Year 12 · Science · Physical sciences
Describing motion with equations
Students use vectors, displacement–time and velocity–time graphs and the equations of uniformly accelerated motion to analyse straight-line motion, including free fall.
Awaiting educator review. This material was drafted by our content team with AI assistance and is waiting for review by a qualified educator. Please check it suits your child before using it. Created 10 October 2026.
These lessons were written around Australian Curriculum strands and matched to New Zealand year levels by age (NZ Year 1 is the first year of school). They are not yet mapped to The New Zealand Curriculum, so check them against your own learning programme.
Learning goals
Learners will
- Distinguish between scalar and vector quantities.
- Use the equations of uniformly accelerated motion to solve problems.
- Interpret the gradient and area of a velocity–time graph.
Success looks like
- I can identify u, v, a, s and t in a problem and choose the right equation.
- I can solve free-fall problems using g = 9.8 m/s².
- I can find acceleration and displacement from a velocity–time graph.
The big idea
Scalars have magnitude only; vectors have magnitude and direction. Distance and speed are scalars. Displacement and velocity are vectors, so we choose a positive direction (for example, up = positive) and use signs consistently.
For uniform acceleration in a straight line, we use five variables: u (initial velocity), v (final velocity), a (acceleration), s (displacement) and t (time).
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
To solve a problem: list the known variables, identify the unknown, choose the equation that links them, substitute with units, then calculate and check that the answer is sensible.
On a velocity–time graph, the gradient gives acceleration and the area under the graph gives displacement.
In free fall near Earth's surface, all objects accelerate downwards at about 9.8 m/s² if air resistance is negligible. A dropped tennis ball is a reasonable approximation over short distances.
Worked examples
Dropping a ball
A ball is dropped from rest from a height of 1.5 m. How long does it take to reach the floor? (Ignore air resistance, g = 9.8 m/s², down positive.)
- Known: u = 0, a = 9.8 m/s², s = 1.5 m. Unknown: t.
- Use s = ut + ½at². With u = 0: 1.5 = ½ × 9.8 × t².
- t² = 1.5 ÷ 4.9 ≈ 0.306.
- t ≈ 0.55 s.
Answer: About 0.55 s.
Braking distance
A car travelling at 20 m/s brakes with a constant deceleration of 5 m/s². How far does it travel before stopping?
- Known: u = 20 m/s, v = 0, a = −5 m/s². Unknown: s.
- Use v² = u² + 2as: 0 = 20² + 2 × (−5) × s.
- 0 = 400 − 10s, so s = 40 m.
Answer: 40 m.
Reading a velocity–time graph
A cyclist's velocity rises steadily from 0 to 6 m/s over 4 s. Find the acceleration and displacement.
- Acceleration = gradient = (6 − 0) ÷ 4 = 1.5 m/s².
- Displacement = area of triangle = ½ × 4 × 6 = 12 m.
Answer: a = 1.5 m/s², s = 12 m.
Practice check
Have a go, then check your answers. Each answer comes with an explanation.
Answer guide for parents
Which is a vector quantity?
Velocity — Velocity has both magnitude and direction.
What does the gradient of a velocity–time graph represent?
Acceleration — Gradient = change in velocity ÷ time = acceleration.
Which equation would you use to find v when you know u, a and s but not t?
v² = u² + 2as — v² = u² + 2as does not include t.
A ball is dropped from rest. Ignoring air resistance, what is its speed after 2.0 s? (g = 9.8 m/s²)
19.6 m/s — v = u + at = 0 + 9.8 × 2.0 = 19.6 m/s.
A car accelerates uniformly from 10 m/s to 30 m/s in 5 s. What is its acceleration?
4 m/s² — a = (30 − 10) ÷ 5 = 4 m/s².
A ball is dropped from rest and falls for 2.0 s. How far does it fall in metres? (g = 9.8 m/s², ignore air resistance)
19.6 — s = ½ × 9.8 × 2.0² = 4.9 × 4 = 19.6 m.
A car at 30 m/s brakes at a constant 6 m/s². How many seconds does it take to stop? (number)
5 — v = u + at: 0 = 30 − 6t, so t = 5 s.
On a velocity–time graph, a train moves at a constant 15 m/s for 20 s. What displacement in metres does the area represent? (number)
300 — Area of rectangle = 15 × 20 = 300 m.
Watch out for
- Students often treat deceleration as a positive number in equations. Choose a positive direction and give acceleration a negative sign when it opposes motion.
- Some think an object at the top of its flight has zero acceleration. Its velocity is momentarily zero, but its acceleration is still 9.8 m/s² downwards.
- Some think heavier objects fall faster in free fall. Without air resistance, all fall with the same acceleration.
Tips for parents
- Encourage a consistent layout: knowns, unknown, equation, substitution, answer with units.
- Record a ball drop in slow motion and compare the measured time with the calculated time; discuss reasons for any difference.
- Safety: drop only soft balls, away from windows and people, and never lean out from heights.
Go further
- Derive s = ut + ½at² from the area under a velocity–time graph.
- Investigate how reaction time adds to stopping distance at 50 km/h and 100 km/h.
- Research how air resistance leads to terminal velocity for skydivers.