Acceleration

A car pulling away from a signal does not jump to its cruising speed — its velocity climbs second by second. A cyclist squeezing the brakes does the opposite: velocity falls away, second by second. Both cars and cyclists are showing acceleration, the quantity that measures how fast velocity itself is changing. It sits right after velocity in the Class 9 chapter on Motion, and it is the bridge to everything that follows: distance-time graphs, velocity-time graphs, and the equations of motion. Get comfortable with one plain idea — acceleration is "change in velocity, per second" — and every numerical in this topic becomes a matter of carefully reading which velocity came first and which came last, then dividing by the time between them.

The core insight

Acceleration is the rate at which velocity changes with time, $a = (v-u)/t$ , positive when speeding up and negative (retardation) when slowing down.

The core idea

Acceleration is defined as the rate of change of velocity with time:

a = \frac{v - u}{t}

where: $u$ is the initial velocity of the object, $v$ is its final velocity, $t$ is the time taken for the velocity to change from $u$ to $v$ , and $a$ is the acceleration produced. The SI unit of acceleration is metre per second squared, written $\text{m/s}^2$ or $\text{m s}^{-2}$ . The sign of $a$ tells the whole story. If velocity increases with time, $a$ comes out positive. If velocity decreases with time, $a$ comes out negative, and this is given a special name — retardation or deceleration. If velocity does not change at all, $a$ is zero, whether the object is at rest or moving at a constant speed. Motion is further sorted by how steadily the velocity changes:

Uniform acceleration — velocity changes by equal amounts in equal intervals of time, such as a body falling freely near Earth's surface (ignoring air resistance).
Non-uniform acceleration — velocity changes by unequal amounts in equal intervals of time, such as a car crawling through traffic, constantly speeding up and braking.

Class 9 does not use vector notation, so direction is carried in words rather than arrows: a common convention treats velocity along the direction of motion as positive, so an object slowing down has an acceleration pointing opposite to its own motion. Thinking of acceleration as "how much speed is gained (or lost), every second" makes the unit obvious too — $5\ \text{m/s}^2$ means $5\ \text{m/s}$ of speed is added with every second that passes, as long as the acceleration stays uniform. This is exactly why the unit is metres per second, per second, compactly written $\text{m/s}^2$ .

Figure

A velocity time graph rising in a straight line from zero at time zero to eighteen metres per second at six seconds, with a labelled rise over run triangle showing the slope equals the acceleration

Derivation AComputing acceleration from initial and final velocity

The goal is to turn three measured quantities — an initial velocity, a final velocity, and the time between them — into a single signed number that tells us both how strongly and in which sense (speeding up or slowing down) an object's motion is changing, using nothing beyond the definition of acceleration itself.

Step 1. Start from the definition. Acceleration is the change in velocity divided by the time taken for that change: $a = \dfrac{\text{change in velocity}}{\text{time taken}}$ . The change in velocity is simply the final velocity minus the initial velocity, so we write $\Delta v = v - u$ .

Step 2. Assemble the formula. Substituting $\Delta v = v - u$ into the definition gives $a = \dfrac{v - u}{t}$ , the working formula for every acceleration numerical in this chapter. Note that $u$ must always be the velocity at the start of the interval being considered, and $v$ the velocity at its end — swapping them flips the sign of the answer.

Step 3. Read the sign once the arithmetic is done. If $v > u$ , the numerator $v - u$ is positive, so $a$ is positive: the object is speeding up. If $v < u$ , the numerator is negative, so $a$ is negative — retardation, the object is slowing down. If $v = u$ , then $a = 0$ : no change in velocity, regardless of how fast the object happens to be moving.

Step 4. Confirm the units. Velocity is measured in $\text{m/s}$ and time in seconds, so acceleration carries units of $\text{m/s}$ divided by $\text{s}$ , giving $\text{m/s}^2$ — a rate of a rate, exactly as the definition promised.

Worked example

Question: A car starts from rest and attains a velocity of $18\ \text{m/s}$ in $6\ \text{s}$ . Assuming the acceleration to be uniform, find its acceleration.

Step 1. List what is given. Initial velocity $u = 0\ \text{m/s}$ (the car "starts from rest"), final velocity $v = 18\ \text{m/s}$ , and time $t = 6\ \text{s}$ .

Step 2. Substitute into $a = \dfrac{v - u}{t}$ :

a = \frac{18 - 0}{6} = 3\ \text{m/s}^2

Step 3. Interpret the result. The car's acceleration is $\mathbf{3\ \text{m/s}^2}$ , meaning its velocity increases by $3\ \text{m/s}$ during every second of the six seconds it takes to reach top speed.

Competitive stretch

Question: A cyclist travelling at $12\ \text{m/s}$ applies the brakes and comes to rest in $4\ \text{s}$ . Find the acceleration and explain its sign.

Step 1. Here $u = 12\ \text{m/s}$ , $v = 0\ \text{m/s}$ (comes to rest), $t = 4\ \text{s}$ .

Step 2. $a = \dfrac{0 - 12}{4} = -3\ \text{m/s}^2$ .

Step 3. The negative sign shows velocity decreasing with time — retardation of magnitude $3\ \text{m/s}^2$ , acting opposite to the cyclist's direction of motion. Notice that both examples use the identical formula; only the roles of $u$ and $v$ changed, which is why sorting out "initial" from "final" before substituting is the single most important step in this topic.

Common traps

Mixing up which velocity is "initial" ( $u$ ) and which is "final" ( $v$ ) — this flips the sign of the whole answer.
Assuming negative acceleration means the object moves backwards — it only means velocity is decreasing; the object can still be moving forward while slowing down.
Assuming zero acceleration means the object is at rest — a body moving at constant, unchanging velocity also has zero acceleration.
Confusing uniform acceleration (equal velocity changes in equal times) with uniform velocity (equal position changes in equal times) — these describe two different situations.

Ask the tutor: Stuck on identifying $u$ and $v$ in a tricky wording, or unsure what a negative acceleration means for a specific motion — ask the AI tutor to work through the problem with you step by step.

Why does a negative answer for acceleration not mean the object is moving backwards?

Acceleration's sign describes how velocity is changing, not the direction of the object's motion. A negative acceleration simply means velocity is decreasing over time — the object can be travelling forward the entire time, just getting slower each second, exactly like a cyclist braking. Only when we also know the object's velocity is negative (moving in the "negative" direction by our sign convention) and its acceleration is positive would the object actually be reversing. At Class 9 level, always check what "positive" was chosen to mean in that specific problem before deciding what a negative acceleration implies physically.

Is zero acceleration the same as being at rest?

No. Zero acceleration only means velocity is not changing — the object could be sitting still, or it could be cruising at a constant $20\ \text{m/s}$ forever. Both situations have $u = v$ , so $a = (v-u)/t = 0$ . What zero acceleration rules out is any change in speed or direction of motion; it says nothing about whether that constant velocity happens to be zero.

How do I decide whether a given time value belongs with a given pair of velocities?

Check that $u$ , $v$ , and $t$ all describe the same stretch of motion — the same object, over the same interval, with no other event happening in between. Word problems sometimes give a longer total time but only want the acceleration during part of it, or mention two separate phases of motion (speeding up, then braking) that must be treated as two separate calculations, each with its own $u$ , $v$ , and $t$ . Reading the question twice before substituting numbers avoids this trap.

Acceleration

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads