GCSE · Physics · AQA · Spec 8463

Acceleration (a = Δv/t)

Every rocket launch, emergency stop, and skydive is defined by one number: acceleration.

What you need to know

  • Acceleration is the rate of change of velocity: a = (v − u) / t
  • Use v^2 − u^2 = 2 a s when time is unknown but distance is given — this equation is on your equation sheet
  • Deceleration is simply a negative acceleration — the same equation applies, the value just comes out negative
  • Near Earth's surface, free-fall acceleration is approximately 9.8 m/s²; terminal velocity occurs when drag equals weight so the resultant force is zero

The big picture

Acceleration measures how quickly velocity changes — not how fast something moves, but how fast its speed or direction is changing. A positive acceleration means speeding up; a negative acceleration means slowing down (deceleration). Near Earth's surface, freely falling objects accelerate at about 9.8 m/s², but drag eventually grows until it balances weight and terminal velocity is reached.

TONIGHT'S REVISION

Acceleration (a = Δv/t)

How quickly velocity changes — and why a falling object eventually stops accelerating

Forces on a falling object — at terminal velocity

Tap each force arrow to see what it does. Notice that the two arrows are equal in size — this is the key to terminal velocity.

SkydiverWeight (W)Drag (D)

Tap a force to see what it does.

Predict, then check

Commit to your answer before revealing — this is the exact thinking the examiner rewards.

A motorbike's velocity changes from 6 m/s to 30 m/s in 8 seconds. A student calculates acceleration as 30 ÷ 8 = 3.75 m/s². What is wrong, and what is the correct answer?

Terminal velocity — reason it through

?

Reason it through

Why does a skydiver eventually stop accelerating and fall at a constant speed?

Link 1 of 4

First link · your turn

What happens to drag as the skydiver falls faster and faster?

2
Locked — reveal the link above first
3
Locked — reveal the link above first
4
Locked — reveal the link above first

Relationship matrix

Tap any cell to reveal it. Tap a column header to read one property down every item.

Resultant forceAccelerationSpeed trend
Free fall (no air resistance)Near Earth's surface
Early stage of fall (air resistance present)Just after release, low speed
Terminal velocityDrag = Weight

Each cell hides a short answer and the reason behind it. Predict before you tap.

AQA Paper 2 · Exam application

Mark scheme practice — AQA style

Read the student answer and see which mark-scheme phrases it hits.

Question

Explain why a skydiver reaches terminal velocity. (4 marks)

Student answer

As the skydiver falls, their speed increases. This means drag increases. The resultant force decreases because drag is getting closer to weight. When drag equals weight, the resultant force is zero, so acceleration is zero and the skydiver falls at a constant speed called terminal velocity.

Method marks0/4

Equations you need

Taken directly from the exam-board specification.

a = (v − u) / t

a = acceleration (m/s^2) · v = final velocity (m/s) · u = initial velocity (m/s) · t = time (s)

Learn it — you must recall this in the exam

v^2 − u^2 = 2 a s

v = final velocity (m/s) · u = initial velocity (m/s) · a = acceleration (m/s^2) · s = distance (m)

Given on the equation sheet — you must know how to use it

Key points

1Acceleration = change in velocity ÷ time — it measures the CHANGE, not the speed itself
2Always subtract initial velocity from final velocity before dividing by time: a = (v − u) / t
3A negative answer means deceleration — the object is slowing down
4Free fall near Earth gives 9.8 m/s² — every second, velocity increases by 9.8 m/s (ignoring air resistance)
5Terminal velocity: when drag equals weight, resultant force = 0 and acceleration = 0, so speed stays constant

Worked example

Problem

A cyclist accelerates from rest to 12 m/s over 8 seconds. Calculate her acceleration.

🧠

Memory hook

Think of acceleration as the 'velocity speedometer's rate of change' — just as speed tells you how fast position changes, acceleration tells you how fast speed changes. Picture a sprinter: they don't instantly hit top speed, they build it second by second — that build-rate is acceleration.

