Forces and motion · GCSE Physics
Acceleration
Teacher-written GCSE Physics revision on acceleration: a = Δv/t in m/s², velocity–time graphs where gradient is acceleration and area is distance, and v² − u² = 2as when time is missing.
Acceleration is the rate of change of velocity: a = Δv / t, in m/s². Gradient of a velocity–time graph is acceleration; area under it is displacement.
The important bits
What you need to know
- 1
Acceleration a = Δv / t = (v − u) / t. It is a vector. Units: metres per second squared (m/s²). u is initial velocity, v is final velocity.
- 2
Slowing down is deceleration, or a negative acceleration if the chosen positive direction is forwards.
- 3
Velocity–time graphs: straight line up means constant acceleration; horizontal means constant velocity (a = 0); straight line down means constant deceleration.
- 4
Gradient of a v–t graph = acceleration. Area between the line and the time axis = displacement (distance travelled if the line stays positive).
- 5
v² − u² = 2as is the motion equation used when time is not given. Keep a in m/s² and s in metres. u and v are in m/s.
- 6
Free-fall acceleration near Earth is g ≈ 9.8 m/s² downwards if air resistance is ignored. At terminal velocity, a = 0 even though speed is large.
- 7
A non-zero resultant force is what causes acceleration (Newton’s second law). Constant velocity means a = 0 and resultant force is zero.
- 8
Count squares on a v–t graph only after you know what one square is worth in metres: (m/s) × (s) = m.
Quotations worth analysing
Short evidence. Real method.
“Acceleration = change in velocity / time taken”
Change in velocity, not change in speed if the question is about a vector, and not velocity divided by time without subtracting u.
“The area under a velocity–time graph is the distance travelled.”
For a straight line that is a triangle or a trapezium. Write the scale of each square first if you are counting.
“v² − u² = 2as”
Use it when you have two speeds, an acceleration and a distance, but no time. Substitute with signs if motion reverses.
Go deeper
Area and gradient are different jobs on the same graph
On a velocity–time graph, a line below the time axis is motion the other way. The area between the line and the axis is still the distance travelled for that interval; subtract areas if you need displacement. Students mix this up with distance–time graphs and quote “gradient is distance”. Write a one-line legend on the sketch: gradient → a, area → s. For a journey that accelerates from rest to 12 m/s in 4.0 s then stays at 12 m/s for 6.0 s, acceleration in the first part is 3.0 m/s² and distance is the triangle plus the rectangle: ½ × 4.0 × 12 + 12 × 6.0 = 24 + 72 = 96 m. That arithmetic is the paper.
Go deeper
v² − u² = 2as is energy in disguise
If a car brakes from 20 m/s to rest with a = −8.0 m/s², 0 − 400 = 2 × (−8.0) × s, so s = 25 m. That matches work done: ½mv² transferred over distance s with F = ma. When time is missing, this equation is faster than finding t first. Keep the sign of a consistent with the direction you called positive. A ball thrown upwards at 15 m/s, taking g = −10 m/s², reaches a top where v = 0: 0 − 225 = 2 × (−10) × s, s = 11.25 m. Using g as +10 without flipping v would break the algebra. Write u, v, a, s, t in a list, tick what you have, then choose the equation. Do not hunt through all of them at once.
See the idea in action
A cyclist accelerates from 4.0 m/s to 10.0 m/s in 6.0 s. a = (10 − 4) / 6.0 = 1.0 m/s². On a velocity–time graph that is a straight line from 4 to 10 in 6 s. Distance = area of the trapezium = ½ × (4 + 10) × 6.0 = 42 m. Checking with v² − u² = 2as: 100 − 16 = 2 × 1.0 × s, so s = 42 m. The two methods agree. If the mass is 80 kg, resultant force F = ma = 80 × 1.0 = 80 N forwards.
Exam technique
Turn knowledge into marks
List u, v, a, s, t before you choose an equation. On a v–t graph, gradient is acceleration and area is distance. Convert km/h to m/s. Keep a negative if the object is slowing in your chosen direction.
Common mistakes
Do not give these marks away
- 01
Reading a distance–time slope as acceleration, or forgetting that area under a v–t graph is distance.
- 02
Using v/t instead of (v − u)/t, or leaving speed in km/h.
- 03
Using v² − u² = 2as with mixed units, or dropping the minus sign on deceleration.
What does the area under a velocity–time graph represent?
AAcceleration
BForce
CDistance travelled
DMass
Show the answer
Distance travelled. Velocity × time is distance when velocity is constant, and the area under a varying v–t graph generalises that product. Gradient would be acceleration.
Quick questions
If this is the bit you searched
How do you find acceleration from a velocity–time graph?
Calculate the gradient: change in velocity divided by change in time. A horizontal line means zero acceleration (constant velocity).
When do I use v² − u² = 2as?
When time is not given and you know two of the speeds, the acceleration and the distance (and need the missing one).
Is deceleration the same as negative acceleration?
If forwards is positive, slowing down is negative acceleration. Always state the sign convention so the examiner can follow.
What is the unit of acceleration?
Metres per second squared (m/s²): how much the velocity, in m/s, changes each second.