If velocity tells you how fast you’re going, acceleration tells you how fast that velocity is changing. Speeding up, slowing down, or even just changing direction — all of these are acceleration. It’s the next step up the ladder: displacement → velocity → acceleration.
📘 What you need to know
Acceleration is the rate of change of velocity — how much the velocity changes each second.
It’s a vector (it has direction), measured in metres per second squared (m s⁻²).
Formula: a = Δv ÷ Δt, where Δv = change in velocity and Δt = time taken.
Change in velocity = final velocity − initial velocity, i.e. Δv = v − u.
Speeding up → positive acceleration. Slowing down → negative acceleration (deceleration).
On a velocity–time graph, the gradient (steepness) is the acceleration.
What acceleration really means
Acceleration is the rate of change of velocity — in plain words, how much your velocity goes up or down every second. If a car’s velocity climbs by 3 m s⁻¹ each second, its acceleration is 3 m s⁻². The unit looks strange at first — “metres per second, per second” — but it’s just telling you how quickly the speed itself is changing.
Average accelerationa = change in velocity (Δv)time taken (Δt)
=
v − uΔt
Here u is the initial velocity (the one you start with) and v is the final velocity (the one you end with). These two letters come back again and again in kinematics, so it’s worth locking them in now.
A quick memory aid: u comes before v in the alphabet, just like the initial velocity comes before the final velocity in time.
Positive vs negative acceleration
Because acceleration is a vector, its sign matters. There are two everyday cases:
Speeding up in the direction you’re already moving gives a positive acceleration. Slowing down gives a negative acceleration — we often call this deceleration. (Watch out though: an object can also have a negative acceleration simply because it’s accelerating in the negative direction, even while speeding up.)
Left: a rocket’s velocity grows each second — positive acceleration. Right: a car’s velocity shrinks as it nears the sign — negative acceleration (deceleration).
One-line summary: acceleration = (final velocity − initial velocity) ÷ time. Positive means speeding up; negative means slowing down (in that direction).
Instantaneous acceleration
Just like with velocity, there’s a difference between average and instantaneous. Average acceleration is worked out over a whole time interval. Instantaneous acceleration is the value at one exact moment.
On a velocity–time graph, the acceleration is the gradient of the line. A straight line means constant acceleration; a curved line means the acceleration itself is changing, and you’d read the instantaneous value by taking the gradient of a tangent at that point.
On a v–t graph the gradient is the acceleration. A straight line = constant acceleration; a curve = changing acceleration.
🧭 Recipe — finding acceleration from a velocity–time graph
Straight line? The acceleration is constant — just take the gradient: rise (change in v)run (change in t)
Curved line? The acceleration is changing. For the value at one instant, draw a tangent and find its gradient.
Watch the sign: a downward-sloping line means the velocity is dropping → negative acceleration.
Units check: velocity is in m s⁻¹ and time in s, so the gradient comes out in m s⁻².
Worked examples
WE 1
A bullet train slowing down
A Japanese bullet train decelerates at a constant rate in a straight line. Its velocity drops from an initial 50 m s⁻¹ to a final 42 m s⁻¹ in 30 seconds. (a) Find the change in velocity. (b) Find the deceleration, and explain how the answer shows the train is slowing down.
(a) Change in velocity: Δv = v − uΔv = 42 − 50Δv = −8 m s⁻¹(b) Acceleration: a = Δv ÷ Δta = −8 ÷ 30a = −0.27 m s⁻²the minus sign shows the velocity is decreasing — the train is slowing down.
WE 2
A car pulling away from rest
A car starts from rest and reaches a velocity of 18 m s⁻¹ in 6.0 s along a straight road. Calculate its average acceleration.
Start from rest, so u = 0Δv = 18 − 0 = 18 m s⁻¹a = Δv ÷ Δta = 18 ÷ 6.0a = 3.0 m s⁻²positive, because the car is speeding up in the direction it’s travelling.
WE 3
A cyclist braking to a stop
A cyclist travelling at 12 m s⁻¹ brakes and comes to a complete stop in 4.0 s. Find the acceleration.
Comes to a stop, so v = 0Δv = 0 − 12 = −12 m s⁻¹a = Δv ÷ Δta = −12 ÷ 4.0a = −3.0 m s⁻²negative because the cyclist is decelerating.
WE 4
Reading acceleration off a graph
On a velocity–time graph, a straight line rises from 4 m s⁻¹ to 20 m s⁻¹ over a time of 8.0 s. Find the acceleration.
Acceleration = gradient = rise ÷ runrise = 20 − 4 = 16 m s⁻¹a = 16 ÷ 8.0a = 2.0 m s⁻²a straight line, so this acceleration is constant the whole way.
💡 Top tips
Always do Δv = v − u in that order (final minus initial). Swapping them flips the sign and ruins the answer.
“From rest” means u = 0; “comes to a stop” means v = 0. Spotting these saves time.
Keep the units as m s⁻² — if the velocity is in km h⁻¹, convert to m s⁻¹ first.
On a v–t graph, gradient = acceleration (not the height of the line — that’s the velocity).
⚠ Common mistakes
Mixing up velocity and acceleration: a fast-moving object can have zero acceleration if its velocity isn’t changing.
Dropping the minus sign on a deceleration — the sign is what shows the object is slowing down.
Reading the height instead of the gradient on a velocity–time graph.
Assuming negative acceleration always means slowing down — it can also mean accelerating in the negative direction.
Forgetting acceleration is a vector — direction matters, so always think about the sign.
Up next: Kinematic Equations (the “SUVAT” equations) — four handy formulas that tie displacement, velocity, acceleration, and time together so you can solve almost any motion problem.
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