This unit is ≈17% of the A-Level Physics, across 7 lessons. Full syllabus
Lesson 2 of 7 · Mechanics and materials
Kinematics, SUVAT and projectile motion
8 min read · about 1 h 55 min with practice3 quick checks≈3% of the testCore: Core: tested on most papers
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Kinematics is the grammar of the mechanics papers. Nearly every AS or Paper 1 series has a 3–6 mark SUVAT or projectile calculation, one-mark MCQs on motion graphs, and a question on the free-fall practical to find g. Later topics (momentum, energy, circular motion, fields) assume this method is automatic.
By the end you’ll be able to
Interpret and sketch displacement–time, velocity–time and acceleration–time graphs using gradients and areas
Select and apply the SUVAT equations to single- and multi-stage motion under constant acceleration
Describe and analyse free-fall experiments to determine g
Solve projectile problems by treating horizontal (constant velocity) and vertical (constant acceleration) motion independently
Explain qualitatively how air resistance changes projectile trajectories
What the exam asks
Graph reading (MCQ and structured): gradients and areas of s–t, v–t and a–t graphs, tangents to curves, and “sketch the corresponding graph”.
SUVAT calculations (2–5 marks): single- and multi-stage motion, usually with a sign trap such as a ball that lands below its starting point.
Projectiles (3–6 marks): horizontal launches from a table or cliff, angled launches from the ground or a height, and “does it clear the wall?”
The free-fall practical (AQA required practical 3, Edexcel core practical 1, OCR A PAG 1; Cambridge Papers 3 and 5): method, linearised graph and systematic errors, often as a 6-mark extended response or a Paper 3 data question.
Air resistance, qualitatively: why the real trajectory is lower, shorter and lopsided.
Core ideas
Definitions you must be able to state
Quantity
Definition
Scalar or vector
Unit
Displacement s
distance moved in a stated direction from a reference point
vector
m
Velocity v
rate of change of displacement,
vii.Check your understanding
3 questions on kinematics, SUVAT and projectile motion. Every option is explained once you answer.
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PromptCard 1 of 3
What does the area under a velocity–time graph represent?
v=ΔtΔs
vector
m s−1
Speed
rate of change of distance
scalar
m s−1
Acceleration a
rate of change of velocity, a=ΔtΔv
vector
m s−2
Instantaneous velocity is the gradient of the tangent to the s–t graph. Average velocity is total displacement ÷ total time; average speed is total distance ÷ total time, and the two differ whenever the object turns back.
Motion graphs
Graph
Gradient gives
Area under the line gives
displacement–time
velocity
nothing useful
velocity–time
acceleration
displacement
acceleration–time
not examined
change in velocity
Area below the time axis of a v–t graph is negative displacement: add magnitudes for distance, add with signs for displacement. A bouncing ball’s v–t graph is a set of parallel lines of gradient −g (up positive), with a near-vertical jump at each bounce.
The constant-acceleration (SUVAT) equations
v=u+ats=2(u+v)ts=ut+21at2v2=u2+2as
Each equation leaves out one of s,u,v,a,t: pick the one that leaves out the quantity you neither know nor need. They are only valid for constant acceleration, so they fail once drag matters. They come from a straight-line v–t graph: the gradient gives v=u+at, the trapezium area gives s=21(u+v)t, and substituting gives the rest.
Signs: choose a positive direction and keep it
Write “up = positive” at the top of your working. Then a=−9.81m s−2 for the whole flight, up and down, and a ball landing 30 m below its launch point has s=−30 m. Mixing conventions mid-question is the most common way to lose the accuracy mark.
Free fall
Without air resistance, every object near the Earth’s surface accelerates at g=9.81m s−2, whatever its mass. From rest the distances fallen in successive equal time intervals are in the ratio 1:3:5:7…, because the total distance grows as t2. This is a favourite MCQ shortcut.
Projectiles: two independent motions linked by time
Horizontal
Vertical (up positive)
Initial velocity
ux=ucosθ
uy=usinθ
Acceleration
0 (no air resistance)
−g
Equation
x=uxt
SUVAT with a=−g
The only quantity the two directions share is time. At the highest point vy=0 but the speed is ux, which is not zero. The velocity at any instant has magnitude vx2+vy2 and makes an angle tan−1(vy/vx) with the horizontal.
For launch and landing at the same height only: time of flight T=g2usinθ, maximum height H=2gu2sin2θ and range R=gu2sin2θ, which is greatest at 45∘. Examiners usually want the components worked out, so derive these rather than quoting them. They are wrong when the landing point is higher or lower than the launch point.
Air resistance and the real trajectory
Drag acts opposite to the velocity and grows with speed. So the horizontal component of velocity falls throughout the flight. On the way up, weight and drag both act downwards, so the ball decelerates faster than g and the rise is shorter in time. On the way down, drag opposes weight, so the fall takes longer. The maximum height and the range are both reduced, the path is asymmetric, the descent is steeper than the climb, and the ball lands more slowly than it was launched.
Measuring g by free fall
A steel ball is released from an electromagnet (or held above a trapdoor) and falls through a light gate, or between two light gates. The drop height h is measured with a metre rule and the time t with an electronic timer.
Released from rest where timing starts: h=21gt2. Plot h against t2, and the gradient is 2g.
Ball already moving at speed u at the first gate: h=ut+21gt, so . Plot against . The gradient is and the intercept is .
If the electromagnet releases the ball late (residual magnetism), every t is too long by the same amount. Plotting h against t gives a straight line with gradient and an intercept on the time axis equal to the delay, so the systematic error drops out of the gradient.
Repeat and average each time, measure h to the bottom of the ball, read the rule with a set square, and use a wide range of heights.
Worked examples
Exam technique
List SUVAT first. Write s,u,v,a,t with values or “?”, then choose the equation that leaves out the unknown you don’t need. This earns method marks even if the arithmetic slips.
Projectiles: two columns. Keep horizontal and vertical working separate and bring them together only through t.
Choose the equation that avoids a quadratic. Find v from v2=u2+2as first, then t from v=u+at.
Use ratios in MCQs. Time to fall ∝h, range ∝u, distances in successive seconds . These save 30–40 seconds an item.
Tangents and gradients. Use a large triangle (hypotenuse at least half the line) and quote the coordinates you read.
“Show that”. Show every substitution and give one more significant figure than the value quoted (5.43 m for “about 5.4 m”).
Sketches. Label axes with quantities and units and mark key values (intercepts, zero gradients).
Common mistakes
Quick recap
Gradient of s–t = velocity; gradient of v–t = acceleration; area under v–t = displacement; area under a–t = change in velocity.
SUVAT needs constant acceleration; each equation omits one variable.
Fix a sign convention; with up positive, a=−9.81m s−2 for the whole flight.
Projectiles: horizontal velocity constant, vertical acceleration g, time links the two.
At the top vy=0 but speed =ucosθ.
Air resistance lowers the height and range and makes the path asymmetric, with a steeper descent.
Free-fall g: plot h against t2 (gradient g/2), or h against when the ball is already moving at the first gate.