This mini-lesson covers the SUVAT equations for constant acceleration, reading displacement–time and velocity–time graphs, vertical motion under gravity, variable acceleration using calculus, and projectiles. Take g = 9.8 m s⁻² throughout.
Where this sits:Unit AS 2: Applied Mathematics — kinematics: motion graphs, constant acceleration formulae, and motion under gravity. Unit A2 2 — variable acceleration using calculus, and projectile motion.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.
Constant acceleration
The SUVAT equations
v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)teach equation LEAVES OUT one variable: s, v, t and a respectively
They apply only when the acceleration is constant. List what you know (s, u, v, a, t), spot the missing one, and pick the equation that omits it.
Worked example — u = 4 m/s, a = 3 m s⁻², t = 5 s
v = u + at = 4 + 3(5) = 19 m/s
s = ut + ½at² = 4(5) + ½(3)(25) = 20 + 37.5 = 57.5 m
Check with s = ½(u + v)t = ½(4 + 19)(5) = 57.5 ✓
Calculate
Final velocity
1A particle starts at 4 m/s and accelerates uniformly at 3 m s⁻² for 5 s. Find its final velocity.
m/s
Hint: v = u + at = 4 + 3 × 5.
Calculate
Displacement
2For the same motion (u = 4 m/s, a = 3 m s⁻², t = 5 s), find the displacement.
m
Hint: s = ut + ½at² = 4(5) + 0.5(3)(5²) = 20 + 37.5.
Graphs
Reading motion graphs
Displacement–time graph: the gradient is the velocity. A horizontal line means at rest.
Velocity–time graph: the gradient is the acceleration; the area under it is the displacement.
Signed areas: area below the time axis counts as negative displacement (moving backwards). For the total distance, add the moduli of the areas.
Worked example
A car accelerates uniformly from rest to 20 m/s in 8 s, then holds 20 m/s for 12 s.
Near the Earth's surface a freely-falling body has constant acceleration g = 9.8 m s⁻² downwards (air resistance ignored). Choose a positive direction and be consistent.
Worked example — dropped from rest, 2 s of fall
Take downwards positive: u = 0, a = 9.8, t = 2.
s = ut + ½at² = 0 + ½(9.8)(4) = 19.6 m
Worked example — thrown up at 20 m/s
Take upwards positive: u = 20, a = −9.8, and at the highest point v = 0.
v² = u² + 2as ⇒ 0 = 400 − 19.6s ⇒ s = 400/19.6 = 20.4 m (3 s.f.)
Calculate
Free fall
3A stone is dropped from rest. Taking g = 9.8 m s⁻², how far does it fall in 2 seconds?
m
Hint: s = ut + ½at² with u = 0: s = 0.5 × 9.8 × 2² = 0.5 × 9.8 × 4.
Calculate
Greatest height
4A ball is thrown vertically upwards at 20 m/s. Taking g = 9.8 m s⁻², find the greatest height reached, to 3 significant figures.
m
Hint: At the top v = 0. Using v² = u² + 2as with a = −9.8: 0 = 400 − 19.6s, so s = 400 ÷ 19.6 = 20.408.
Quick check
At the highest point
?A ball is thrown straight up. At its highest point, what are its velocity and acceleration?
A2 skill
Variable acceleration — use calculus
When acceleration is not constant, SUVAT is illegal. Use calculus instead:
v = ds/dt · a = dv/dt = d²s/dt²and backwards: s = ∫v dt · v = ∫a dt (do not forget the constant of integration)
Worked example — s = t³ − 6t² + 9t
v = ds/dt = 3t² − 12t + 9 · a = dv/dt = 6t − 12
The particle is instantaneously at rest when v = 0: 3(t − 1)(t − 3) = 0 ⇒ t = 1 s and t = 3 s.
At t = 1: a = −6 m s⁻² (decelerating).
A2 skill
Projectiles
Treat the horizontal and vertical motions separately. They share only the time.
Horizontal: no acceleration ⇒ x = (u cos θ) t.
Vertical: acceleration −g ⇒ y = (u sin θ) t − ½gt².
Worked example — u = 25 m/s at 30° above the horizontal
u sin θ = 25 × 0.5 = 12.5 m/s upwards; u cos θ = 25 × 0.866 = 21.65 m/s horizontally.
Time of flight (back to the same height): t = 2u sin θ / g = 25/9.8 = 2.55 s (3 s.f.)
Range = horizontal speed × time = 21.65 × 2.551 = 55.2 m (3 s.f.)
Modelling: we ignore air resistance and treat the projectile as a particle. Real range would be less.
Calculate
Time of flight
5A projectile is launched at 25 m/s at 30° above the horizontal from ground level. Taking g = 9.8 m s⁻², find the time of flight to 3 significant figures.
s
Hint: Vertically: u sin30 = 12.5 m/s up, and it returns to y = 0 when t = 2u sinθ ÷ g = 25 ÷ 9.8 = 2.551 s.
Sort it
Where does it come from on a graph?
Tap a quantity, then tap the graph feature that gives it.
📈 Gradient of a v–t graph
🔲 Area under a v–t graph
📉 Gradient of an s–t graph
Quick check
SUVAT legal?
?A particle has acceleration a = 6t − 12 m s⁻². Can you use v = u + at?
Match it
Which SUVAT equation?
Tap the equation on the left, then the variable it leaves out.
Statement
Answer
Quick check
Projectile independence
?Two identical balls: one is dropped from a height, the other is fired horizontally from the same height at the same instant. Which lands first?
Quick check
Deceleration sign
?A car travelling at +20 m/s decelerates uniformly to rest in 5 s. What is its acceleration?
Examiner traps
Kinematics pitfalls
SUVAT only for CONSTANT acceleration. If a depends on t, integrate.
Sign of g. If up is positive, a = −9.8. Mixing signs mid-question is fatal.
"Distance" vs "displacement" on a v–t graph: areas below the axis are negative displacement but positive distance.
At the highest point v = 0, but a = 9.8 m s⁻² still.
Projectiles: resolve the initial speed into u cos θ (horizontal) and u sin θ (vertical) — do not use u itself in the vertical equation.
Constants of integration are the initial velocity and displacement — find them.
Recap
The big ideas to know
SUVAT (constant a only): v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t
Graphs: s–t gradient = velocity · v–t gradient = acceleration · area under v–t = displacement
Gravity: a = g = 9.8 m s⁻² downwards; choose a positive direction and keep signs consistent
Variable acceleration: differentiate s → v → a, integrate a → v → s (+ constant)
Projectiles: horizontal x = (u cos θ)t · vertical y = (u sin θ)t − ½gt² · time links them
Kinematics describes the motion; Newton's laws explain WHY it happens. Press Finish to see your score.
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