This mini-lesson covers AQA section Q: the language of motion (position, displacement, distance, velocity, speed, acceleration), s–t and v–tgraphs, the constant-acceleration (SUVAT) equations, variable acceleration using calculus, and projectiles under gravity.
On a v–t graph the gradient is acceleration and the area 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. We take g = 9.8 m s⁻² downwards and ignore air resistance unless told otherwise. Press Start when you're ready.
Section Q1–Q2 · language & graphs
Displacement, velocity, and graphs
Vectors vs scalars:displacement (vector) is how far you are from the start and in which direction; distance travelled (scalar) counts every metre, including doubling back. Likewise velocity is a vector and speed is its magnitude.
Displacement–time graph:gradient = velocity. A horizontal line means at rest. A negative gradient means moving back towards the origin.
Velocity–time graph:gradient = acceleration, and the area under the graph = displacement. Area below the axis counts as negative displacement.
Careful: to get the distance travelled from a v–t graph you must add the areas as positive values; to get the displacement you subtract areas below the axis.
Section Q3 · SUVAT
The constant acceleration equations
When acceleration is constant, the five quantities s, u, v, a and t are linked by:
v = u + at · s = ut + ½at²v² = u² + 2as · s = ½(u + v)t · s = vt − ½at²
Each equation is missing one variable. List what you know, decide what you want, and pick the equation that misses the one you neither know nor want.
Worked example
A particle starts with u = 4 m s⁻¹ and accelerates 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
Check with s = ½(u + v)t = ½(4 + 16)(6) = 60 ✓
Signs: choose a positive direction and stick to it. For a body thrown upwards, taking up as positive gives a = −9.8 m s⁻².
Calculate
SUVAT: find v
1A particle has initial velocity u = 4 m s⁻¹ and constant acceleration a = 2 m s⁻². Find its velocity after t = 6 s.
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.
m
Hint: s = ut + ½at² = 4×6 + ½×2×6² = 24 + 36.
Calculate
Free fall
3A stone is dropped from rest and falls 20 m. Taking g = 9.8 m s⁻² and ignoring air resistance, find its speed on landing, to 1 d.p.
m s⁻¹
Hint: v² = u² + 2as = 0 + 2 × 9.8 × 20 = 392, so v = √392.
Quick check
Reading an s–t graph
?On a displacement–time graph, what does the gradient represent?
Quick check
Area under a v–t graph
?On a velocity–time graph, what does the area between the graph and the time axis represent?
Section Q4 · calculus
Variable acceleration
If the acceleration is not constant, SUVAT is illegal. Use calculus instead:
v = ds/dt a = dv/dt = d²s/dt²and back the other way: v = ∫a dt · s = ∫v dt (don't lose the constant of integration!)
Worked example
A particle moves so that s = t³ − 4t² + 3t (metres, t in seconds).
v = ds/dt = 3t² − 8t + 3
At t = 4: v = 3(16) − 8(4) + 3 = 48 − 32 + 3 = 19 m s⁻¹
a = dv/dt = 6t − 8, so at t = 4, a = 24 − 8 = 16 m s⁻² — clearly not constant.
Instantaneously at rest means v = 0 — solve 3t² − 8t + 3 = 0. Maximum velocity means a = dv/dt = 0.
Calculate
Calculus: find v
4A particle has displacement s = t³ − 4t² + 3t metres. Find its velocity when t = 4 s.
m s⁻¹
Hint: v = ds/dt = 3t² − 8t + 3. Substitute t = 4: 48 − 32 + 3.
Sort it
Which tool?
Tap a task, then tap the technique you would use.
📉 Differentiate
📈 Integrate
🅢 Use a SUVAT equation
Section Q5 · projectiles
Projectiles
Model a projectile as a particle with no air resistance. Then the horizontal and vertical motions are independent:
Horizontally: acceleration is zero, so the horizontal velocity u cos θ is constant and x = (u cos θ)t.
Vertically: acceleration is g = 9.8 m s⁻² downwards, with initial vertical velocity u sin θ. Use SUVAT vertically.
Worked example — a ball projected at 20 m s⁻¹ at 30° above the horizontal
Vertical component: u sin 30° = 20 × 0.5 = 10 m s⁻¹. Horizontal: u cos 30° = 20 × 0.8660 = 17.32 m s⁻¹.
Time of flight (back to the same height, so vertical displacement = 0): 0 = 10t − ½(9.8)t² → t(10 − 4.9t) = 0 → t = 10 ÷ 4.9 = 2.04 s (3 s.f.)
Range = 17.32 × 2.041 = 35.3 m (3 s.f.)
Greatest height: v² = u² + 2as with v = 0 vertically → 0 = 10² − 2(9.8)H → H = 100 ÷ 19.6 = 5.10 m
At the top of the flight the vertical velocity is zero, but the acceleration is still 9.8 m s⁻² downwards — and the horizontal velocity is unchanged.
Calculate
Projectile: time of flight
5A ball is projected at 20 m s⁻¹ at 30° above the horizontal from ground level (g = 9.8 m s⁻²). Find its time of flight, to 2 d.p.
s
Hint: Vertical: u sin30° = 10 m s⁻¹. Solve 0 = 10t − 4.9t², giving t = 10 ÷ 4.9.
Match it
Which SUVAT equation?
Tap an equation on the left, then the variable it leaves out on the right.
Equation
Missing variable
Quick check
At the top of the flight
?A projectile is at the highest point of its path. Ignoring air resistance, what is its acceleration there?
Quick check
From v to a
?A particle has velocity v = 3t² − 8t + 3. How do you find its acceleration?
Section Q2 · areas
Finding displacement from a v–t graph
The area under a velocity–time graph is the displacement. Split the graph into triangles, rectangles and trapezia.
area of a trapezium = ½ (a + b) ha and b are the two parallel sides — here, the two horizontal lengths
Worked example
A train accelerates uniformly from rest to 20 m s⁻¹ in 5 s, travels at 20 m s⁻¹ for 10 s, then decelerates uniformly to rest in 5 s.
The v–t graph is a trapezium with parallel sides 20 s (total time) and 10 s (the constant-speed section), and height 20 m s⁻¹.
Displacement = ½ (20 + 10) × 20 = ½ × 30 × 20 = 300 m