Acceleration — Changing Velocity

In Lesson 3 velocity told us how fast position was changing. But velocity itself rarely stays fixed — a train pulls away from a platform, a cyclist coasts to a halt, a dropped phone falls faster every instant. Whenever velocity is changing, something is accelerating. Acceleration is simply the answer to the question: "how quickly is the velocity changing?"

Acceleration is the rate of change of velocity. If the velocity changes by $\Delta v$ over a time interval $\Delta t$, the (average) acceleration is the change in velocity divided by the time it took:

Read that fraction carefully. The top is a velocity (units m/s); the bottom is a time (units s). So acceleration is "a velocity, divided by a time" — a measure of how much velocity is gained or lost in each second.

Acceleration is the rate of change of velocity: $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v-u}{\Delta t}$, where $u$ is the initial and $v$ the final velocity. It measures how much velocity is gained or lost each second.

Pause — copy the definition and the formula $a = (v-u)/\Delta t$, labelling $u$ and $v$, into your book before moving on.

We just saw that acceleration is a velocity divided by a time. That raises a question: if velocity is already "metres per second", what does dividing it by another "second" actually mean? This card answers it → it builds the unit m s⁻² from the formula and gives it a plain-English reading.

Units are not decoration — they fall straight out of the formula. The unit of acceleration tells you exactly what acceleration is, if you read it slowly.

Put the units of each quantity into $a = \dfrac{\Delta v}{\Delta t}$. Velocity is measured in metres per second (m/s, or m s⁻¹) and time in seconds (s). So:

The plain-English reading is the key to understanding it: an acceleration of 3 m s⁻² means the velocity increases by 3 m/s every second. After 1 s the object is moving 3 m/s faster; after 2 s it is 6 m/s faster; after 3 s, 9 m/s faster. The unit literally counts "how many metres-per-second of speed are added each second".

The SI unit of acceleration is m s⁻² ("metres per second, per second"). An acceleration of 3 m s⁻² means the velocity changes by 3 m/s every second. Never write it as m/s — that is a velocity.

Pause — copy the unit m s⁻², the working that derives it, and the plain-English reading ("velocity changes by 3 m/s each second") into your book before moving on.

We just saw how the unit m s⁻² counts the velocity gained each second. That raises a question: velocity is a vector with a direction — so does the velocity change, and the acceleration, carry a direction too? This card answers it → yes: acceleration is a vector, and on a line it gets a + or − sign.

Velocity is a vector, so its change $\Delta v = v - u$ is also a vector — it points somewhere. Dividing a vector ($\Delta v$) by a positive scalar (time) leaves the direction unchanged. Therefore acceleration is a vector: it has magnitude and direction.

In one-dimensional (rectilinear) motion we use the same +/− sign convention from Lesson 1. Choose a positive direction; then a positive acceleration points the positive way and a negative acceleration points the negative way. The direction of the acceleration is the direction in which the velocity is changing — the direction of $\Delta v$.

To find $\Delta v$ you subtract vectors, exactly as in Lesson 1: $\Delta v = v - u = v + (-u)$. Keep every sign. The sign of the final answer is the direction of the acceleration.

Acceleration is a vector — it has direction. The direction of acceleration is the direction of the velocity change ($\Delta v = v - u$), not the direction of motion. In 1D it takes a + or − sign from your chosen convention.

Pause — copy the rule "direction of acceleration = direction of $\Delta v$, not of motion" and the worked $\Delta v = (+8)-(+20) = -12$ m/s example into your book before moving on.

We just saw that acceleration takes a sign from the direction of the velocity change. That raises a question: students often read a minus sign as "slowing down" — is that safe? This card answers it → no: the sign is a direction, and whether an object speeds up or slows down depends on how $a$ compares with $v$.

Here is the single most-tested idea in this lesson. The sign of acceleration tells you its direction — nothing more. Whether an object is speeding up or slowing down depends on whether the acceleration points the same way as the velocity or the opposite way.

• Same signs ($v$ and $a$ both + or both −) → the object speeds up. The push is in the direction it is already moving.
• Opposite signs (one +, one −) → the object slows down. The push opposes the motion.

So a negative acceleration can mean either. A car moving east (+) with acceleration west (−) is slowing down. But a car already moving west (−) with acceleration west (−) is speeding up — both are negative, so the speed grows. The minus sign never decided that on its own.

The sign of acceleration is a direction, not "slowing down". If $v$ and $a$ have the SAME sign the object speeds up; if they have OPPOSITE signs it slows down. A negative acceleration speeds up an object that is already moving in the negative direction.

Pause — copy the rule "same signs → speed up, opposite signs → slow down" plus one example of a negative acceleration that speeds an object up, into your book before moving on.

We just saw how the signs of $v$ and $a$ decide speeding up versus slowing down. That raises a question: can we now put real numbers in and calculate an acceleration cleanly? This card answers it → yes, once the acceleration is uniform (constant), $a = (v-u)/\Delta t$ gives one clean number.

Uniform (constant) acceleration means the acceleration keeps the same magnitude and direction throughout the interval. The velocity then changes by equal amounts in equal times — gain 5 m/s in the first second, another 5 m/s the next, and so on. This is the special case the whole module is built on, and the case where $a = (v-u)/\Delta t$ gives a single, exact answer.

To calculate a uniform acceleration: write $u$, $v$ and $\Delta t$ with their signs, substitute into $a = (v-u)/\Delta t$, and read the sign of the answer as the direction.

Uniform acceleration is constant in magnitude and direction, so velocity changes by equal amounts in equal times. To calculate it, substitute signed $u$, $v$, $\Delta t$ into $a = (v-u)/\Delta t$ and read the sign as the direction. Always quote magnitude, direction and the unit m s⁻².

Pause — copy the definition of uniform acceleration and BOTH worked examples (sprinter +3.0 m s⁻²; braking car −3.0 m s⁻²) into your book before moving on.