★ Exam tip

On AQA Paper 2, if a question gives you two velocities and a distance but no time, reach straight for v^2 − u^2 = 2 a s from the equation sheet — identify it before you start working, not halfway through.

⚠ Watch out

Forgetting to subtract the initial velocity — if an object is already moving, you MUST use (v − u) in a = (v − u) / t, not just v. Using only the final velocity silently drops the 'change in' meaning.

Check yourself

Without looking: what is the acceleration of an object whose velocity changes from 25 m/s to 10 m/s in 5 seconds, and is this value positive or negative?

Flashcards

(25)
What does acceleration measure?
The rate of change of velocity — how quickly velocity increases or decreases per second.
Write the equation for uniform acceleration.
a = (v − u) / t
What are the units of acceleration?
Metres per second squared (m/s²).
What is deceleration in terms of acceleration?
Deceleration is a negative acceleration — the object is slowing down, so the value of a is negative.
What is the approximate acceleration due to gravity near Earth's surface?
9.8 m/s².
What does terminal velocity mean?
The constant velocity reached when drag force equals the weight of a falling object, so the resultant force is zero and acceleration is zero.
Why is resultant force zero at terminal velocity?
Drag (upward) has grown to equal weight (downward) — the two forces are balanced, so there is no net force and no acceleration.
An object accelerates from 5 m/s to 25 m/s in 4 s. What is a?
a = (25 − 5) / 4 = 20 / 4 = 5 m/s²
Which equation links velocity, acceleration and distance when time is unknown?
v^2 − u^2 = 2 a s
Is v^2 − u^2 = 2 a s recalled or given on the AQA equation sheet?
Given on the physics equation sheet — you apply it, you don't need to memorise it.
What does 'u' represent in the acceleration equations?
u = initial velocity in m/s — the velocity at the START of the time interval.
What does 'v' represent in the acceleration equations?
v = final velocity in m/s — the velocity at the END of the time interval.
A car brakes from 30 m/s to 0 m/s in 6 s. What is the acceleration?
a = (0 − 30) / 6 = −5 m/s². Negative — this is deceleration.
Why does a skydiver accelerate at first after jumping?
Weight is greater than drag, so there is a resultant downward force — the skydiver accelerates downward.
Why does a skydiver's acceleration decrease as they fall faster?
As speed increases, drag increases, reducing the resultant force — so acceleration gets smaller even though they are still speeding up.
At terminal velocity, what is the value of acceleration?
Zero — the resultant force is zero, so there is no acceleration and speed is constant.
A ball is dropped and falls freely (ignore air resistance). What is its acceleration?
9.8 m/s² downward — the acceleration of free fall near Earth's surface.
What does a negative value of acceleration always mean physically?
The object is slowing down (decelerating) — velocity is decreasing over time.
What practical (RP7) is linked to acceleration at GCSE?
RP7 — investigating the effect of varying force and varying mass on acceleration, using a trolley and light gates.
If u = 0 (starts from rest), how does v^2 − u^2 = 2 a s simplify?
v² = 2as — the u² term becomes zero and drops out.
What is the change in velocity (Δv) for an object going from 8 m/s to 20 m/s?
Δv = 20 − 8 = 12 m/s.
A stone is thrown upward. What is its acceleration during flight (ignore air resistance)?
9.8 m/s² downward throughout — gravity acts constantly, decelerating it on the way up and accelerating it on the way down.
Which AQA paper is acceleration assessed on?
Paper 2.
An object travelling at 10 m/s accelerates at 3 m/s² for 5 s. What is its final velocity?
Rearrange a = (v − u) / t: v = u + at = 10 + (3 × 5) = 25 m/s.
What happens to drag force as a falling object speeds up?
Drag increases — air resistance grows with speed, so the faster the object falls, the greater the upward drag force.

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Learn Acceleration (a = Δv/t) properly — interactive practice, marked questions and flashcards.

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