This mini-lesson covers Topic 7 of Edexcel A-level Mathematics (9MA0): the language of kinematics (position, displacement, distance, velocity, speed, acceleration); s–t and v–t graphs; the constant-acceleration (suvat) formulae; calculus for variable acceleration; and projectiles — motion under gravity in a vertical plane, with g = 9.8 m s⁻².
On a v–t graph the gradient is acceleration and the area under the graph is displacement. On an s–t graph the gradient is velocity.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s−2 unless a question says otherwise. Press Start when you are ready.
Kinematics · language & graphs
Displacement, velocity, acceleration & graphs
Distance and speed are scalars and are always positive. Displacement, velocity and acceleration are vectors and may be negative.
On a displacement–time (s–t) graph: gradient = velocity. A horizontal line means the object is at rest.
On a velocity–time (v–t) graph: gradient = acceleration, and the area under the graph = displacement. A horizontal line means constant velocity.
Area below the axis counts as negative displacement. To find the total distance travelled from a v–t graph, add the areas as positive values; to find displacement, subtract the parts below the axis.
Quick check
Reading a v–t graph
?On a velocity–time graph, what does the area between the graph and the time axis represent?
Kinematics · constant acceleration
The suvat formulae
These five formulae apply only when acceleration is constant. Each one leaves out exactly one of the five quantities.
v = u + at · s = ut + ½at² · v² = u² + 2asalso s = ½(u + v)t and s = vt − ½at²
Worked example
A particle starts with u = 4 m s⁻¹ and accelerates uniformly at a = 2 m s⁻² for t = 6 s.
v = u + at = 4 + 2 × 6 = 16 m s⁻¹
s = ut + ½at² = 4 × 6 + ½ × 2 × 6² = 24 + 36 = 60 m
Method: write down u, v, a, s, t — list what you know and what you want, then pick the formula that misses out the one quantity you neither know nor want.
Calculate
suvat — find v
1A particle starts at u = 4 m s⁻¹ and accelerates uniformly at a = 2 m s⁻² for t = 6 s. Find its final velocity v.
m s⁻¹
Hint: v = u + at = 4 + 2 × 6.
Calculate
suvat — find s
2For the same particle (u = 4 m s⁻¹, a = 2 m s⁻², t = 6 s), find the displacement s.
m
Hint: s = ut + ½at² = 4 × 6 + 0.5 × 2 × 6².
Match it
suvat formula → the quantity it leaves out
Tap an item on the left, then its partner on the right.
Formula
Missing quantity
Kinematics · vertical motion
Motion under gravity in a straight line
For vertical motion, the acceleration is g = 9.8 m s⁻² downwards. Choose a positive direction and stick to it.
Worked example — a ball thrown straight up
Take upwards as positive: u = +14.7 m s⁻¹, a = −9.8 m s⁻².
At the highest point v = 0. Using v = u + at: 0 = 14.7 − 9.8t → t = 14.7 ÷ 9.8 = 1.5 s.
Greatest height: v² = u² + 2as → 0 = 14.7² − 2 × 9.8 × s → s = 216.09 ÷ 19.6 = 11.025 m ≈ 11.0 m.
By symmetry it takes another 1.5 s to fall back, so it is in the air for 3 s.
Sign discipline: if up is positive, then g is −9.8, and a downward displacement is negative. Most lost marks in this topic are sign errors, not method errors.
Calculate
Straight up
3A ball is thrown vertically upwards at 14.7 m s⁻¹. Taking g = 9.8 m s⁻², find the time to reach its highest point.
s
Hint: At the top v = 0. Using v = u + at: 0 = 14.7 − 9.8t, so t = 14.7 ÷ 9.8.
Quick check
When is suvat legal?
?The suvat formulae can only be used when...
Kinematics · variable acceleration
Calculus when the acceleration varies
If a is not constant, suvat is useless. Use calculus instead — differentiate to go down the chain, integrate to go back up.
v = ds/dt · a = dv/dt = d²s/dt²and going back: v = ∫ a dt · s = ∫ v dt
Worked example — v = 3t² − 4t + 2
Acceleration at t = 2: a = dv/dt = 6t − 4, so a = 6 × 2 − 4 = 8 m s⁻².
Displacement from t = 0 to t = 3: s = ∫₀³ (3t² − 4t + 2) dt = [t³ − 2t² + 2t]₀³
= (27 − 18 + 6) − 0 = 15 m
Watch out: "at rest" means v = 0, while "maximum velocity" means a = dv/dt = 0. And if v changes sign inside the interval, the distance travelled is not the same as the displacement — split the integral at the instant v = 0.
Calculate
Differentiate for a
4A particle has velocity v = 3t² − 4t + 2 (m s⁻¹). Find its acceleration when t = 2 s.
m s⁻²
Hint: a = dv/dt = 6t − 4. Substitute t = 2.
Calculate
Integrate for s
5The same particle has v = 3t² − 4t + 2. Find its displacement between t = 0 and t = 3.
m
Hint: s = ∫₀³ v dt = [t³ − 2t² + 2t]₀³ = 27 − 18 + 6.
Sort it
Which tool tells you what?
Tap a fact, then tap the tool it belongs to.
📈 s–t graph
📉 v–t graph
∫ Calculus
Kinematics · projectiles
Projectiles: motion under gravity in a plane
A projectile launched with speed u at angle θ above the horizontal is modelled as a particle with no air resistance. Split the velocity into components and treat the directions separately:
Horizontal: initial velocity u cos θ, acceleration 0, so the horizontal velocity is constant and x = (u cos θ)t.
Vertical: initial velocity u sin θ, acceleration −g = −9.8 m s⁻² — use suvat here.
The two directions share only one thing: the time t.
Treat the two directions separately. The only link between them is the time, t.
Worked example — u = 20 m s⁻¹ at 30° above the horizontal (level ground)
Components: horizontal 20 cos 30° = 17.32 m s⁻¹, vertical 20 sin 30° = 10 m s⁻¹.
Time of flight (vertical displacement 0): 0 = 10t − ½ × 9.8 × t² → t(10 − 4.9t) = 0 → t = 10 ÷ 4.9 = 2.04 s (3 s.f.). Equivalently T = 2u sin θ ÷ g.
Greatest height: H = (u sin θ)² ÷ (2g) = 10² ÷ 19.6 = 5.10 m (3 s.f.).
Range: R = 17.32 × 2.0408 = 35.3 m (3 s.f.).
Calculate
Projectile — time of flight
6A projectile is launched at 20 m s⁻¹ at 30° above the horizontal from level ground. Taking g = 9.8 m s⁻², find the time of flight, to 3 significant figures.
s
Hint: Vertical component = 20 sin 30° = 10 m s⁻¹. Time of flight T = 2u sin θ ÷ g = 20 ÷ 9.8.
Quick check
Projectile facts
?For a projectile modelled with no air resistance, which statement is true?
Quick check
At the top
?At the highest point of a projectile's flight, which is zero?