9.8 | 9.8 |
| 2 | 19.6 | 19.6 − 9.8 = 9.8 | 9.8 |
| 3 | 29.4 | 29.4 − 19.6 = 9.8 | 9.8 |
| 4 | 39.2 | 39.2 − 29.4 = 9.8 | 9.8 |
In every one-second interval, the velocity increases by exactly 9.8 m s⁻¹. Therefore, the acceleration is constant and equal to:
$$g = 9.8 \text{ m s}^{-2} \text{ (downward)}$$The value of $g$ is not exactly the same everywhere on Earth. It is slightly larger at the poles (9.83 m s⁻²) and slightly smaller at the equator (9.78 m s⁻²) due to Earth's shape and rotation. For Class 9 calculations, we use $g = 9.8$ m s⁻² (or sometimes $10$ m s⁻² for approximate calculations).
Galileo Galilei famously demonstrated (around 1590) that all objects fall at the same rate regardless of their mass — a discovery that challenged centuries of Aristotelian belief. On the Moon (where there is no air resistance and weaker gravity), a hammer and a feather dropped simultaneously do hit the ground at the same time, as demonstrated by Apollo 15 astronaut David Scott in 1971.
Students often assume that a fast-moving object must have a large acceleration, or that a slow-moving object cannot have large acceleration. Both assumptions are wrong.
The truth: Acceleration depends only on the change in velocity, not on the velocity itself.
Key insight: Acceleration is about the rate of change of velocity. An object moving at constant velocity — no matter how fast — has zero acceleration.
This is why the NCERT textbook carefully uses the word "change" in the definition of acceleration. Speed alone tells you nothing about acceleration. Always look for whether velocity is changing.
NCERT Activity 4.2 asks students to collect data on how quickly different cars can accelerate from 0 to 100 km h⁻¹ and then calculate their corresponding acceleration in m s⁻².
Step 1: Convert 100 km h⁻¹ to m s⁻¹:
$$100 \text{ km h}^{-1} = 100 \times \frac{5}{18} \approx 27.78 \text{ m s}^{-1}$$Step 2: Use the acceleration formula with $u = 0$, $v = 27.78$ m s⁻¹, and $t$ = time taken (in seconds):
$$a = \frac{27.78}{t} \text{ m s}^{-2}$$| Car | 0–100 km h⁻¹ Time (s) | Acceleration (m s⁻²) | Comparison |
|---|---|---|---|
| Sports Car A | 4.0 s | $27.78 / 4.0 \approx$ 6.9 m s⁻² | Nearly 0.7 g — very high |
| Family Sedan B | 9.0 s | $27.78 / 9.0 \approx$ 3.1 m s⁻² | Moderate |
| Economy Hatchback C | 14.0 s | $27.78 / 14.0 \approx$ 2.0 m s⁻² | Modest |
| City Bus D | 25.0 s | $27.78 / 25.0 \approx$ 1.1 m s⁻² | Low (heavy vehicle) |
Q. What is retardation and is it different from acceleration?
Retardation (also called deceleration) is acceleration that opposes the direction of motion — it slows the object down. Mathematically, it is simply negative acceleration.
Retardation is not a separate physical quantity. It is acceleration with a negative sign when the chosen direction of motion is positive.
| Parameter | Acceleration | Retardation |
|---|---|---|
| Effect on speed | Increases speed (if in direction of motion) | Decreases speed |
| Direction relative to velocity | Same direction as velocity (when speeding up) | Opposite direction to velocity |
| Sign (taking motion as positive) | Positive | Negative |
| Is it a separate quantity? | Yes — acceleration | No — it is negative acceleration |
| Example | Car pressing accelerator: +0.5 m s⁻² | Car braking: −3 m s⁻² |
When an exam question asks for "retardation", give the magnitude of the acceleration (positive number) and state the direction separately. For example: "The retardation is 3 m s⁻² directed opposite to the motion" — rather than writing "acceleration = −3 m s⁻²". Both are correct physics, but the first phrasing avoids sign confusion in the answer.
The acceleration we have studied so far — calculated over a finite time interval using $a = (v-u)/t$ — is the average acceleration. In reality, the acceleration of an object can change from moment to moment, just as the instantaneous speed shown on a speedometer differs from average speed.
Instantaneous acceleration is the acceleration of an object at a specific instant of time. It is defined as the limiting value of average acceleration as the time interval approaches zero:
$$a_{\text{inst}} = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}$$This is the derivative of velocity with respect to time — a concept from calculus that you will study in detail in Class 11. For Class 9, the NCERT Exploration chapter introduces the idea qualitatively, asking you to recognise that average and instantaneous acceleration can differ for non-uniform acceleration.
For a deeper study of instantaneous acceleration, visit: Instantaneous Velocity and Acceleration →
Q1. Define acceleration. What is its SI unit and what type of quantity is it?
Acceleration is the rate of change of velocity. It is defined as the change in velocity divided by the time interval: $a = (v-u)/t$.
SI unit: metre per second squared (m s⁻²). Type: Vector quantity — it has both magnitude and direction.
Q2. What is retardation? Give two examples from everyday life.
Retardation (or deceleration) is the negative acceleration of a body — it occurs when the acceleration is directed opposite to the velocity, causing the object to slow down. It is not a separate physical quantity; it is simply negative acceleration.
Examples: (1) A car applying brakes — it slows down as the braking force opposes forward motion. (2) A ball thrown vertically upward — gravity decelerates it at 9.8 m s⁻² until it momentarily stops at the top.
Q3. Can an object moving at high speed have zero acceleration? Give an example.
Yes. Acceleration depends on the change in velocity, not on the magnitude of velocity. If an object moves at constant velocity (same speed, same direction), its acceleration is zero regardless of how fast it is moving.
Example: A spacecraft travelling through empty space at 10,000 m/s with no forces acting on it — zero acceleration (Newton's first law). Or a bullet travelling at 800 m/s after it leaves the gun, if air resistance is negligible.
Q4. What is acceleration due to gravity? What is its value near Earth's surface?
Acceleration due to gravity ($g$) is the constant acceleration with which all objects fall freely towards Earth's surface when only gravity acts on them (no air resistance).
Near Earth's surface: $g \approx 9.8$ m s⁻², directed downward (towards Earth's centre).
It means a freely falling object gains 9.8 m/s of downward velocity for every second of fall.
Q5. Distinguish between uniform acceleration and non-uniform acceleration.
Uniform acceleration: Velocity changes by equal amounts in equal time intervals. The magnitude and direction of acceleration remain constant. Example: free fall near Earth's surface.
Non-uniform acceleration: Velocity changes by unequal amounts in equal time intervals. The magnitude or direction (or both) of acceleration varies. Example: a car in city traffic — it accelerates, brakes, and turns; the acceleration changes continuously.
N1. A motorcycle starting from rest attains a velocity of 18 m s⁻¹ in 6 seconds. Calculate its acceleration.
Given: $u = 0$ m s⁻¹, $v = 18$ m s⁻¹, $t = 6$ s
$$a = \frac{v - u}{t} = \frac{18 - 0}{6} = \mathbf{3 \text{ m s}^{-2}}$$The motorcycle accelerates at 3 m s⁻² in the direction of motion.
N2. A train moving at 72 km h⁻¹ applies brakes and comes to rest in 10 seconds. Find its acceleration. Is this acceleration positive or negative?
Step 1: Convert 72 km h⁻¹ to m s⁻¹:
$$u = 72 \times \frac{5}{18} = 20 \text{ m s}^{-1}$$Step 2: Apply the formula ($v = 0$, $t = 10$ s):
$$a = \frac{0 - 20}{10} = \mathbf{-2 \text{ m s}^{-2}}$$The acceleration is negative (retardation) — directed opposite to the direction of motion. The train decelerates at 2 m s⁻².
N3. An object has an initial velocity of 5 m s⁻¹. After 4 seconds it has a velocity of 13 m s⁻¹. Calculate (a) the acceleration and (b) the velocity after another 3 seconds if the same acceleration continues.
(a) Acceleration:
$$a = \frac{13 - 5}{4} = \frac{8}{4} = \mathbf{2 \text{ m s}^{-2}}$$(b) Velocity after another 3 s: Now $u = 13$ m s⁻¹, $a = 2$ m s⁻², $t = 3$ s
$$v = u + at = 13 + (2 \times 3) = 13 + 6 = \mathbf{19 \text{ m s}^{-1}}$$N4. A stone is dropped from a height. (a) What is its velocity after 3 s? (b) What is its velocity after 5 s? Take $g = 9.8$ m s⁻².
Given: $u = 0$ (dropped from rest), $a = g = 9.8$ m s⁻² (downward)
Using $v = u + at$:
(a) After 3 s:
$$v = 0 + 9.8 \times 3 = \mathbf{29.4 \text{ m s}^{-1} \text{ (downward)}}$$(b) After 5 s:
$$v = 0 + 9.8 \times 5 = \mathbf{49 \text{ m s}^{-1} \text{ (downward)}}$$Navigate to related topics and practice resources for Chapter 4: Describing Motion Around Us.